That fog example was frightening. In dense fog, the only safe way to keep driving is to reduce speed, so that you can stop within the distance you can see (in fact, that's true of driving in general).
> ... so that you can stop within the distance you can see ... Yes, true, plus the fact that the "three-second rule" won't work if you're driving toward a prior crash and can't rely on the expectation that the car in front of you is still braking to a stop. For that case you need even more time to react. This makes fog more dangerous than most people realize. Thanks for writing!
The problem then is with people behind you driving dangerously fast. (Which is probably most drivers in foggy conditions, considering even with 100ft of visibility you would need to drive 25mph to stop safely) It is probably more dangerous to drive slowly in foggy conditions due to the likelyhood that someone else reckless behind you will make a mistake. The only safe option in foggy conditions is just to wait it out, as the video says.
Some cars have for a while thermal camera superimposing view on the digital dashboard and nowadays perhaps even a head-up display not to mention that you have also had radar-assisted automatic emergency breaking for over a decade but generally, it's better to slow down than to trust with your and other's lives on tech, especially as many emergency breaking assistants like one in newer Tesla cars given up on the radar that made them pretty useless in fog while radar or even LIDAR can experience malfunction or interference that cannot be easily spot. So, until high-speed thermal cameras with sufficient resolution in cars proliferate, and traffic signs are redesigned to be clearly visible in the infrared spectrum, it's better to plan ahead and avoid driving in an area with heavy fog if you can...
@@lutusp There is also the issue that people driving in fog underestimate their speed, resulting in them going faster. I'd add that drivers should be using their hazard flashers in low visibility conditions to extend how far the driver behind them can see you from.
@@IonorRea The technology of assisted automatic emergency breaking may have been in beta for a decade and I feel like its still there. I've had my car suddenly slam to a stop because of a shadow on an otherwise sunny day. It can't tell the difference between a two dimensional object or a three dimensional one. Also my daughter just collided with someone who was at a stop sign (fully stopped) when the emergency assist didn't react fast enough. My daughter felt it took over because when she hit the breaks the car didn't slow down even faster. Were waiting for the investigation on that. So our theory is the vehicle prevented her from applying the breaks in an emergency force vs the computer thinking it just needed to slow. The collision was slow enough for minor damage and hopefully the car can say how far it traveled while the computer took over breaking. The investigator will compare it to it's normal stopping distance at the speed that the computer says she was going (48mph). I personally would drop the title "emergency" assist and just label it "assist if it wants to even if its wrong".
@@whoakshatsharma That is inappropriate anthropomorphism. Physics doesn't do what it is told. Objects behave predictably. Physics accurately describes that predictable behavior.
@@marc21256 > "Physics doesn't do what it is told. " True. One could say it conforms to the underlying laws, and even that sounds like anthropomorphizing, but most people follow the logic. This is why the only true physics is mathematics. I don't like putting equations in my videos (it loses too many viewers) but sometimes I must. :)
I was expecting a bunch of fear-mongering nonsense, but ended up watching a VERY well-informed video. Thank you for the strictly scientific approach. Too many people talk with emotions when they get into driving and safety, so I’m glad to see somebody not do that for a change.
One must also remember that the road surface has an effect on braking distance. Both the obvious friction question, and also elasticity and things like rolling particles (gravel for example).
This strikes me as a good example of how to believe your own crap doesn't stink. Not only do we ignore some of the most important factors, tire grip, suspension to accommodate weight transfer during braking, and overall brake performance...but we also get to hear a fantasy land world where it takes three seconds to apply a brake. If it took three seconds to apply the brake, there would be an accident at every intersection and parking lot every few minutes.
> ... but we also get to hear a fantasy land world where it takes three seconds to apply a brake. But that's not the claim. It doesn't take three seconds to apply the brakes, it takes three seconds to react to a situation that requires braking. Then the actual braking begins.
3:34 This example doesn't take into account that the 'monster vehicle' has 5 times more tire contact patch as well, since it has 20 tires. Mass per tire contact patch is also important. Otherwise, overinflated tires would lead to the same stopping distance as correctly inflated ones.
Both of you apparently stopped watching to post an ignorant reply. He covered this very subject moments later. Showing how the additional tire friction did not change anything due to the fact that each wheel ends up sharing a smaller percentage of vehicle weight. Which in turns reduces the amount of friction at wheel. It's why a dulley pickup trick has less traction in the rear on wet roads then it would if it only had 2 rear wheels. The vehicle weight is spread out across 4 instead of 2 wheels. The 2 rear wheel truck will have a better time pressing down to the pavement then the 4 rear wheel truck will.
@@williamallen7836 Same net downward force on a smaller contact patch will result in more contact pressure, which is another factor affecting traction and braking distance. He assumes that the relation between the net downward force and the friction force is linear, namely Fricition = μN, which is not the case as the contact patch gets smaller.
> This example doesn't take into account that the 'monster vehicle' has 5 times more tire contact patch as well, since it has 20 tires. Actually I did cover that issue with a comparison table that trades off contact area and downward force, I also mentioned that actual tires stop following the rules if exposed to loads they cannot handle. But to a first-order approximation, fewer tires with more loading provides the same overall braking friction. Thanks for writing!
@@ohanneskamerkoseyan3157 > " which is not the case as the contact patch gets smaller." Fewer tires exert more force per contact area, so the correspondence I describe remains approximately true until the materials begin to break down. The posted equation describes an ideal correspondence, not an exact one. BTW the HyperPhysics page I linked to agrees with my analysis. Nevertheless, when safety agencies post stopping distances, they don't include vehicle mass in the calculations -- there's no need.
> ... to the point of being misleading itself. Not as misleading as the many online sources that try to say that mass directly increases stopping distance, as with my quoted examples.
@@patrickbuick5459that's likely because their brakes failed, which is by the way not uncommon for poorly maintained trucks. This can also happen if the truck is overloaded (i.e. loaded more than the certified maximum allowed mass). The video assumes the brakes are working properly and the truck is properly loaded.
@@streettrialsandstuffIt also assumes everyone knows how to drive - following distance, reaction times, and everyone knows how to "threshold brake". I know professional racecar drivers who can't threshold brake. (if they didn't have ABS, they'd be dead.) For the simulations to work, everyone has to be braking at maximum. Nobody does that in real life. The reality is as mass increases you need to use more brake to stop in the expected distance, and people don't expect that. We're used to a certain feel (g-force) under braking that's way below maximum. As the mass increases, that force will not stop you in the same distance.
This is about a physics principle, which is correct, not about a badly engineered vehicle that shouldn't be on the road in the first place. The point of this kind of science is to point out, not how things are, but how they ought to be.
Whilst you are absolutely correct in 95% of what you say, your analysis of mass vs stopping distance does not take into account the brakes themselves. When breaking, there are two sets of friction at play. The first, which you take great care to analyze in detail, is the friction between the wheel and the road. As you point out, increasing the mass results in increased maximum available friction, thus cancelling out the increased force required to stop the vehicle. The second, however, is the friction within the brakes themselves. This is unaffected by the mass, since the force is applied solely by the pressure of the brake pads on the disks / rotors/ drums. This force comes from the driver's pedal (possibly boosted by power-assisted brakes) but will not automatically be increased with an increase in mass. Don't forget, in this case we actually don't want the friction to be high enough to lock the brakes up, which further complicates the calculations The whole situation gets even more complicated when ABS is brought into play. Your example of the cars joined together works when it comes to road friction, but fails to take into account that the five joined vehicles have five times as many sets of brakes. Overall, I think that as long as a car is operating within it's design specifications, it probably doesn't make much difference, but we all know that many people overload their cars (either directly, or indirectly) and in these cases assuming the mass has no overall effect can be dangerous.
> ... your analysis of mass vs stopping distance does not take into account the brakes themselves. My analysis assumes that tire-pavement friction resists the car's forward motion. The brakes must allow this to happen, so they're not part of the analysis, which focuses on that one factor. In science, one must control all factors in order to be able to study just one.
@@lutusp I agree with your statement about Science, which is why I posted here and why you are incorrect. Braking relies on two types of friction, each of which must work correctly and, if either changes to the other, you will skid. Static friction is defined as the frictional force which acts between the two surfaces when they are in the rest position with respect to each other. This is what you have / want between the tire and the road surface. If this friction breaks down, you have a skid. As your analysis correctly explains, part of the calculation of the maximum force that can be applied before this friction fails involves the pressure between the two surfaces - caused by the mass of the vehicle. Therefor, as you say, as the mass increases, so does the maximum frictional force and one cancels the other out. BUT Dynamic friction is defined as the frictional force which is created between any two surfaces when they are in a moving position. It is also called Kinetic Friction. This is what you have within the brakes themselves, between the pads and the rotors / disks / drums. It is this friction that actually dissipates the energy that the moving vehicle is losing (basically, converting the vehicles kinetic energy primarily into heat.) If enough force is applied to the brakes, then this friction will increase to become static friction, the brakes lock up, and you have a skid (see above.) It is this situation that ABS is designed to prevent, by removing the braking force as the brakes approach locking up. So, to avoid a skid, you need for the static friction between the wheels and the road to be more than the force being applied to stop the vehicle (which is what you analyzed) AND the dynamic friction within the brakes to be high enough to dissipate all the energy. Now, once again, force and area come into play here. The smaller the total area of the brake pads, the greater the force required to dissipate the vehicles kinetic energy in the same time. There are two risks here. If the amount of energy to be dissipated is too high (due to the mass of the vehicle) then the brakes will overheat and either lock up, or fail completely. If the force applied to the brakes is too high, then the brakes will lock up. Either form of locking up will mean a skid. (I'm not going to go into why a skid is bad here, but it revolves around the transition between static and dynamic friction, and back again, if you want to look it up.) In your example, with multiple vehicles, and therefore wheels and, as a result, sets of brakes, the total area of the pads is greater, meaning less pressure being required to provide the braking force and the heat being dissipated over a greater area. When you reduce the number of wheels, you also reduce the brake pad area, which means each must dissipate more energy to stop the vehicle in the same distance. This means more heat and a greater pressure. This is not automatically compensated for as it is with the static friction you analyze. So, if you increase the mass of the vehicle, you either rely on the brakes doing more work without failing (assuming you are able to apply sufficient pressure) or you have to apply the same, lower, force for a longer period to stop the vehicle, which increases stopping distance. As you say, "In science, one must control all factors in order to be able to study just one." In this case, you have focused (and correctly analyzed) one (the static friction between the wheel and road) but ignored the other. I should also point out that we are not talking about the extra mass caused by having an extra person in the car, which is relatively minor, but rather massive differences. If you don't believe me, try speaking to truck drivers who transport heavy loads of steel and tell them they don't need more stopping distance when fully loaded! They will laugh you out of the room.
> If you don't believe me, try speaking to truck drivers who transport heavy loads of steel and tell them they don't need more stopping distance when fully loaded! This is not about existing trucks, which represent engineering and economic compromises. This is about physics. In physics, a more massive vehicle presses down with more force, and if economic considerations are set aside, if two vehicles of different masses have been provided with appropriate tires and brakes, they will stop in the same distance. This ought to be self-evident. You see, in science, to study factor Z, one must first control for factors A-Y, otherwise the result is tainted by uncontrolled or unknown issues. Let me put it this way -- for two cars of differing masses, they can be engineered to stop in the same distance -- different tires, different brakes. For two cars traveling at different speeds, they cannot. That was the point I was making in the video. > ... which is why I posted here and why you are incorrect. Some advice. Never address a forum's participants as though they are the topic. Instead, address the topic itself. One can say, "Your position on issue X is incorrect." But don't say, "You are incorrect." Of course, this depends on whether one intends to generate heat, or light.
@@lutusp OK, "Your position on this is incorrect". Happy now? The video doesn't talk about re-engineering and having a new vehicle with upgraded brakes. It simply challenges the notion that a heavier vehicle will require a greater stopping distance. The video states this isn't the case, based on the static friction between the wheel and the road. That part I can appreciate, and have acknowledged, but it ignores the brakes themselves and the dynamic friction element, which is my point. You talk (above) about studying science and fixing all but one variable, to allow you to study the effect of that variable, yet you fail to acknowledge that there are at least two variables at play here - which is the point I am making! In fact, the claim of upgrading brakes actually introduces an additional variable! If you want to argue that the brakes can be made more efficient, then one can also argue that the tires or the road surface can be improved to make the static friction better - negating the whole basis of the point being made! You say that "for two cars of differing masses, they can be engineered to stop in the same distance", this is true, but is not the same as the point made in the video, with the analogy of linking cars together. For the most part, the video was informative and helpful, but in this aspect the video presents an argument that does not work. I see other commenters have drawn attention to the "computer simulation" aspect. Any simulation is only as good as the way it is written and will only include the factors the programmer considered. There comes a point where you have to do it "for real" to get complete data. This is why manufacturers use Crash-Test dummies and actually crash real cars. Simulations can be useful, but in and of themselves, they don't prove anything. All the simulations in the world will not account for every factor. This is why I mentioned truckers. Their real-world experience and expertise shows the fallacy of the "weight doesn't matter" argument. You mention, in your comments, that this is about science. True, but science is, at it's heart, an attempt to explain and understand the real world. We have ideas, propose hypotheses create experiments to test those hypotheses and use the results to refine our models and inform our understanding. Yes, physics and Math are closely related, but any mathematical analysis will inevitably be an approximation of reality.
@@paulsimpson6290 > but any mathematical analysis will inevitably be an approximation of reality. Not if we understand reality, because modern physical theories are mathematical -- no guesswork, no unevaluated gray areas. There are two kinds of scientific theories -- mathematical, and incomplete. No one in science accepts "approximations of reality" as though that is the goal. The reason we know the Standard Model isn't complete is because there are experimental outcomes that the math don't explain. And no one is willing to say that our scientific theories are only educated guesses. Leave that to the psychologists.
1:22 Braking distance isn't about energy (that comes into play only in mega-heavy vehicles, or repeated stops) Braking distance isn't governed by E=mv^2/2, but by x=1/2at^2+vot+xo where t=(Vo-Vf)/a, and a=F/m~g*mu, mu~1 for ideal conditions, and down from there. So a small mu (like ice) will increase the stopping distance proportionally. So simplifies to x=3*V^2/(2*a); x~ v^2/22 (for feet per second) So 20 mi/h ~ 45 ft to stop So 40 mi/h ~ 158 ft to stop So 80 mi/h ~ 622 ft to stop yes, I skipped the step to convert mi/h to ft/s. Because I'd be using m/s for all of it... Big trucks stop slow because of cost. No laws of physics would be broken making an 80,000 lb loaded truck stop in the same distance as a Porsche 911. It would just be expensive, and maybe heavy.
> Braking distance isn't about energy ... But it is. If there were no energy, there would be no need to dissipate that energy as heat in the brakes. Thanks for writing!
@lutusp and repeatedly stopping demishies the stopping distance as the brakes get hot from repeated heat cycles. There is a point of failure, a point where the materials no longer can perform like intended. Everything is based on ideal conditions. So when weather changes so does your ability to stop. Even the condition of the components on said vehicles. From swelling brake lines to water saturated brake fluid. To many variables to make assumptions. Hot Temperatures are bad for brakes, they are less effective in those conditions. So more mass means more heat, which demishies the effectiveness of the braking system. So mass does cause a vehicle to stop further if fully loaded than not. You are not including the heat soaking of the braking components. Do a heat test with a thermal camera on a fully loaded vehicle and empty and get back to me on how the temperature of the brakes are the same. 😅
Gotta love a mythbusting video that none the less perpetuates other myths. The U.S. in not the only country using MPH, a quick online search verifies this.
This video contains some great demonstrations and good points, but I noticed a mistake. 9:03 Where did you get the formula for E from? It is completely wrong, and dimensional analysis shows it doesn't even give you energy. The "wrong!->" formula is completely correct for calculating the kinetic energy of a body. Your next slide you bring up to "show the correct formula" even shows you that same formula for the kinetic energy that you labeled "wrong!->" (Work = mgμd = m*v^2/2, where the third element is the kinetic energy). In ideal conditions, I think you have shown clearly that braking distance doesn't vary depending on the mass of the vehicle. In reality, there exist some factors which could change this however. For instance, if weak brakes are used, a human could feasibly not be able to generate enough braking force to max out the tyre friction. If a trailer with bad brakes is used, it would carry its weight on its own axle, meaning its weight doesn't contribute to the braking force of the vehicle, and stopping distance could become dependent on mass. This will happen in any vehicle whenever weight is loaded above an axle with a bad/ no set of brakes. Also, hot brakes can become less effective, and the more massive the vehicle, the more energy you have to put into the brakes, making them heat more. Hypothetically, this means mass could affect braking distance.
> Where did you get the formula for E from? Which equation? There are several. In one, I provide the canonical form: E = 1/2 m v^2. In another, I accommodate the effect of friction, which agrees with field tests that show little to no effect from increased vehicle mass and also agrees with my monster-car example, which proves the point in a simple way. In a slide I call wrong, the author uses the canonical form shown above, then concludes that this means more massive vehicles require more stopping distance, which is wrong. That's my meaning for "wrong" -- it misapplies the canonical energy equation and fails to take enough factors into account. My replacement equation does do this. The remainder of your comments refer to extraneous factors, all true, but all about economics and engineering, not physics. My point is that the physics makes mass nearly irrelevant, and many people don't understand this.
@@lutusp Thanks for the response. I don't completely agree the other factors I bring up are extraneous, nor that they aren't fundamentally concepts that have to do with physics, however I completely agree that if we disregard those factors, stopping distance should not depend on mass. The formula for E I was referring to was the equation labeled as "right" at the moment of the time stamp 9:03, and this the remainder of this comment will be specifically about what's on screen in that moment. You label the correct equation for kinetic energy as "wrong" while saying "In this case, I include the correct equation just for contrast". At the same time, you show the equation E=mv^2/2μg, which uses the variable name E, which is convetionally used for energy. You then say that the frictional term in the denominator and the mass in the numerator cancels out, which would give the final units of the equation being energy if true. It is however not true that μg cancels out with m, as μg has units of acceleration, while m has units of mass, meaning they cannot cancel out. So the formula is not correct for calculating energy if this was its intended use. If the formula was meant to calculate stopping distance, it would also be wrong, as the formula for stopping distance is mv^2/2mμg=v^2/2μg. It is true that the inference made about stopping distance in the sentence below the formula for kinetic energy was wrong. However the formula for kinetic energy wan't wrong, so I don't think you should have labeled it as such. It is in fact the formula you include of E=mv^2/2μg that seems to be wrong, as it neither gives you stopping distance, energy, nor do the terms m and μg cancel out as you say.
@@HenrikMyrhaug > You label the correct equation for kinetic energy as "wrong" ... Yes, because in the page's context, it's being misapplied as though it can be used to correlate stopping distance with vehicle mass, which is certainly wrong. The original page's author then draws a classic wrong conclusion based on an equation that, in a different context, would be correct. My equation on the same slide tries to correct for this error, and though only approximate, it's a big improvement -- it at least addresses the context of stopping distance.
In theory, yes, heavier vehicles should have the same stopping distance as lighter vehicles. In practice that's not the case. Now which matters more for driving safely...
> In practice that's not the case. In practice, with properly designed vehicles, it is the case. My point is economics decides this, not physics. Trucks that can stop within the theoretically established distance are more expensive. > Now which matters more for driving safely... If driving safety were the only issue, all vehicles would stop in the same, optimal, distance. And they would be more expensive. Thanks for writing!
Thank you Paul. Very well done. I'll have my 16-year-old daughter watch it!. A related topic worth examining is "How long should the traffic light yellow be illuminated before turning red -- for various speed limits?"
You did not prove anything. Higher speed vehicles stop in distances proportional to speed squared, but this is not because enery. Brakes apply generally constant pressure and this makes higher them more powerful at higher speeds ( remember formula E=F*X) . Twice the speed twice the stopping time, but average speed for that time is also higher for faster vehicle. So it is proportional to speed squared. Heavy vehicles stop harder because tire and road surface interface is not a simple thing to be represented as only friction coefficient. Heavier vehicles use harder tires to be economical, or their tires simply wil be worn out very soon. This makes trucks and heavy vehicles stop in longer distances. Also the brake systems are different, air brakes are generally less responsive than hydrolic brakes of a small vehicle.
> You did not prove anything. I proved that properly designed braking systems make all vehicles equal. > This makes trucks and heavy vehicles stop in longer distances. No, it makes better braking systems more expensive. What I proved is that trucks take more distance to stop only because they would otherwise be more expensive. > Also the brake systems are different, air brakes are generally less responsive than hydrolic brakes of a small vehicle. Yes -- economics, not physics. Thanks for writing!
Thank you! I've always thought the idea that "it takes longer for heavier vehicles to stop" idea was wrong. Consumer Reports does stopping distance tests on hundreds of vehicles. I'm going to take a sample of them from compact cars and those over 6000 pounds and see how they vary.
@@fvl548 These are the average they list for each category of car: Category Average dry braking 60-0 mph, ft. Sports/sporty cars 120 Luxury midsized cars 128 Luxury small SUVs 129 Small cars 130 Luxury small cars 131 Small SUVs 132 Luxury large cars 132 Large cars 132 Midsized cars 133 Midsized SUVs 134 Luxury midsized SUVs 134 Compact pickups 136 Minivans 138 Compact electric/hybrid cars 138 Luxury large SUVs 138 Full-sized pickups 140 Large SUVs 143 Average of all tested vehicles 132
Yet another physicist forgets about the elasticity modulus of the tyre and how it makes the friction equation non-linear. This one is so easy to demonstrate safely that it honestly astonishes me that people don't test it BEFORE publishing erroneous videos about it.
> Yet another physicist forgets about the elasticity modulus of the tyre and how it makes the friction equation non-linear. I clearly mentioned this -- several times -- as the practical limit of my example. Nevertheless, most people get this issue entirely wrong, an error my five-car example is meant to address and falsify. > it honestly astonishes me that people don't test it BEFORE publishing erroneous videos about it. Watch the video again. Locate where I don't qualify my results as approximate.
Actually, the coefficient of traction of a tire IS quite linear, until the sheer strength of the rubber compound is exceeded. Tires with the appropriate inflation pressure, size and compound for the weight of the vehicle under test is assumed. It is also assumed, that the brake system compensates for weight transfer, so the axle torque is exactly proportionate to the load on each tire. (An ideal antilock braking system) You would be surprised, how good the braking systems on newer modern cars actually are!@@lutusp
Interesting and informative video Paul. Misconception, assumptions, partial knowledge can lead to bad outcomes. Thanks for sharing. This should be useful for almost everyone.
I learned to drive double decker buses in trhe 1960s in London. I soon learned the fallacy of big vehicle = longer stopping distance by experience. We were takenn to a closed track at Chiswick where we were allowed to slam on at various speeds, culminating in the famous "skid pan" where we drove over a constant spray of water on a smooth surface with worn rear tyres. The instructor would command a certain course, either straight ot curved then yank the handbrake (rear wheels only) to induce a skid. We then had the hair-raising experience of trying to correct the skid and bring the bus to a safe stop. A few years later I had the need to perform an emergency stop with a loaded bus. The vehicle stopped promptly within what I would have expected for a car at the same speed. Strong brakes and wide tyres added up to loads of anti-traction (friction). We were taught the "cadence" breaking method of quickly alternating between full braking and less breaking to avoid the wheels locking - skidding tyres offer no traction. Modern ABS brakes work that way.
I think I'm missing something. Do I understand correctly if I say that the calculations show that mass has no effect on stopping distance but "there are many practical and economic issues that explain why a heavy car or truck requires more stopping distance." (6:00) So tradeoffs in engineering, materials, cost, etc. are such that heavier vehicles DO require more stopping distance. It seems to me that the correct answer depends on the context of the question. If you're asking about simulated vehicles on simulated roads, then the answer is no, heavier vehicles don't require more stopping distance. If you're asking about real vehicles on real roads, then the answer is yes, heavy vehicles require more stopping distance. Maybe I'm just dense, but if I'm can be confused, other people can be too. I think it's dangerous to be at all ambiguous about this point unless you want people getting behind the wheel of a large heavy vehicle, thinking that they can stop just as quickly as a smaller lighter vehicle and driving accordingly. If the focus is on safety, then the answer should pertain to how a real driver should treat a real vehicle in the real world. From watching this video, it looks like people should walk away with the knowledge that yes, heavier vehicles do require more stopping distance, which would mean it's not a myth, but just has some nuance.
> If you're asking about real vehicles on real roads, then the answer is yes, heavy vehicles require more stopping distance. Not if the tires and brakes are designed properly. This is the point of my five-car monster vehicle example -- it weighs five times as much, but it stops in exactly the same distance as one car. Everything after that is about engineering -- which tires, which brakes, and so forth. Not physics.
well friction does not depend on mass in this case, masses should cancel out if two cars with different masses stop by making tires skid, and guess what they do not, their friction depends on pressure applied on the brake disk/drum, so mass does not cancel out and braking distance increases with mass through physics. there is a slight increase in friction because heavier mass increases the tire contact area of tire on road but its way too low to decrease braking distance
I just opened online calculator for car stopping distance, I entered the 903 km/h speed that average A320 lands at on a standard 10,000ft runway, it told me that I'd need a 17,000+ft runway. Of course, it didn't ask for the weight of my vehicle... :)
@@shinobuoshino5066 First off: Look up how planes land! Also, 903km/h is just below the cruisingspeed of an A320. Landing speed is around 200km/h. Off course it won't ask you how much your vehicle weighs, since most people don't know. Post this calculator as a refference so we can factcheck, you're "factchecking"
> "there is a slight increase in friction because heavier mass increases the tire contact area of tire on road but its way too low to decrease braking distance" Not really. In practice, mass is nearly eliminated as a factor unless the brakes cannot handle the load or the tires melt. Even in ideal circumstances it's not an exact cancellation, but sufficient that safety agencies don't bother factoring in vehicle mass in calculating stopping distance.
the math/sim your using is a hugeass oversimplification. according to some reddit comment: Freshman year physics - F(friction) = u(coefficient) x Fn(normal force) Senior year physics - Yeah, remember that simple equation? Throw that out the window when you're working with a non-linear elastic, pneumatic wheel applying torque. We're going to need to use this partial differential equation that's going to take up most of the page and you'd need several computers to solve for. Non-smartass answer: Because the traditional Ff=u x Fn is an oversimplification when it comes to wheels and tires. It works great for something like a wood block where it's just sliding and doesn't deform. But a tire is a complex thing. You're transmitting force through this spinning wheel that's under pressure from air inside, generating tension in steel belts in multiple directions, and rubber, all of which is going to flex and move. In that case contact patch becomes more relevant to how it's going to grip and behave.
You may notice that I qualify every mention of this relationship, to take these points into account without losing my viewers by citing the technical details. But the point my video makes is that more massive vehicles do not require more stopping distance, unless they're improperly designed. All those sources I quoted got it wrong precisely because they lost the forest for the trees -- they assumed that E = 1/2 m v^2 must mean that more massive vehicles require more stopping distance proportional to mass, which is false.
Heavier vehicles will always require more stopping distance regardless of design (assuming everything aside from mass is identical). In real-world scenarios, there is often some degree of slippage between the tire and the road. This slip introduces factors like tire load sensitivity, where the increase in friction with higher load is less than proportional. Heavier vehicles do tend to have more load on their tires, but the traction increase is not directly proportional. This is due to the nuances of tire behavior, including factors like tire composition, road conditions, and the presence of anti-lock braking systems (ABS) that can modulate braking to prevent wheel lockup.
@@Venous_fx > Heavier vehicles will always require more stopping distance regardless of design (assuming everything aside from mass is identical). My video's five-car monster vehicle example falsifies this claim. Same tires, same brakes, same driver actions, everything but the total mass. But then, that was the purpose of the example -- to force people to see that the commonly heard claim is false. In everyday reality, this simple truth is hidden by bad engineering.
@@lutusp As I stated before, your simulation is not an accurate representation of reality. Heavier vehicles will always require more stopping distance regardless of design (assuming everything aside from mass is identical). In real-world scenarios, there is often some degree of slippage between the tire and the road. This slip introduces factors like tire load sensitivity, where the increase in friction with higher load is less than proportional. There are many complex mechanisms at play in the tire and road interaction. Your simulation does not account for this.
@@Venous_fx > As I stated before, your simulation is not an accurate representation of reality. I proved this commonly heard claim false in the simplest possible way -- I welded five two-ton vehicles together, creating one ten-ton vehicle, and that much heavier vehicle stops in the same distance as one two-ton vehicle. How is this in any way unclear? A heavy vehicle with adequate tires and brakes, stops in the same distance as a small, light vehicle with its own appropriate tires and brakes -- assuming both vehicles are traveling at the same speed. Assume also that ABS is in use, which means the tires are at the edge of breaking traction with the roadway, which is the most effective braking friction point. This is why safety agencies list speeds, but DO NOT list masses, in their officially authorized braking distance charts -- they do this because MASS DOES NOT MATTER assuming tires and brakes are correctly chosen for the vehicle's mass. Speed matters, and speed is listed as a factor in stopping distance. Reaction time matters, and is also listed. MASS DOES NOT MATTTER. "Vehicle Stopping Distance and Time" from UPenn: nacto.org/docs/usdg/vehicle_stopping_distance_and_time_upenn.pdf No mention of mass or weight. Automotive Fleet: KNOW YOUR STOPPING DISTANCE -- www.automotive-fleet.com/driver-care/239402/driver-care-know-your-stopping-distance No mention of mass or weight. Dozens more examples, but you know what? You could do your own research and save me some time answering an inquiry I have answered hundreds of time over the years. Thanks for writing.
A better way to explain the stopping distance being independent of mass is that the total energy needs to be dissipated in braking, so ΣE= (1/2)m(v)^2 - μmgx = 0, where μ is coefficient of friction between tires and road and x is stopping distance (1/2)m(v)^2 = μmgx x=(v^2)/(2μg) However, in practice this assumes that the loading and the braking force is evenly distributed, where in reality you will probably have one axle lock up before the others in a large truck, which was mentioned in the video as an issue when trucks are empty, since the weight shifts forward and unloads the rear axles.
Yes, but when doing science, it's necessary to control for all extraneous factors in order to properly study just one. Otherwise no useful science is possible.
Alright I gave this thumbnail the benefit of the doubt. I'm not very disappointed, despite the claims being slightly exaggerated 05:13 the way I like to remember this, is that the extra "stopping" energy has already been already "spent" when accelerating that car. ie. you already cancelled the extra work (watts) you need to spend to stop the heavier car when you accelerated to that speed in reality, a heavier car will be sliightly faster to stop due to material science reasons if it's the only added factor. the weight distribution is a tiny but not insignificant factor too, again due to material science (the way the rubber interacts with the asphalt). That is until you hit a "breaking point", when the car becomes so heavy the breaks can't break enough, and overheat
> the way I like to remember this, is that the extra "stopping" energy has already been already "spent" when accelerating that car. Well, energy has two primary forms -- kinetic and potential. When you accelerate, you're turning kinetic energy into a mixture of potential and kinetic energy, which are represented by the car's mass and speed. When you hit the brakes, you're converting that energy mix into kinetic energy again. I hope this helps, and thanks for posting.
An empty semi truck with a 48' trailer will take more time to stop when fully loaded even with more friction to the pavement. An 80000lb vehicle DOES take more distance to stop than the same vehicle empty. I know you're only talking about cars, but more weight means more potential energy to slow and stop the vehicle. As a former truck driver, I know this is true. Like most things, they look good on paper/graphics, but in real life things vary greatly. I do agree that driver reaction time plays a HUGE part in stopping of any vehicle. I like the fact that you point out some real physics in this video showing the stopping distance once brakes are applied goes up 4 times for double the speed. Thanks a million for understanding this and conveying it too. Overall, I'd have to say this was a good video, thank you.
trucker here as well and you are correct. an empty truck will stop faster than a fully loaded one. they might have gotten them to stop in 30 feet in lab condition with perfect tyres and pavement but that shit just doesn't happen in reality.
Completely agree. When hauling a trailer behind my pickup with my race car on the trailer it definitely takes much more distance to stop. And the trailer has brakes
> I know you're only talking about cars, but more weight means more potential energy to slow and stop the vehicle. Yes, and more tire friction. If the tires are brakes are designed to accommodate all operating modes, the mass factor is nearly eliminated (not precisely eliminated, as I point out in the video several times). And the truck design would be very expensive. My sole point with this video to to stop people from claiming that more massive vehicles must require more stopping distance solely based on their mass. This is false. Real trucks do require more stopping distance proportional to mass, because of how they're designed, not because of physics.
@@trusconi81 > When hauling a trailer behind my pickup with my race car on the trailer it definitely takes much more distance to stop ... Yes, because of the overall setup and the difference between the loading on the different axles, not because of basic concepts. My video is meant to falsify the widely held wrong assumption that more massive vehicles require more stopping distance as a direct function of their mass, which is false. In your hauling setup, unless the trailer brakes are extraordinarily well-designed, they will produce either more or less braking force that is required to produce zero net longitudinal force on the hitch. If the latter were true (nearly never), then the truck and the trailer become independent braking systems, not a candidate for a jackknife, the usual case, and which certainly will take more distance to stop, as you experience. A perfect setup would create zero longitudinal force on the hitch during braking, so the hitch doesn't either pull or push the truck (during braking, not while hauling). That would be the basis for a valid comparison test, and it would make all stopping distances the same to the degree allowed by loading of the tires and brakes.
It would be interesting to see if you would stand in 321 feet in front of a fully loaded Australian 2AB Quad road train doing 100kmph to prove that all vehicles on the road stop in the same distance. Theory is good but in theory people pay attention to the road. Watch any tyre test and the same car will stop in a different distance based on the brand currently on. I agree that from a physics perspective they should be the same. However from a reality, accounting and engineering compromise perspective this doesn't work in practice. The people using physics to prove why this is the case should be called out.
@boomerau • 4 hours ago > It would be interesting to see if you would stand in 321 feet in front of a fully loaded Australian 2AB Quad road train doing 100kmph to prove that all vehicles on the road stop in the same distance. Yes, except that's not what I said. I said that trucks don't stop in the same distance as cars, and the reason is economics.
So, your 10t car needs 20 wheels and 20 sets of brakes. Stack those 5 cars on each other, so there's only 4 wheels and 4 brakes, and you will _indeed_ have a much harder time stopping. That's the problem most people face as they load their car/truck. Driving around with nothing in the trunk/bed is much different than driving around with a heavy load. The braking force you're used to will not be enough to stop in the expected distance. (and if your trailer brakes go out, then you have a problem... the mass of 5 cars, without the brakes of 5 cars.) As mass increases, so does the work necessary to stop -- inertia. The brakes can easily overcome tire traction - aka. a skid. (I'll leave the murderous nature of ABS for a different time.) Ultimately, the tires are what does all the work - keeping you connected to the ground. That is a finite amount of traction (grip), and it does depend on weight. It's a balancing act between inertia and traction. It's as much a "science" issue as it is engineering. Tires and brakes can only get so big, and there will only be so many of them. When you double the mass, does it double the traction? Usually, it doesn't. If you're curious, look at airplane brakes. That's a huge mass, with tremendous traction, on a surface designed to maximize traction, so they have equally massive multi-pack brakes. They, too, have more brake than traction, and they have an anti-skid system like a car's ABS.
> So, your 10t car needs 20 wheels and 20 sets of brakes. Stack those 5 cars on each other, so there's only 4 wheels and 4 brakes, and you will indeed have a much harder time stopping. Yes, as explained in the video, but let me put it in simpler terms: for two cars of different masses, engineering changes can produce the same stopping distance. But for two cars traveling at different speeds, no engineering change can equalize their stopping distances. I hope this helps, and thanks for writing.
I'd like some more explanation in the numbers being used . Also, did the computer simulations take into consideration: - tyrecircumferance - wheelweight - suspension (and flex) Are there other models beside the SUV? There is also NO sag, they stay level. This sounds like a theory I want to learn more about as it goes against what I've been thaught. Even tho, I still think it will be false. That's why I inquere the numbers used and to see how you corelate them. Thanks
No. He also didn't consider the tyre's modulus of elasticity which is the single biggest influence on friction differences between a laden and unladen vehicle.
@@SwervingLemon Indeed! Good one, there are so many parameters that contribute to braking performance that never were mentioned. After I wrote the comment, I also wonderded about the wheight shift and its effect on grip front and rear. And max braking capacity per tyre. To me this is using the most basic rule of physics to explain something much more complicated than percieved.
The primary thing you need to understand is that vehicle mass doesn't determine stopping distance. My five-car monster vehicle proves that -- its ten tons of mass stops in exactly the same distance as a two ton car. If you reduce the number of tires from 20 to 4, this degrades braking effectiveness, but if the tires and brakes are designed properly to accept the expected loads, the change will be small. Not zero, but small. That's it, in a nutshell.
This video incorrectly compares cars, which have 4 tires rigidly tied together to 18 wheel trucks, which typically have 8 tires rigidly connected together and 10 more wheels rigidly tied together, but those 2 sets are only connected by an imperfect pivot point. If you could eliminate the pivot point and only drive in a straight line, then, in theory, that truck can brake just like a car. Furthermore, since some trucks are empty and some trucks are full, there can be no way to insure that all the tires will brake evenly according to their load. So, in the real world, trucks need a greater stopping distance than cars. Any truck driver who knows their stuff will tell you that. Furthermore, if your tires are bouncing off the road due to an empty truck and large potholes, your stopping distance will increase. This guy also has obviously never driven in a congested area. If you EVER leave more than a car length to the car in front of you, someone WILL cut out in front of you, taking up that space. Unfortunately, tailgating is seldomly enforced where I used to live, which is New Jersey. Again, real world vs theory.
> So, in the real world, trucks need a greater stopping distance than cars. This is why I presented the five-car monster vehicle example -- it proves that this claim isn't true. And when trucks take more distance to stop (which they certainly do) it's because of design, not because of physics. > Again, real world vs theory. The distinction between real world and theory in this case is about economics. A very expensive truck would stop in the same distance as a car, but it would be very expensive. Not because some natural law that requires trucks to stop slowly.
@@lutusp In New Jersey, where I spent most of my life, most truck accidents are caused by car drivers, and their bad behavior around trucks, and this is based partly on their assumption that a truck's stopping distance is the same as a car's. You are being dangerous and irresponsible by perpetuating this myth, and I can only hope that the influence of this video doesn't kill somebody. The fact is that facts turn into falsehoods when they are presented with incomplete information. The only difference between theory and the real world is that theory fails to account for every parameter. Your example of many cars rigidly tied together fails to account for uneven load distribution, widely varying loads, and varying road conditions. Even with today's technology, it's impossible to design a truck that can reliably stop as well as a car. A box truck is only thing I can think of that comes close. The ONLY thing I can agree with you here, is that maybe in the future if someone can design a truck that can stop like a car, it would be far too expensive.
It's misleading to say mass has no effect on braking distance. The peak traction of a tire is dependent on a lot of things like the contact patch area, surface roughness (both in terms of static friction and for the tread to physically deform into), temperature (rubber deforms differently depending on temperature), surface conditions, etc. A counter example would be wet conditions. As you scale up the contact patch with mass, the water between the tire and the road surface has to displace the water a longer distance. Granted it has more force to do so with, but fluid dynamics do not scale linearly (to double the water particle speed you need four times the force). Neither does the contact patch area scale linearly with mass even if you kept the same tire (there's no simple equation for this as it depends on the tire wall stiffness, tire pressure, compound stiffness, cord stiffness, cord pattern, and so forth). This is one of the many reasons truck tires have a much larger diameter than a car tire while the width is still in the same ballpark as a wide sports car tire.
> It's misleading to say mass has no effect on braking distance ... There are second-order effects in everyday practice, but from a physics standpoint the statement is correct. And my example section shows many educational sites that try to claim that mass causes a direct, linear increase in stopping distance, which is certainly wrong.
Your mass having now affect on stopping distance is a classic engineer vs mechanic argument. Is the model and math correct? maybe, In the context of the model. But that isn't real life. you have 5X the cars, 5x the tires, 5x the braking the force, 5x the rolling resistance, 5x the heat dissipation, 5x everything. Of course 5x the mass won't change the stopping distance if you increase everything equally 5x. Take 2 identical vehicles. Fill one up with bricks until it's 5x the mass of the other one. You're really going to try to say they will stop in the same distance? I bet your model with 5x the cars will also prove mass has no affect on acceleration aswell. Because those 5 cars will also accelerate exactly the same as one.
> You're really going to try to say they will stop in the same distance? Yes, and I proved it. Consider that five cars that aren't connected to each other individually stop as though they were one car, in the expected distance for one car. And those five cars, when connected, all traveling at the same speed, stop in the same distance as for one car. How could it be otherwise? The example is really simple, intuitive, and the result is borne out in field tests. Thanks for writing!
@@lutusp I agree 5 cars connected will stop in the distance of 1 car. Just like 5 cars connected will accelerate at the same rate as 1 car. I'm just saying that isn't the same as saying weight has no effect on stopping distance. A more accurate model for that statement would be taking one car and increasing only the weight. eventually the weight will exceed the stopping power of the brakes and the distance will increase. Then you will need to increase the stopping power of the brakes. this is why large trucks have significantly larger braking systems than small cars. The increased weight will increase grip of the tires on the road as you said, to an extent, but won't increase the grip of the pads on the rotors or the rotors ability to dissipate the extra heat produced. There is also increased weight transfer to take into account. but that depends on where the extra weight is added, limits of the suspension ect. I guess if your point is a large truck the weight of 5 cars, with the stopping power of 5 cars, and the tire contact patch of 5 cars can stop in the same distance as 1 of those cars then we agree. because that truck is using (converting to heat) 5 times the energy to stop 5 times the weight. So using 1/5 the energy will stop 1/5 the weight in the same distance with all things equal. Thanks for responding! I appreciate a good youtube video that can get people thinking and discussing stuff like this.
@@cargasm383 > I'm just saying that isn't the same as saying weight has no effect on stopping distance. Yes, but ... wait for it ... no one said that. My claim is that a car's mass doesn't normally increase its stopping distance as is commonly believed, and in cases where this is true, it's because of bad engineering, not physics. Thanks for writing!
@@cargasm383 > I'm just saying that isn't the same as saying weight has no effect on stopping distance. Yes! But I never said that. I said that two properly engineered cars, same speed, different masses, will stop in the same distance. And I proved this in the video, in a way that's nigh impossible not to understand.
@@lutusp At 3:15 in the video you said "a few minutes from now you will know why mass should have no effect on stopping distance" Thats the part I was commenting on. Mostly because it was followed by the comment about shattering beliefs. came off as a "I'm smarter than you and this is why" type comment. Which I found funny because we seem to agree mass does have an effect, but as you said proper engineering can overcome that effect of increased mass.
Too simplified to simulate 5 connected cars to show the same stopping distance. Film a real world example. Drive an empty vehicle, measure the stop at a 60mph or 100km/h. Load vehicle excessively and retest.
Not the same example. My example makes it self-evident that mass doesn't increase stopping distance. In real tests, mass has a small effect on stopping distance unless the responsible issues are accommodated in the design of the tires and brakes. In other words, it's an engineering issue, not a conceptual one.
He has stated many times throughout the video, he's assuming perfect brake bias, which produces axle torque perfectly proportionate to the load on each tire. (A really good antilock brake system) He is also assuming, that no tire is loaded to its sheer strength, which is a limiting factor to stopping force. Coefficient of traction is actually pretty constant, until you exceed the sheer strength of the rubber compound.
Automotive engineer here. Generally good information, but sorry, your formula at 9:00 is wrong. Kinetic energy is universal and has no friction or gravity term. Your claim of 4x stopping distance at twice the speed is physically true, though, just the equations you show have an error. Maybe the term you showed is something else, but it is not energy. Make the unit test: kg * (m/s)^2 / (1 * m/s^2) = kg m is not Joule. In reality, at higher speeds, the tires have less time to grip into the road and the car (not a racecar usually) experiences aerodynamic lift, so the stopping distance is way more than 4x at 2x speed. And do not start with aquaplaning (drop in friction above a certain speed), this will make the comparison completely scary.
> Kinetic energy is universal and has no friction or gravity term. Yes, that's correct, but the equation you refer to describes the dynamics of braking, therefore it must go beyond kinetic+potential energy. For completeness I included the source -- the HyperPhysics mathematical analysis included later in that section (and see below for more on this topic). >> Make the unit test: kg * (m/s)^2 / (1 * m/s^2) This lacks a friction term, which I included in my equation to explain why mass doesn't affect stopping distance in a first-order analysis (second-order effects prevent this from being the whole story). EDIT: The equation you referenced and correctly identified as wrong, is meant only as an illustration -- it requires at least one more term to function as intended. Here is the correct form: E = m v^2 / 2 (1+mu g) This form behaves itself with differing friction amounts, and if no friction, becomes the classic energy equation E = m v^2 / 2. Again, the equation in the video is only meant to convey an idea and won't behave as shown. Thanks for your correction!
@@doczooc > It is stopping distance d = v^2 / (2 mu g) Well, you did correctly identify a problem in my equation -- it fails as anything but an illustration and needs at least one more term to function as intended. If I wanted to include a mass term, it can be written this way, which unfortunately would have left most viewers confused: E = m v^2 / 2 (1+ mu g) This prevents an infinite result in the event of no friction (mu) -- the equation in the video has that rather serious drawback. :) And in the case of no friction, it becomes the classic energy equation E = m v^2 / 2. Again, I thank you for your helpful posts.
> brake quality was more important than speed. When the tires lose their grip on the pavement, brake quality is no longer an issue. Ultimately, the tire friction limit is the deciding factor.
#1. I've never heard anyone (at least anyone of authority) claim braking distance increased proportionally with speed. It has always been square proportional (2x speed = 4x distance, 3x speed = 9x distance, etc...). That is basic middle/high school physics (Ek = mv^2/2). #2 is splitting hairs. While it may be theoretically possible to engineer a truck that stops as fast as a small car (there's a video out there of a kid dashing across the road and a well engineered truck stopping on a dime avoiding tragedy), most trucks in the US at least are not engineered to stop as quickly. In practice, a larger (more mass) vehicle will take longer to stop. The difference is glaring if you've ever driven a truck. #4 should be obvious, though from personal experience having lived in an area with wildfires and hearing about pileups because people insisted on driving through smoke so thick you can't see 10 feet in front of you...maybe its not as obvious to some.
@LC-uh8if • 19 minutes ago > #1. I've never heard anyone (at least anyone of authority) claim braking distance increased proportionally with speed. That's why I posted a few sources verbatim for those kinds of claims. They are legion. > ... The difference is glaring if you've ever driven a truck. Also covered in the video, where I say that trucks really do take more distance to stop, and the reason is economics. > ... maybe its not as obvious to some. We regularly hear pileup stories involving hundreds of vehicles, so clearly someone isn't getting the message. Thanks for writing!
Stopping distance is doubled for a mass that is doubled when equal amount of force is used. Larger mass must use higher force to stop in same distance. With Road Vehicles, slip must also be considered since even if brakes had infinite force, it means nothing when wheel slides. Drivers will then opt to use less brake force so there's less wheel slip, but doing so will increase stopping distance. So yes mass isn't the only factor in stopping distance for road vehicles, but it isn't irrelevant.
> Stopping distance is doubled for a mass that is doubled when equal amount of force is used ... Yes, but that's not the claim. The erroneous claim is to say that a heavier vehicle requires more stopping distance. That's wrong -- such a vehicle requires better equipment to deal with more energy. There is no physical principle that mandates a greater stopping distance for heavier cars, when other factors are properly controlled for.
Sorry, I'm on the road and can't see the specific equation you're referring to. If it includes the friction coefficient μ, then yes, that factor is dependent on mass -- more mass, more friction. The conversion isn't perfect in practice, but it's the reason there is nearly equal stopping distances for light and heavy vehicles, but only if they're engineered to manage the different energies. Thanks for writing!
So five cars with five sets of tires break the same as one vehicle with the same mass and number of cars. Twenty tires have more traction than four tires. The principle may be sound but there is a traction limit. Four tires simply do not have the same tractive effort as twenty tires. Longer stopping distances are required to avoid the limitations of the equipment. Maybe if we put more sets of tires on a big rig we could get it to stop at the same distance as a car, but that's just not practical.
> Twenty tires have more traction than four tires. Traction depends on a combination of surface area and downward force, so ... no. This doesn't mean a tiny tire surface will bear the load, but friction force increases as area decreases. Smaller area, greater frictional force per unit area.
Self-driving cars will still crash, but hopefully less often than we do as they can understand the physics and engineering issues you teach here, if not as a concept as a practical implementation.
Yes, and as people are discovering, self-driving cars are a much tougher problem that we thought a few years ago. Two companies have recently stopped their self-driving pilot programs after injuries and deaths, and Tesla's "self-driving" claims are being challenged by regulators. But eventually we will solve this one. Thanks for writing!
While I do agree there is nothing in physics that says heavy vehicles should stop slower (or faster), the real world is slightly more complicated than the simple dry friction model taught at school, F=mu*N. In practice, as the friction force gets so large that it causes stresses in the materials involved approaching the strength of the material it self, the coefficient of friction starts to taper off, as the material starts to give. In the case of rubber, it is also a viscoelastic material, which causes the onset of degradation of grip to set in even sooner. The result is that vehicles with lots of weight compared to the contact patch of the tires do have worse stopping distances. Such vehicles of course also have better rolling resistance, and are thus more economical to drive. So yes economics are indeed the main driving factor of elongated stopping distances of utility vehicles.
We can take the energy approach and see that during braking _all_ the kinetic energy has to be converted to something else for the vehicle to become stopped. Most of the energy transfers into heat; in the brakes, the tyres and I suppose the road as well. The original equation for KE at 9:04 is the correct one, it just doesn't directly give the distance. The "corrected" one relates to distance, but is _NOT_ the energy (rather closer to a momentum). So the amount of heat generated is directly proportional to the mass, and a 40t vehicle would need to dissipate 20 times as much heat as a 2t one with the same pre-brake speed. For the distance to remain the same, the time must remain the same, which means that in practice the brakes might have to become so hot that they might break any reasonable material and design limitation. Therefore they must be designed to extend the time instead, and consequently will have a longer stopping distance. (Of course it also helps to have larger and more wheels&brakes to distribute the heat dissipation among.)
> The result is that vehicles with lots of weight compared to the contact patch of the tires do have worse stopping distances. Yes, but that is an engineering problem, not a physics problem. If you control for extraneous factors, the result agrees with the common-sense expectation. I did say that an expensive make-over would allow the truck to stop in the same distance as a small car. And it's true -- too expensive for everyday reality, but I was addressing a principle, not a budget. Thanks for writing!
Whilst I agree from a physics point if view on stopping distance being not dependent on mass - You are assuming ideal brakes working efficiently , and this being an engineering issue means in the real world stopping distance can depend to some degree on mass, and how slowly some vehicles are allowed be designed to stop
> ... and this being an engineering issue means in the real world stopping distance can depend to some degree on mass Yes, and I made that point. Real trucks really do take (much) more distance to stop than a small car, and this results from how they're designed.
Your clinical values of pavement friction are ignoring tire slip on the rim. At such great frictional loads that fewer tires would have you're more likely to increase rim slip. These are fractional increases in stopping distance . . . that is until you blow out a tire. This is the reason the 18 wheeler needs so many wheels, it needs the reduction of frictional loading. Observe a cement truck that has a drop down extra pair of tires in the back. These tires are for traveling when the truck is full and can stay stowed out of the way when the truck is empty. Your point that a mega vehicle stops at the same distance 5 individual vehicles would stop is sound except you'd need specialized tires.
> Your clinical values of pavement friction are ignoring tire slip on the rim. If the tires slip along the wheel rim, something went wrong in the tire shop and the company is responsible. But this isn't a realistic scenario if the wheels and tires are appropriate to the vehicle. Thanks for writing!
@@lutusp >appropriate to the vehicle. So just math then and not real world physics. Have some special tires been invented that would be appropriate on both a 20,000 ton car and a 2 ton car? Your math checks out, but something is going to give because material science isn't keeping up.
@@djdickey > Have some special tires been invented that would be appropriate on both a 20,000 ton car and a 2 ton car? No one made that claim. The right tires and brakes can be designed for each example vehicle, not some universal solution for any vehicle.
> How drunk or decrepit do you have to be to have a 1.5 second reaction time? Cell phones. Car radios. Children in the car. ... Children in the car will cell phones. Dogs and cats. In science, the idea is to control for all extraneous factors in order to be able to study just one.
@@lutusp If you use a cell phone while driving, if you don't have the presence of mind to focus on the road while adjusting the radio, if you did not adequately prepare to accomodate children or pets OR are one of the people who "Can't walk and chew gun at the same time" or "can't talk and walk at the same time", any accident as a result of your failure to hit the brakes in time is 100% your fault and should not even be considered here.
1:10 It is *true* that the braking distance goes up by the square of the starting speed. But your explanation is *also* a myth. You assumed that the rate at which we can dissipate energy will be constant. Even though in the case of dissipating energy as heat via friction, the rate increases linearly with speed. Which actually cancels out the squaring of the energy. And so it will only take _twice_ as long to get rid of all the energy when our starting speed is _twice_ as fast. (If the dissipation rate was constant, then the deceleration would increase as speed decreases. Which is not what we observe in reality, nor in your own simulation.) BUT! That is time, not distance! The total distance the car will travel is the same as the average speed multiplied by time. And when we double the starting speed, (assuming liniar deceleration) we also double the average speed. And that is why the braking distance goes up by the square of the starting speed! We spend *twice* as long getting rid of that speed. _And_ we go *twice* as fast (on average) while doing it! (2time * 2speed) = 4distance! If you want a visual example. Draw a graph with the vertical axis as speed, and the horizontal axis as time. (0, 0 preferably where the two axis's intersect.) Draw a line that starts at some amount of speed and has some downwards slope. Now draw a second line with the same slope, but have it start at double the speed. You now have two triangles, and the area of these triangles will be (speed * time). Also known as distance. Visually, 4 of the first will fit in the second. And the two edges that form the right angle at (0, 0) have independantly doubled from the first to the second, just from changing the starting point. By doubling one value (or starting speed), we have actually doubled two seperate values (Average speed and time), which are then multiplied together for our result (braking distance), turning our liniar increase in starting speed, into the exponential increase in braking distance. And that is *actually* why the braking distance goes up by the square of the starting speed! Your fault was to look at the quadratic increase in energy with speed, and the quadratic increase in braking distance, and assume they were related. Even though they are not.
> Your fault was to look at the quadratic increase in energy with speed ... The values I showed are based on physics, then verified in field trials, which agree. Thanks for writing!
I don't think it would be completely inaccurate to say that mass increases the braking distance as there is a noticeable difference to my carry truck when I have a 170KG generator in the back. Though I get that bigger cars could probably stop quicker with the hefty brakes they have on them, especially high performance spec cars.
That should not be the case. The car's on board computer would have detected more weight and adjusted the braking system accordingly so there is no noticeable difference when braking and accelerating. If the added load means the car is taking noticeably longer to stop then there is either something wrong with the electronics/braking system or you have dangerously overloaded the vehicle, and in either case you must not be operating it on the road in such unsafe state.
@@bethhughes559 It is a little compact pickup tuck from 1990, it has no electronics other than the lights, gauges, fans, ignition, starter, and the bed tilt pump. I am also not overloading it as the SWL for the bed is 500KG.
@@arianaashworth3091 From 1990? A 33 year old vehicle shouldn't really be on the road anyway, they are unsafe and give off dangerous emissions. Perhaps consider replacing it with an EV.
@@bethhughes559 I am not getting rid of it just because it is older, there's nothing even wrong with it! It passed the MOT with no issues and it has had a catalytic converter retrofitted. A few people I know who own electric cars have been having problems with the charging network, which is actually one of the reasons the generator ends up being transported around on it in the first place.
@arianaashworth3091 • 8 days ago (edited) > I don't think it would be completely inaccurate to say that mass increases the braking distance as there is a noticeable difference to my carry truck when I have a 170KG generator in the back. Yes, but my video addresses the basic physics, not specific examples that have extraneous, uncontrolled factors. In science one must eliminate all factors but one -- the factor being studied. Thanks for writing!
Of course, assuming humans will be humans, whatever you do may be completely irrelevant: you may get into an accident and it could still be someone else's fault. Happened to my mom twice.
I assume you mean actual physical cars, not computer graphic cars. It's not easy to get the desired forces and effects without full-size cars, or computer models of full-size cars, which is what I chose. Thanks for writing!
@@lutusp Yea, I meant full size real cars, obviously it is very easy to lie and cheat with computers, you could give them any stopping distance you want. And I don't mean to accuse you of anything, my point is only that even by accident it is easy to mislead with computers.
@@MegaLokopo I should explain something -- the numbers in the computer simulation result from a combination of theory and simple confirmation through field tests using real cars. The field tests came first, then the theoretical explanation. Put more simply, it says, "You have the field measurements with real cars, now here's the explanation." Also, in science, a theory than cannot be confirmed in nature must be discarded. This one is amply confirmed by experiment. I hope this helps.
@@lutusp so why not show the ample confirmation made by the physical experiment. In the video you make a claim but you don't prove it or show real world evidence, if you have the evidence you should show it.
@@MegaLokopo That corpus of evidence is large and readily available. My video is meant to summarize that evidence. Consider -- if someone performed a test that contradicted modern theory -- the theory my video explains -- that person would become famous overnight. But hey -- knock yourself out: www.google.com/search?q=car+stopping+distances
> "Braking 60-0 in the dry is more like 130 ft in sedans." In field tests, and consistent with theory, and assuming adequate brakes, tires and equal axle loading, all vehicles stop in the same distance for a given speed. This is why safety agencies list vehicle speed and correlate that with stopping distance, but omit vehicle mass. But if a vehicle doesn't have brakes and tires consistent with its mass, or is loaded improperly, this relationship goes out the window. > Many can do it in even less depending on ABS and driver skill. No, not so. Once ABS kicks in, the stopping distance is fixed, and beyond a driver's ability to change it -- except, of course, to make it longer by releasing the brakes.
@@lutusp 130 feet is not unrealistic. Motor trend reports 2013 Audi A5 coupé to have stopping distance 60-0 of 101 feet in the dry. 3500 lb is typical sedan weight. Therefore by your logic it should be more like 100 ft 60-0 right? Not 180? ABS gives up braking distance to allow steering. If the tire was using literally 100% of its grip to stop, you wouldn't be able to turn. ABS pulses the brake, when the brake is released, turning capability is restored. Therefore ABS is technically not the greatest stopping distance possible. However I will say this is a minor caveat I wouldn't hold against your analysis. ABS is good enough and is often super close to the theoretical limit. Just a quirk of how it works if it's using PWM
@@lutusp just wanted to add, I don't dispute your conclusions at all. Just the actual realistic braking distance. It's fine to present a worst case scenario but you should at least mention that 180 ft is just about a worst case 60-0. I have watched so many dashcam videos where people act like they've never even heard of a brake. Probably at least half of drivers lack the necessary understanding of car handling enough to be safe at modern highway speeds. So, it's genuinely not that bad to assume worst case distances. But, attentive and skilled drivers will do far better than the numbers you are presenting in this video.
@@HarryTheOwlcat > But, attentive and skilled drivers will do far better than the numbers you are presenting in this video. Yes, that's true, mostly having to do with reaction time, which varies greatly from driver to driver. Other factors are road surface conditions, brake design, and differential axle loading, all of which can undermine the effectiveness of a stop. In many ways my numbers are worst-case, for example the three-second rule and an imagined 1.5 second reaction time. But safety agencies agree with these conservative limits and precautions. Thanks for writing!
@@HarryTheOwlcat > It's fine to present a worst case scenario but you should at least mention that 180 ft is just about a worst case 60-0. Yes, true, but in a publicly released video that might change a young driver behavior, it's best to look at published field-test stopping distances and choose the worst case. Consider the alternative -- a distraught mother decides to hold me to account for her son's avoidable accident. Weirder things have happened. :) Thanks for posting!
I'm confused where the line is between physics and engineering. If a car had the weight of the moon, of course its braking distance would increase above that of a regular car. But I guess that is because the load has too little contact (only 4 tyres). Braking force isn't increasing proportionally to its mass. So I guess the solution is engineering more wheels?
In realistic circumstances, increasing vehicle mass doesn't increase braking distance -- that was the point of my five-car monster vehicle example. As you reduce the tire contact area, the relationship begins to break down, but not in a linear way, until the brakes cannot handle the energy or the tires melt under load.
@@lutusp You keep referencing "until...the tires melt under load" - this is not the only limit, mechanical grip of the tire can easily be exceeded without them 'melting'. Braking on ice or loose surfaces shows this. This limit of mechanical grip is certainly (in the real world) a significant factor in the reality of heavier vehicles requiriing more stopping distance (eg large SUV or semi vs a compact). Tires can not be manufactured to have 'infiinite grip' especially on poor surfaces like dirt and ice. While we like to simplify problems to get at the 'base' understanding in physics, we can't lose sight of the 'real-world'. In reality, heavier vehicles take more distance to stop - that is the fact that exists and people need to know, most do not need to know the 'ideals' underpinning forces and motion. (Yes, I do have a physics degree - currently working on my PhD but a ways to go yet).
All your points are correct and important, bur in a video meant for non-specialists I try to avoid too much technical detail. If the point is to alert people to driving dangers, there's no point in creating a perfectly accurate video that no one watches. Thanks for posting!
@@lutusp Alert drivers to what? The fact of the matter is, your model of a constant deceleration is flawed enough it could be considered misleading. Big trucks simply do take longer to stop due to material constraints, and limited contact area between disk and brake pad. Drivers should therefore take more caution around them. Saying that we shouldn't model with mass in respect to stopping distance is flawed.
@@BrainiacManiac142 > Saying that we shouldn't model with mass in respect to stopping distance is flawed. Yes, it would be better to instead provide a graphic, easily understood example that falsifies the presumption. But then, I did that in the video. Everything after that is about engineering, not theory. Falsifying a commonly held wrong belief was the primary aim. Thanks for writing!
So in theory I could fit the brakes, wheels and tyres from a classic 690kg Mini to my 1525kg Jaguar XJ and it would stop just as well as a classic Mini? I don't believe that for 1 second and clearly neither did Jaguar.
> So in theory I could fit the brakes, wheels and tyres from a classic 690kg Mini to my 1525kg Jaguar XJ and it would stop just as well as a classic Mini? Nope. But then, I never made that claim. More massive vehicles require more massive braking systems. Thanks for writing!
@@matthewjenkins1161 > Well you did make the claim the weight makes little to n difference though. No, I proved graphically that heavy vehicles can be engineered to stop in the same distance as light vehicles. The five-car monster vehicle example demonstrates this point in a way that's impossible to dismiss. In everyday reality, if appropriate tires and brakes are provided, the same principle applies.
@@matthewjenkins1161 > They need to be because weight DOES make a difference. Yes, as I explain in the video. That difference can be engineered away, but only if we are willing to do it.
> Uh.. Kinetic energy of a car certainly is E=mv^2 and not whatever formula you showed at 9:05 ... Yes, that's true, but the discussion at that point is stopping distance, which requires us to take friction into account. Thanks for writing!
The formula he showed is actually similar to the acceleration of the force of gravity on a falling object. Braking is the reverse, much like the Deceleration of an object, that has been tossed straight up. The formula for kinetic energy applies to the total heat dissipated by the brakes, bringing the car to a stop from the various speeds. Yes, the car doing 80 takes 4 X the time to stop as the car doing 20, but, since the velocity is 4 X faster, it indeed covers 16 times more distance, given the higher velocity, multiplied by the longer time it takes to bring it to a stop.
I often ask people why we wear a seatbelt, and most people never think about the glass cutting you open from the windshield. With simple propagandized safety messages, people never learn WHY we do what we do.
That fog example was frightening. In dense fog, the only safe way to keep driving is to reduce speed, so that you can stop within the distance you can see (in fact, that's true of driving in general).
> ... so that you can stop within the distance you can see ...
Yes, true, plus the fact that the "three-second rule" won't work if you're driving toward a prior crash and can't rely on the expectation that the car in front of you is still braking to a stop. For that case you need even more time to react. This makes fog more dangerous than most people realize.
Thanks for writing!
The problem then is with people behind you driving dangerously fast. (Which is probably most drivers in foggy conditions, considering even with 100ft of visibility you would need to drive 25mph to stop safely) It is probably more dangerous to drive slowly in foggy conditions due to the likelyhood that someone else reckless behind you will make a mistake. The only safe option in foggy conditions is just to wait it out, as the video says.
Some cars have for a while thermal camera superimposing view on the digital dashboard and nowadays perhaps even a head-up display not to mention that you have also had radar-assisted automatic emergency breaking for over a decade but generally, it's better to slow down than to trust with your and other's lives on tech, especially as many emergency breaking assistants like one in newer Tesla cars given up on the radar that made them pretty useless in fog while radar or even LIDAR can experience malfunction or interference that cannot be easily spot. So, until high-speed thermal cameras with sufficient resolution in cars proliferate, and traffic signs are redesigned to be clearly visible in the infrared spectrum, it's better to plan ahead and avoid driving in an area with heavy fog if you can...
@@lutusp There is also the issue that people driving in fog underestimate their speed, resulting in them going faster. I'd add that drivers should be using their hazard flashers in low visibility conditions to extend how far the driver behind them can see you from.
@@IonorRea The technology of assisted automatic emergency breaking may have been in beta for a decade and I feel like its still there. I've had my car suddenly slam to a stop because of a shadow on an otherwise sunny day. It can't tell the difference between a two dimensional object or a three dimensional one. Also my daughter just collided with someone who was at a stop sign (fully stopped) when the emergency assist didn't react fast enough. My daughter felt it took over because when she hit the breaks the car didn't slow down even faster. Were waiting for the investigation on that. So our theory is the vehicle prevented her from applying the breaks in an emergency force vs the computer thinking it just needed to slow. The collision was slow enough for minor damage and hopefully the car can say how far it traveled while the computer took over breaking. The investigator will compare it to it's normal stopping distance at the speed that the computer says she was going (48mph). I personally would drop the title "emergency" assist and just label it "assist if it wants to even if its wrong".
The great things about computer simulations, the computer still only does what it's told.
And great thing about physics is that it only does what is told by its laws.
@@whoakshatsharmaAnd the great thing about physicists is their misplaced self-confidence...
@@whoakshatsharma That is inappropriate anthropomorphism. Physics doesn't do what it is told.
Objects behave predictably. Physics accurately describes that predictable behavior.
The great thing about most drivers is that they think they are the exception and are invincible until they aren't.
@@marc21256 > "Physics doesn't do what it is told. " True. One could say it conforms to the underlying laws, and even that sounds like anthropomorphizing, but most people follow the logic.
This is why the only true physics is mathematics. I don't like putting equations in my videos (it loses too many viewers) but sometimes I must. :)
I was expecting a bunch of fear-mongering nonsense, but ended up watching a VERY well-informed video. Thank you for the strictly scientific approach. Too many people talk with emotions when they get into driving and safety, so I’m glad to see somebody not do that for a change.
Thank you for your kind comments!
One must also remember that the road surface has an effect on braking distance. Both the obvious friction question, and also elasticity and things like rolling particles (gravel for example).
Yes, all true. When doing science it's a good idea to focus on one factor at a time, to avoid creating results that have more than one explanation.
This strikes me as a good example of how to believe your own crap doesn't stink. Not only do we ignore some of the most important factors, tire grip, suspension to accommodate weight transfer during braking, and overall brake performance...but we also get to hear a fantasy land world where it takes three seconds to apply a brake. If it took three seconds to apply the brake, there would be an accident at every intersection and parking lot every few minutes.
> ... but we also get to hear a fantasy land world where it takes three seconds to apply a brake.
But that's not the claim. It doesn't take three seconds to apply the brakes, it takes three seconds to react to a situation that requires braking. Then the actual braking begins.
FYI Britain also still use MPH and miles
Thanks! I'm sorry to hear that, but thanks anyway.
The formula is the same, no matter what unit of measurement is used. (MPH X 1.609344.=KMH)
3:34 This example doesn't take into account that the 'monster vehicle' has 5 times more tire contact patch as well, since it has 20 tires. Mass per tire contact patch is also important. Otherwise, overinflated tires would lead to the same stopping distance as correctly inflated ones.
Yes. He's forgotten to factor the modulus of elasticity of the tyre itself.
Both of you apparently stopped watching to post an ignorant reply. He covered this very subject moments later. Showing how the additional tire friction did not change anything due to the fact that each wheel ends up sharing a smaller percentage of vehicle weight. Which in turns reduces the amount of friction at wheel. It's why a dulley pickup trick has less traction in the rear on wet roads then it would if it only had 2 rear wheels. The vehicle weight is spread out across 4 instead of 2 wheels. The 2 rear wheel truck will have a better time pressing down to the pavement then the 4 rear wheel truck will.
@@williamallen7836 Same net downward force on a smaller contact patch will result in more contact pressure, which is another factor affecting traction and braking distance. He assumes that the relation between the net downward force and the friction force is linear, namely Fricition = μN, which is not the case as the contact patch gets smaller.
> This example doesn't take into account that the 'monster vehicle' has 5 times more tire contact patch as well, since it has 20 tires.
Actually I did cover that issue with a comparison table that trades off contact area and downward force, I also mentioned that actual tires stop following the rules if exposed to loads they cannot handle. But to a first-order approximation, fewer tires with more loading provides the same overall braking friction.
Thanks for writing!
@@ohanneskamerkoseyan3157 > " which is not the case as the contact patch gets smaller." Fewer tires exert more force per contact area, so the correspondence I describe remains approximately true until the materials begin to break down. The posted equation describes an ideal correspondence, not an exact one. BTW the HyperPhysics page I linked to agrees with my analysis.
Nevertheless, when safety agencies post stopping distances, they don't include vehicle mass in the calculations -- there's no need.
I agree, this is theoretical to the point of being misleading itself.
> ... to the point of being misleading itself.
Not as misleading as the many online sources that try to say that mass directly increases stopping distance, as with my quoted examples.
@@lutusp Tell that to those who have been crushed by heavy vehicles in crashes because they could not stop.
@@patrickbuick5459that's likely because their brakes failed, which is by the way not uncommon for poorly maintained trucks. This can also happen if the truck is overloaded (i.e. loaded more than the certified maximum allowed mass).
The video assumes the brakes are working properly and the truck is properly loaded.
@@streettrialsandstuffIt also assumes everyone knows how to drive - following distance, reaction times, and everyone knows how to "threshold brake". I know professional racecar drivers who can't threshold brake. (if they didn't have ABS, they'd be dead.) For the simulations to work, everyone has to be braking at maximum. Nobody does that in real life. The reality is as mass increases you need to use more brake to stop in the expected distance, and people don't expect that. We're used to a certain feel (g-force) under braking that's way below maximum. As the mass increases, that force will not stop you in the same distance.
This is about a physics principle, which is correct, not about a badly engineered vehicle that shouldn't be on the road in the first place. The point of this kind of science is to point out, not how things are, but how they ought to be.
Whilst you are absolutely correct in 95% of what you say, your analysis of mass vs stopping distance does not take into account the brakes themselves. When breaking, there are two sets of friction at play. The first, which you take great care to analyze in detail, is the friction between the wheel and the road. As you point out, increasing the mass results in increased maximum available friction, thus cancelling out the increased force required to stop the vehicle. The second, however, is the friction within the brakes themselves. This is unaffected by the mass, since the force is applied solely by the pressure of the brake pads on the disks / rotors/ drums. This force comes from the driver's pedal (possibly boosted by power-assisted brakes) but will not automatically be increased with an increase in mass. Don't forget, in this case we actually don't want the friction to be high enough to lock the brakes up, which further complicates the calculations The whole situation gets even more complicated when ABS is brought into play. Your example of the cars joined together works when it comes to road friction, but fails to take into account that the five joined vehicles have five times as many sets of brakes.
Overall, I think that as long as a car is operating within it's design specifications, it probably doesn't make much difference, but we all know that many people overload their cars (either directly, or indirectly) and in these cases assuming the mass has no overall effect can be dangerous.
> ... your analysis of mass vs stopping distance does not take into account the brakes themselves.
My analysis assumes that tire-pavement friction resists the car's forward motion. The brakes must allow this to happen, so they're not part of the analysis, which focuses on that one factor.
In science, one must control all factors in order to be able to study just one.
@@lutusp I agree with your statement about Science, which is why I posted here and why you are incorrect.
Braking relies on two types of friction, each of which must work correctly and, if either changes to the other, you will skid.
Static friction is defined as the frictional force which acts between the two surfaces when they are in the rest position with respect to each other. This is what you have / want between the tire and the road surface. If this friction breaks down, you have a skid. As your analysis correctly explains, part of the calculation of the maximum force that can be applied before this friction fails involves the pressure between the two surfaces - caused by the mass of the vehicle. Therefor, as you say, as the mass increases, so does the maximum frictional force and one cancels the other out.
BUT
Dynamic friction is defined as the frictional force which is created between any two surfaces when they are in a moving position. It is also called Kinetic Friction. This is what you have within the brakes themselves, between the pads and the rotors / disks / drums. It is this friction that actually dissipates the energy that the moving vehicle is losing (basically, converting the vehicles kinetic energy primarily into heat.) If enough force is applied to the brakes, then this friction will increase to become static friction, the brakes lock up, and you have a skid (see above.) It is this situation that ABS is designed to prevent, by removing the braking force as the brakes approach locking up.
So, to avoid a skid, you need for the static friction between the wheels and the road to be more than the force being applied to stop the vehicle (which is what you analyzed) AND the dynamic friction within the brakes to be high enough to dissipate all the energy. Now, once again, force and area come into play here. The smaller the total area of the brake pads, the greater the force required to dissipate the vehicles kinetic energy in the same time. There are two risks here. If the amount of energy to be dissipated is too high (due to the mass of the vehicle) then the brakes will overheat and either lock up, or fail completely. If the force applied to the brakes is too high, then the brakes will lock up. Either form of locking up will mean a skid. (I'm not going to go into why a skid is bad here, but it revolves around the transition between static and dynamic friction, and back again, if you want to look it up.)
In your example, with multiple vehicles, and therefore wheels and, as a result, sets of brakes, the total area of the pads is greater, meaning less pressure being required to provide the braking force and the heat being dissipated over a greater area. When you reduce the number of wheels, you also reduce the brake pad area, which means each must dissipate more energy to stop the vehicle in the same distance. This means more heat and a greater pressure. This is not automatically compensated for as it is with the static friction you analyze.
So, if you increase the mass of the vehicle, you either rely on the brakes doing more work without failing (assuming you are able to apply sufficient pressure) or you have to apply the same, lower, force for a longer period to stop the vehicle, which increases stopping distance.
As you say, "In science, one must control all factors in order to be able to study just one." In this case, you have focused (and correctly analyzed) one (the static friction between the wheel and road) but ignored the other.
I should also point out that we are not talking about the extra mass caused by having an extra person in the car, which is relatively minor, but rather massive differences.
If you don't believe me, try speaking to truck drivers who transport heavy loads of steel and tell them they don't need more stopping distance when fully loaded! They will laugh you out of the room.
> If you don't believe me, try speaking to truck drivers who transport heavy loads of steel and tell them they don't need more stopping distance when fully loaded!
This is not about existing trucks, which represent engineering and economic compromises. This is about physics. In physics, a more massive vehicle presses down with more force, and if economic considerations are set aside, if two vehicles of different masses have been provided with appropriate tires and brakes, they will stop in the same distance. This ought to be self-evident.
You see, in science, to study factor Z, one must first control for factors A-Y, otherwise the result is tainted by uncontrolled or unknown issues.
Let me put it this way -- for two cars of differing masses, they can be engineered to stop in the same distance -- different tires, different brakes. For two cars traveling at different speeds, they cannot. That was the point I was making in the video.
> ... which is why I posted here and why you are incorrect.
Some advice. Never address a forum's participants as though they are the topic. Instead, address the topic itself. One can say, "Your position on issue X is incorrect." But don't say, "You are incorrect." Of course, this depends on whether one intends to generate heat, or light.
@@lutusp OK, "Your position on this is incorrect". Happy now?
The video doesn't talk about re-engineering and having a new vehicle with upgraded brakes. It simply challenges the notion that a heavier vehicle will require a greater stopping distance. The video states this isn't the case, based on the static friction between the wheel and the road. That part I can appreciate, and have acknowledged, but it ignores the brakes themselves and the dynamic friction element, which is my point.
You talk (above) about studying science and fixing all but one variable, to allow you to study the effect of that variable, yet you fail to acknowledge that there are at least two variables at play here - which is the point I am making! In fact, the claim of upgrading brakes actually introduces an additional variable! If you want to argue that the brakes can be made more efficient, then one can also argue that the tires or the road surface can be improved to make the static friction better - negating the whole basis of the point being made! You say that "for two cars of differing masses, they can be engineered to stop in the same distance", this is true, but is not the same as the point made in the video, with the analogy of linking cars together.
For the most part, the video was informative and helpful, but in this aspect the video presents an argument that does not work. I see other commenters have drawn attention to the "computer simulation" aspect. Any simulation is only as good as the way it is written and will only include the factors the programmer considered. There comes a point where you have to do it "for real" to get complete data. This is why manufacturers use Crash-Test dummies and actually crash real cars. Simulations can be useful, but in and of themselves, they don't prove anything. All the simulations in the world will not account for every factor. This is why I mentioned truckers. Their real-world experience and expertise shows the fallacy of the "weight doesn't matter" argument.
You mention, in your comments, that this is about science. True, but science is, at it's heart, an attempt to explain and understand the real world. We have ideas, propose hypotheses create experiments to test those hypotheses and use the results to refine our models and inform our understanding. Yes, physics and Math are closely related, but any mathematical analysis will inevitably be an approximation of reality.
@@paulsimpson6290 > but any mathematical analysis will inevitably be an approximation of reality.
Not if we understand reality, because modern physical theories are mathematical -- no guesswork, no unevaluated gray areas.
There are two kinds of scientific theories -- mathematical, and incomplete. No one in science accepts "approximations of reality" as though that is the goal.
The reason we know the Standard Model isn't complete is because there are experimental outcomes that the math don't explain. And no one is willing to say that our scientific theories are only educated guesses. Leave that to the psychologists.
1:22 Braking distance isn't about energy (that comes into play only in mega-heavy vehicles, or repeated stops)
Braking distance isn't governed by E=mv^2/2, but by x=1/2at^2+vot+xo where t=(Vo-Vf)/a, and a=F/m~g*mu, mu~1 for ideal conditions, and down from there. So a small mu (like ice) will increase the stopping distance proportionally.
So simplifies to x=3*V^2/(2*a); x~ v^2/22 (for feet per second)
So 20 mi/h ~ 45 ft to stop
So 40 mi/h ~ 158 ft to stop
So 80 mi/h ~ 622 ft to stop
yes, I skipped the step to convert mi/h to ft/s. Because I'd be using m/s for all of it...
Big trucks stop slow because of cost. No laws of physics would be broken making an 80,000 lb loaded truck stop in the same distance as a Porsche 911. It would just be expensive, and maybe heavy.
> Braking distance isn't about energy ...
But it is. If there were no energy, there would be no need to dissipate that energy as heat in the brakes.
Thanks for writing!
@lutusp and repeatedly stopping demishies the stopping distance as the brakes get hot from repeated heat cycles. There is a point of failure, a point where the materials no longer can perform like intended. Everything is based on ideal conditions. So when weather changes so does your ability to stop. Even the condition of the components on said vehicles. From swelling brake lines to water saturated brake fluid. To many variables to make assumptions. Hot Temperatures are bad for brakes, they are less effective in those conditions. So more mass means more heat, which demishies the effectiveness of the braking system. So mass does cause a vehicle to stop further if fully loaded than not. You are not including the heat soaking of the braking components. Do a heat test with a thermal camera on a fully loaded vehicle and empty and get back to me on how the temperature of the brakes are the same. 😅
Gotta love a mythbusting video that none the less perpetuates other myths. The U.S. in not the only country using MPH, a quick online search verifies this.
Yes, too bad but true, fewer countries than I thought are fully metric. Thanks for posting!
This video contains some great demonstrations and good points, but I noticed a mistake.
9:03 Where did you get the formula for E from? It is completely wrong, and dimensional analysis shows it doesn't even give you energy. The "wrong!->" formula is completely correct for calculating the kinetic energy of a body. Your next slide you bring up to "show the correct formula" even shows you that same formula for the kinetic energy that you labeled "wrong!->" (Work = mgμd = m*v^2/2, where the third element is the kinetic energy).
In ideal conditions, I think you have shown clearly that braking distance doesn't vary depending on the mass of the vehicle.
In reality, there exist some factors which could change this however. For instance, if weak brakes are used, a human could feasibly not be able to generate enough braking force to max out the tyre friction.
If a trailer with bad brakes is used, it would carry its weight on its own axle, meaning its weight doesn't contribute to the braking force of the vehicle, and stopping distance could become dependent on mass. This will happen in any vehicle whenever weight is loaded above an axle with a bad/ no set of brakes.
Also, hot brakes can become less effective, and the more massive the vehicle, the more energy you have to put into the brakes, making them heat more. Hypothetically, this means mass could affect braking distance.
> Where did you get the formula for E from?
Which equation? There are several. In one, I provide the canonical form: E = 1/2 m v^2. In another, I accommodate the effect of friction, which agrees with field tests that show little to no effect from increased vehicle mass and also agrees with my monster-car example, which proves the point in a simple way.
In a slide I call wrong, the author uses the canonical form shown above, then concludes that this means more massive vehicles require more stopping distance, which is wrong. That's my meaning for "wrong" -- it misapplies the canonical energy equation and fails to take enough factors into account. My replacement equation does do this.
The remainder of your comments refer to extraneous factors, all true, but all about economics and engineering, not physics. My point is that the physics makes mass nearly irrelevant, and many people don't understand this.
@@lutusp Thanks for the response. I don't completely agree the other factors I bring up are extraneous, nor that they aren't fundamentally concepts that have to do with physics, however I completely agree that if we disregard those factors, stopping distance should not depend on mass. The formula for E I was referring to was the equation labeled as "right" at the moment of the time stamp 9:03, and this the remainder of this comment will be specifically about what's on screen in that moment.
You label the correct equation for kinetic energy as "wrong" while saying "In this case, I include the correct equation just for contrast". At the same time, you show the equation E=mv^2/2μg, which uses the variable name E, which is convetionally used for energy. You then say that the frictional term in the denominator and the mass in the numerator cancels out, which would give the final units of the equation being energy if true.
It is however not true that μg cancels out with m, as μg has units of acceleration, while m has units of mass, meaning they cannot cancel out. So the formula is not correct for calculating energy if this was its intended use. If the formula was meant to calculate stopping distance, it would also be wrong, as the formula for stopping distance is mv^2/2mμg=v^2/2μg.
It is true that the inference made about stopping distance in the sentence below the formula for kinetic energy was wrong. However the formula for kinetic energy wan't wrong, so I don't think you should have labeled it as such. It is in fact the formula you include of E=mv^2/2μg that seems to be wrong, as it neither gives you stopping distance, energy, nor do the terms m and μg cancel out as you say.
@@HenrikMyrhaug > You label the correct equation for kinetic energy as "wrong" ...
Yes, because in the page's context, it's being misapplied as though it can be used to correlate stopping distance with vehicle mass, which is certainly wrong. The original page's author then draws a classic wrong conclusion based on an equation that, in a different context, would be correct. My equation on the same slide tries to correct for this error, and though only approximate, it's a big improvement -- it at least addresses the context of stopping distance.
In theory, yes, heavier vehicles should have the same stopping distance as lighter vehicles. In practice that's not the case. Now which matters more for driving safely...
> In practice that's not the case.
In practice, with properly designed vehicles, it is the case. My point is economics decides this, not physics. Trucks that can stop within the theoretically established distance are more expensive.
> Now which matters more for driving safely...
If driving safety were the only issue, all vehicles would stop in the same, optimal, distance. And they would be more expensive.
Thanks for writing!
@@lutusp How often are people driving properly designed vehicles? What counts as a properly designed vehicle? Has anyone ever made one?
Driving safely counters all imperfect functions of a vehicle.
@@MegaLokopo The expression "properly designed vehicle" is like citing "a reasonable person" in a legal proceeding. It's a fiction, but a useful one.
@@lutusp That's true. I guess you did use properly and not the word perfect.
Thank you Paul. Very well done. I'll have my 16-year-old daughter watch it!.
A related topic worth examining is "How long should the traffic light yellow be illuminated before turning red -- for various speed limits?"
I agree with that idea, and thanks for writing!
You did not prove anything. Higher speed vehicles stop in distances proportional to speed squared, but this is not because enery. Brakes apply generally constant pressure and this makes higher them more powerful at higher speeds ( remember formula E=F*X) . Twice the speed twice the stopping time, but average speed for that time is also higher for faster vehicle. So it is proportional to speed squared.
Heavy vehicles stop harder because tire and road surface interface is not a simple thing to be represented as only friction coefficient. Heavier vehicles use harder tires to be economical, or their tires simply wil be worn out very soon. This makes trucks and heavy vehicles stop in longer distances. Also the brake systems are different, air brakes are generally less responsive than hydrolic brakes of a small vehicle.
> You did not prove anything.
I proved that properly designed braking systems make all vehicles equal.
> This makes trucks and heavy vehicles stop in longer distances.
No, it makes better braking systems more expensive. What I proved is that trucks take more distance to stop only because they would otherwise be more expensive.
> Also the brake systems are different, air brakes are generally less responsive than hydrolic brakes of a small vehicle.
Yes -- economics, not physics.
Thanks for writing!
Thank you! I've always thought the idea that "it takes longer for heavier vehicles to stop" idea was wrong. Consumer Reports does stopping distance tests on hundreds of vehicles. I'm going to take a sample of them from compact cars and those over 6000 pounds and see how they vary.
Now I'm curious to what you found? ....
@@fvl548 These are the average they list for each category of car:
Category Average dry braking 60-0 mph, ft.
Sports/sporty cars 120
Luxury midsized cars 128
Luxury small SUVs 129
Small cars 130
Luxury small cars 131
Small SUVs 132
Luxury large cars 132
Large cars 132
Midsized cars 133
Midsized SUVs 134
Luxury midsized SUVs 134
Compact pickups 136
Minivans 138
Compact electric/hybrid cars 138
Luxury large SUVs 138
Full-sized pickups 140
Large SUVs 143
Average of all tested vehicles 132
@@aintquitewright1480 Thanks for posting this! It confirms the idea that mass doesn't increase stopping distance proportionally, as so many believe.
Yet another physicist forgets about the elasticity modulus of the tyre and how it makes the friction equation non-linear.
This one is so easy to demonstrate safely that it honestly astonishes me that people don't test it BEFORE publishing erroneous videos about it.
> Yet another physicist forgets about the elasticity modulus of the tyre and how it makes the friction equation non-linear.
I clearly mentioned this -- several times -- as the practical limit of my example. Nevertheless, most people get this issue entirely wrong, an error my five-car example is meant to address and falsify.
> it honestly astonishes me that people don't test it BEFORE publishing erroneous videos about it.
Watch the video again. Locate where I don't qualify my results as approximate.
Actually, the coefficient of traction of a tire IS quite linear, until the sheer strength of the rubber compound is exceeded. Tires with the appropriate inflation pressure, size and compound for the weight of the vehicle under test is assumed. It is also assumed, that the brake system compensates for weight transfer, so the axle torque is exactly proportionate to the load on each tire. (An ideal antilock braking system) You would be surprised, how good the braking systems on newer modern cars actually are!@@lutusp
Interesting and informative video Paul. Misconception, assumptions, partial knowledge can lead to bad outcomes. Thanks for sharing. This should be useful for almost everyone.
Thank you for your kind comments! I hope to make more informative videos of this kind.
Irony?
I learned to drive double decker buses in trhe 1960s in London. I soon learned the fallacy of big vehicle = longer stopping distance by experience. We were takenn to a closed track at Chiswick where we were allowed to slam on at various speeds, culminating in the famous "skid pan" where we drove over a constant spray of water on a smooth surface with worn rear tyres. The instructor would command a certain course, either straight ot curved then yank the handbrake (rear wheels only) to induce a skid. We then had the hair-raising experience of trying to correct the skid and bring the bus to a safe stop.
A few years later I had the need to perform an emergency stop with a loaded bus. The vehicle stopped promptly within what I would have expected for a car at the same speed. Strong brakes and wide tyres added up to loads of anti-traction (friction). We were taught the "cadence" breaking method of quickly alternating between full braking and less breaking to avoid the wheels locking - skidding tyres offer no traction. Modern ABS brakes work that way.
Thanks for posting this real-world example -- much appreciated!
I think I'm missing something. Do I understand correctly if I say that the calculations show that mass has no effect on stopping distance but "there are many practical and economic issues that explain why a heavy car or truck requires more stopping distance." (6:00) So tradeoffs in engineering, materials, cost, etc. are such that heavier vehicles DO require more stopping distance.
It seems to me that the correct answer depends on the context of the question. If you're asking about simulated vehicles on simulated roads, then the answer is no, heavier vehicles don't require more stopping distance. If you're asking about real vehicles on real roads, then the answer is yes, heavy vehicles require more stopping distance.
Maybe I'm just dense, but if I'm can be confused, other people can be too. I think it's dangerous to be at all ambiguous about this point unless you want people getting behind the wheel of a large heavy vehicle, thinking that they can stop just as quickly as a smaller lighter vehicle and driving accordingly.
If the focus is on safety, then the answer should pertain to how a real driver should treat a real vehicle in the real world. From watching this video, it looks like people should walk away with the knowledge that yes, heavier vehicles do require more stopping distance, which would mean it's not a myth, but just has some nuance.
> If you're asking about real vehicles on real roads, then the answer is yes, heavy vehicles require more stopping distance.
Not if the tires and brakes are designed properly. This is the point of my five-car monster vehicle example -- it weighs five times as much, but it stops in exactly the same distance as one car. Everything after that is about engineering -- which tires, which brakes, and so forth. Not physics.
@@lutusp I see. Thanks for setting me straight. Loved the video.
Hi Paul, your Website and RUclips-Channel inspire me regulary. Thank you very much.
Well, thanks! I'm glad people are still watching and reading. Especially reading, which seems to have fallen out of fashion. :)
well friction does not depend on mass in this case, masses should cancel out if two cars with different masses stop by making tires skid, and guess what they do not, their friction depends on pressure applied on the brake disk/drum, so mass does not cancel out and braking distance increases with mass through physics.
there is a slight increase in friction because heavier mass increases the tire contact area of tire on road but its way too low to decrease braking distance
I just opened online calculator for car stopping distance, I entered the 903 km/h speed that average A320 lands at on a standard 10,000ft runway, it told me that I'd need a 17,000+ft runway.
Of course, it didn't ask for the weight of my vehicle... :)
@@shinobuoshino5066 First off: Look up how planes land! Also, 903km/h is just below the cruisingspeed of an A320. Landing speed is around 200km/h.
Off course it won't ask you how much your vehicle weighs, since most people don't know. Post this calculator as a refference so we can factcheck, you're "factchecking"
> "there is a slight increase in friction because heavier mass increases the tire contact area of tire on road but its way too low to decrease braking distance"
Not really. In practice, mass is nearly eliminated as a factor unless the brakes cannot handle the load or the tires melt. Even in ideal circumstances it's not an exact cancellation, but sufficient that safety agencies don't bother factoring in vehicle mass in calculating stopping distance.
Fog example, get as far off of the road as you can manage as some people move to the shoulder and yet they keep driving instead of stopping.
the math/sim your using is a hugeass oversimplification. according to some reddit comment:
Freshman year physics - F(friction) = u(coefficient) x Fn(normal force)
Senior year physics - Yeah, remember that simple equation? Throw that out the window when you're working with a non-linear elastic, pneumatic wheel applying torque. We're going to need to use this partial differential equation that's going to take up most of the page and you'd need several computers to solve for.
Non-smartass answer: Because the traditional Ff=u x Fn is an oversimplification when it comes to wheels and tires. It works great for something like a wood block where it's just sliding and doesn't deform. But a tire is a complex thing. You're transmitting force through this spinning wheel that's under pressure from air inside, generating tension in steel belts in multiple directions, and rubber, all of which is going to flex and move. In that case contact patch becomes more relevant to how it's going to grip and behave.
You may notice that I qualify every mention of this relationship, to take these points into account without losing my viewers by citing the technical details. But the point my video makes is that more massive vehicles do not require more stopping distance, unless they're improperly designed.
All those sources I quoted got it wrong precisely because they lost the forest for the trees -- they assumed that E = 1/2 m v^2 must mean that more massive vehicles require more stopping distance proportional to mass, which is false.
Heavier vehicles will always require more stopping distance regardless of design (assuming everything aside from mass is identical). In real-world scenarios, there is often some degree of slippage between the tire and the road. This slip introduces factors like tire load sensitivity, where the increase in friction with higher load is less than proportional.
Heavier vehicles do tend to have more load on their tires, but the traction increase is not directly proportional. This is due to the nuances of tire behavior, including factors like tire composition, road conditions, and the presence of anti-lock braking systems (ABS) that can modulate braking to prevent wheel lockup.
@@Venous_fx > Heavier vehicles will always require more stopping distance regardless of design (assuming everything aside from mass is identical).
My video's five-car monster vehicle example falsifies this claim. Same tires, same brakes, same driver actions, everything but the total mass.
But then, that was the purpose of the example -- to force people to see that the commonly heard claim is false. In everyday reality, this simple truth is hidden by bad engineering.
@@lutusp As I stated before, your simulation is not an accurate representation of reality. Heavier vehicles will always require more stopping distance regardless of design (assuming everything aside from mass is identical). In real-world scenarios, there is often some degree of slippage between the tire and the road. This slip introduces factors like tire load sensitivity, where the increase in friction with higher load is less than proportional. There are many complex mechanisms at play in the tire and road interaction. Your simulation does not account for this.
@@Venous_fx > As I stated before, your simulation is not an accurate representation of reality.
I proved this commonly heard claim false in the simplest possible way -- I welded five two-ton vehicles together, creating one ten-ton vehicle, and that much heavier vehicle stops in the same distance as one two-ton vehicle. How is this in any way unclear?
A heavy vehicle with adequate tires and brakes, stops in the same distance as a small, light vehicle with its own appropriate tires and brakes -- assuming both vehicles are traveling at the same speed. Assume also that ABS is in use, which means the tires are at the edge of breaking traction with the roadway, which is the most effective braking friction point.
This is why safety agencies list speeds, but DO NOT list masses, in their officially authorized braking distance charts -- they do this because MASS DOES NOT MATTER assuming tires and brakes are correctly chosen for the vehicle's mass.
Speed matters, and speed is listed as a factor in stopping distance.
Reaction time matters, and is also listed.
MASS DOES NOT MATTTER.
"Vehicle Stopping Distance and Time" from UPenn: nacto.org/docs/usdg/vehicle_stopping_distance_and_time_upenn.pdf
No mention of mass or weight.
Automotive Fleet: KNOW YOUR STOPPING DISTANCE -- www.automotive-fleet.com/driver-care/239402/driver-care-know-your-stopping-distance
No mention of mass or weight.
Dozens more examples, but you know what? You could do your own research and save me some time answering an inquiry I have answered hundreds of time over the years.
Thanks for writing.
A better way to explain the stopping distance being independent of mass is that the total energy needs to be dissipated in braking, so
ΣE= (1/2)m(v)^2 - μmgx = 0, where μ is coefficient of friction between tires and road and x is stopping distance
(1/2)m(v)^2 = μmgx
x=(v^2)/(2μg)
However, in practice this assumes that the loading and the braking force is evenly distributed, where in reality you will probably have one axle lock up before the others in a large truck, which was mentioned in the video as an issue when trucks are empty, since the weight shifts forward and unloads the rear axles.
Yes, but when doing science, it's necessary to control for all extraneous factors in order to properly study just one. Otherwise no useful science is possible.
Alright I gave this thumbnail the benefit of the doubt. I'm not very disappointed, despite the claims being slightly exaggerated
05:13 the way I like to remember this, is that the extra "stopping" energy has already been already "spent" when accelerating that car. ie. you already cancelled the extra work (watts) you need to spend to stop the heavier car when you accelerated to that speed
in reality, a heavier car will be sliightly faster to stop due to material science reasons if it's the only added factor. the weight distribution is a tiny but not insignificant factor too, again due to material science (the way the rubber interacts with the asphalt). That is until you hit a "breaking point", when the car becomes so heavy the breaks can't break enough, and overheat
Oh, also ABS might be a factor in this too
07:30 preach!
08:43 would be a good idea to test fancy new ai, bard, chatgpt etc. see what they give as the answer
> the way I like to remember this, is that the extra "stopping" energy has already been already "spent" when accelerating that car.
Well, energy has two primary forms -- kinetic and potential. When you accelerate, you're turning kinetic energy into a mixture of potential and kinetic energy, which are represented by the car's mass and speed.
When you hit the brakes, you're converting that energy mix into kinetic energy again.
I hope this helps, and thanks for posting.
@@lutusp Yes indeed :) Thanks. But also, make it in beamng please I'm begging you
An empty semi truck with a 48' trailer will take more time to stop when fully loaded even with more friction to the pavement. An 80000lb vehicle DOES take more distance to stop than the same vehicle empty. I know you're only talking about cars, but more weight means more potential energy to slow and stop the vehicle. As a former truck driver, I know this is true. Like most things, they look good on paper/graphics, but in real life things vary greatly. I do agree that driver reaction time plays a HUGE part in stopping of any vehicle. I like the fact that you point out some real physics in this video showing the stopping distance once brakes are applied goes up 4 times for double the speed. Thanks a million for understanding this and conveying it too. Overall, I'd have to say this was a good video, thank you.
trucker here as well and you are correct. an empty truck will stop faster than a fully loaded one. they might have gotten them to stop in 30 feet in lab condition with perfect tyres and pavement but that shit just doesn't happen in reality.
think at trucks as barely road worthy vehicles.
Completely agree. When hauling a trailer behind my pickup with my race car on the trailer it definitely takes much more distance to stop. And the trailer has brakes
> I know you're only talking about cars, but more weight means more potential energy to slow and stop the vehicle.
Yes, and more tire friction. If the tires are brakes are designed to accommodate all operating modes, the mass factor is nearly eliminated (not precisely eliminated, as I point out in the video several times). And the truck design would be very expensive.
My sole point with this video to to stop people from claiming that more massive vehicles must require more stopping distance solely based on their mass. This is false. Real trucks do require more stopping distance proportional to mass, because of how they're designed, not because of physics.
@@trusconi81 > When hauling a trailer behind my pickup with my race car on the trailer it definitely takes much more distance to stop ...
Yes, because of the overall setup and the difference between the loading on the different axles, not because of basic concepts. My video is meant to falsify the widely held wrong assumption that more massive vehicles require more stopping distance as a direct function of their mass, which is false.
In your hauling setup, unless the trailer brakes are extraordinarily well-designed, they will produce either more or less braking force that is required to produce zero net longitudinal force on the hitch. If the latter were true (nearly never), then the truck and the trailer become independent braking systems, not a candidate for a jackknife, the usual case, and which certainly will take more distance to stop, as you experience.
A perfect setup would create zero longitudinal force on the hitch during braking, so the hitch doesn't either pull or push the truck (during braking, not while hauling). That would be the basis for a valid comparison test, and it would make all stopping distances the same to the degree allowed by loading of the tires and brakes.
It would be interesting to see if you would stand in 321 feet in front of a fully loaded Australian 2AB Quad road train doing 100kmph to prove that all vehicles on the road stop in the same distance. Theory is good but in theory people pay attention to the road. Watch any tyre test and the same car will stop in a different distance based on the brand currently on. I agree that from a physics perspective they should be the same. However from a reality, accounting and engineering compromise perspective this doesn't work in practice. The people using physics to prove why this is the case should be called out.
@boomerau • 4 hours ago
> It would be interesting to see if you would stand in 321 feet in front of a fully loaded Australian 2AB Quad road train doing 100kmph to prove that all vehicles on the road stop in the same distance.
Yes, except that's not what I said. I said that trucks don't stop in the same distance as cars, and the reason is economics.
So, your 10t car needs 20 wheels and 20 sets of brakes. Stack those 5 cars on each other, so there's only 4 wheels and 4 brakes, and you will _indeed_ have a much harder time stopping. That's the problem most people face as they load their car/truck. Driving around with nothing in the trunk/bed is much different than driving around with a heavy load. The braking force you're used to will not be enough to stop in the expected distance. (and if your trailer brakes go out, then you have a problem... the mass of 5 cars, without the brakes of 5 cars.)
As mass increases, so does the work necessary to stop -- inertia. The brakes can easily overcome tire traction - aka. a skid. (I'll leave the murderous nature of ABS for a different time.) Ultimately, the tires are what does all the work - keeping you connected to the ground. That is a finite amount of traction (grip), and it does depend on weight. It's a balancing act between inertia and traction. It's as much a "science" issue as it is engineering. Tires and brakes can only get so big, and there will only be so many of them. When you double the mass, does it double the traction? Usually, it doesn't.
If you're curious, look at airplane brakes. That's a huge mass, with tremendous traction, on a surface designed to maximize traction, so they have equally massive multi-pack brakes. They, too, have more brake than traction, and they have an anti-skid system like a car's ABS.
> So, your 10t car needs 20 wheels and 20 sets of brakes. Stack those 5 cars on each other, so there's only 4 wheels and 4 brakes, and you will indeed have a much harder time stopping.
Yes, as explained in the video, but let me put it in simpler terms: for two cars of different masses, engineering changes can produce the same stopping distance. But for two cars traveling at different speeds, no engineering change can equalize their stopping distances.
I hope this helps, and thanks for writing.
I'd like some more explanation in the numbers being used .
Also, did the computer simulations take into consideration:
- tyrecircumferance
- wheelweight
- suspension (and flex)
Are there other models beside the SUV? There is also NO sag, they stay level.
This sounds like a theory I want to learn more about as it goes against what I've been thaught. Even tho, I still think it will be false. That's why I inquere the numbers used and to see how you corelate them. Thanks
No. He also didn't consider the tyre's modulus of elasticity which is the single biggest influence on friction differences between a laden and unladen vehicle.
@@SwervingLemon Indeed! Good one, there are so many parameters that contribute to braking performance that never were mentioned.
After I wrote the comment, I also wonderded about the wheight shift and its effect on grip front and rear. And max braking capacity per tyre.
To me this is using the most basic rule of physics to explain something much more complicated than percieved.
The primary thing you need to understand is that vehicle mass doesn't determine stopping distance. My five-car monster vehicle proves that -- its ten tons of mass stops in exactly the same distance as a two ton car.
If you reduce the number of tires from 20 to 4, this degrades braking effectiveness, but if the tires and brakes are designed properly to accept the expected loads, the change will be small. Not zero, but small.
That's it, in a nutshell.
the engineering follows physics
This video incorrectly compares cars, which have 4 tires rigidly tied together to 18 wheel trucks, which typically have 8 tires rigidly connected together and 10 more wheels rigidly tied together, but those 2 sets are only connected by an imperfect pivot point. If you could eliminate the pivot point and only drive in a straight line, then, in theory, that truck can brake just like a car. Furthermore, since some trucks are empty and some trucks are full, there can be no way to insure that all the tires will brake evenly according to their load. So, in the real world, trucks need a greater stopping distance than cars. Any truck driver who knows their stuff will tell you that. Furthermore, if your tires are bouncing off the road due to an empty truck and large potholes, your stopping distance will increase. This guy also has obviously never driven in a congested area. If you EVER leave more than a car length to the car in front of you, someone WILL cut out in front of you, taking up that space. Unfortunately, tailgating is seldomly enforced where I used to live, which is New Jersey. Again, real world vs theory.
> So, in the real world, trucks need a greater stopping distance than cars.
This is why I presented the five-car monster vehicle example -- it proves that this claim isn't true. And when trucks take more distance to stop (which they certainly do) it's because of design, not because of physics.
> Again, real world vs theory.
The distinction between real world and theory in this case is about economics. A very expensive truck would stop in the same distance as a car, but it would be very expensive. Not because some natural law that requires trucks to stop slowly.
@@lutusp In New Jersey, where I spent most of my life, most truck accidents are caused by car drivers, and their bad behavior around trucks, and this is based partly on their assumption that a truck's stopping distance is the same as a car's. You are being dangerous and irresponsible by perpetuating this myth, and I can only hope that the influence of this video doesn't kill somebody. The fact is that facts turn into falsehoods when they are presented with incomplete information. The only difference between theory and the real world is that theory fails to account for every parameter. Your example of many cars rigidly tied together fails to account for uneven load distribution, widely varying loads, and varying road conditions. Even with today's technology, it's impossible to design a truck that can reliably stop as well as a car. A box truck is only thing I can think of that comes close. The ONLY thing I can agree with you here, is that maybe in the future if someone can design a truck that can stop like a car, it would be far too expensive.
It's misleading to say mass has no effect on braking distance. The peak traction of a tire is dependent on a lot of things like the contact patch area, surface roughness (both in terms of static friction and for the tread to physically deform into), temperature (rubber deforms differently depending on temperature), surface conditions, etc.
A counter example would be wet conditions. As you scale up the contact patch with mass, the water between the tire and the road surface has to displace the water a longer distance. Granted it has more force to do so with, but fluid dynamics do not scale linearly (to double the water particle speed you need four times the force). Neither does the contact patch area scale linearly with mass even if you kept the same tire (there's no simple equation for this as it depends on the tire wall stiffness, tire pressure, compound stiffness, cord stiffness, cord pattern, and so forth). This is one of the many reasons truck tires have a much larger diameter than a car tire while the width is still in the same ballpark as a wide sports car tire.
> It's misleading to say mass has no effect on braking distance ...
There are second-order effects in everyday practice, but from a physics standpoint the statement is correct. And my example section shows many educational sites that try to claim that mass causes a direct, linear increase in stopping distance, which is certainly wrong.
Your mass having now affect on stopping distance is a classic engineer vs mechanic argument. Is the model and math correct? maybe, In the context of the model. But that isn't real life. you have 5X the cars, 5x the tires, 5x the braking the force, 5x the rolling resistance, 5x the heat dissipation, 5x everything. Of course 5x the mass won't change the stopping distance if you increase everything equally 5x. Take 2 identical vehicles. Fill one up with bricks until it's 5x the mass of the other one. You're really going to try to say they will stop in the same distance? I bet your model with 5x the cars will also prove mass has no affect on acceleration aswell. Because those 5 cars will also accelerate exactly the same as one.
> You're really going to try to say they will stop in the same distance?
Yes, and I proved it. Consider that five cars that aren't connected to each other individually stop as though they were one car, in the expected distance for one car. And those five cars, when connected, all traveling at the same speed, stop in the same distance as for one car. How could it be otherwise?
The example is really simple, intuitive, and the result is borne out in field tests.
Thanks for writing!
@@lutusp I agree 5 cars connected will stop in the distance of 1 car. Just like 5 cars connected will accelerate at the same rate as 1 car.
I'm just saying that isn't the same as saying weight has no effect on stopping distance.
A more accurate model for that statement would be taking one car and increasing only the weight. eventually the weight will exceed the stopping power of the brakes and the distance will increase. Then you will need to increase the stopping power of the brakes. this is why large trucks have significantly larger braking systems than small cars. The increased weight will increase grip of the tires on the road as you said, to an extent, but won't increase the grip of the pads on the rotors or the rotors ability to dissipate the extra heat produced.
There is also increased weight transfer to take into account. but that depends on where the extra weight is added, limits of the suspension ect.
I guess if your point is a large truck the weight of 5 cars, with the stopping power of 5 cars, and the tire contact patch of 5 cars can stop in the same distance as 1 of those cars then we agree. because that truck is using (converting to heat) 5 times the energy to stop 5 times the weight. So using 1/5 the energy will stop 1/5 the weight in the same distance with all things equal.
Thanks for responding! I appreciate a good youtube video that can get people thinking and discussing stuff like this.
@@cargasm383 > I'm just saying that isn't the same as saying weight has no effect on stopping distance.
Yes, but ... wait for it ... no one said that. My claim is that a car's mass doesn't normally increase its stopping distance as is commonly believed, and in cases where this is true, it's because of bad engineering, not physics.
Thanks for writing!
@@cargasm383 > I'm just saying that isn't the same as saying weight has no effect on stopping distance.
Yes! But I never said that. I said that two properly engineered cars, same speed, different masses, will stop in the same distance. And I proved this in the video, in a way that's nigh impossible not to understand.
@@lutusp At 3:15 in the video you said "a few minutes from now you will know why mass should have no effect on stopping distance" Thats the part I was commenting on. Mostly because it was followed by the comment about shattering beliefs. came off as a "I'm smarter than you and this is why" type comment. Which I found funny because we seem to agree mass does have an effect, but as you said proper engineering can overcome that effect of increased mass.
Too simplified to simulate 5 connected cars to show the same stopping distance. Film a real world example. Drive an empty vehicle, measure the stop at a 60mph or 100km/h. Load vehicle excessively and retest.
Not the same example. My example makes it self-evident that mass doesn't increase stopping distance. In real tests, mass has a small effect on stopping distance unless the responsible issues are accommodated in the design of the tires and brakes. In other words, it's an engineering issue, not a conceptual one.
He has stated many times throughout the video, he's assuming perfect brake bias, which produces axle torque perfectly proportionate to the load on each tire. (A really good antilock brake system) He is also assuming, that no tire is loaded to its sheer strength, which is a limiting factor to stopping force. Coefficient of traction is actually pretty constant, until you exceed the sheer strength of the rubber compound.
Automotive engineer here. Generally good information, but sorry, your formula at 9:00 is wrong. Kinetic energy is universal and has no friction or gravity term. Your claim of 4x stopping distance at twice the speed is physically true, though, just the equations you show have an error. Maybe the term you showed is something else, but it is not energy. Make the unit test: kg * (m/s)^2 / (1 * m/s^2) = kg m is not Joule.
In reality, at higher speeds, the tires have less time to grip into the road and the car (not a racecar usually) experiences aerodynamic lift, so the stopping distance is way more than 4x at 2x speed. And do not start with aquaplaning (drop in friction above a certain speed), this will make the comparison completely scary.
I think I figured out the equation you wanted to write at 9:00. It is stopping distance d = v^2 / (2 mu g).
> Kinetic energy is universal and has no friction or gravity term.
Yes, that's correct, but the equation you refer to describes the dynamics of braking, therefore it must go beyond kinetic+potential energy. For completeness I included the source -- the HyperPhysics mathematical analysis included later in that section (and see below for more on this topic).
>> Make the unit test: kg * (m/s)^2 / (1 * m/s^2)
This lacks a friction term, which I included in my equation to explain why mass doesn't affect stopping distance in a first-order analysis (second-order effects prevent this from being the whole story).
EDIT: The equation you referenced and correctly identified as wrong, is meant only as an illustration -- it requires at least one more term to function as intended. Here is the correct form:
E = m v^2 / 2 (1+mu g)
This form behaves itself with differing friction amounts, and if no friction, becomes the classic energy equation E = m v^2 / 2.
Again, the equation in the video is only meant to convey an idea and won't behave as shown.
Thanks for your correction!
@@doczooc > It is stopping distance d = v^2 / (2 mu g)
Well, you did correctly identify a problem in my equation -- it fails as anything but an illustration and needs at least one more term to function as intended.
If I wanted to include a mass term, it can be written this way, which unfortunately would have left most viewers confused:
E = m v^2 / 2 (1+ mu g)
This prevents an infinite result in the event of no friction (mu) -- the equation in the video has that rather serious drawback. :)
And in the case of no friction, it becomes the classic energy equation E = m v^2 / 2.
Again, I thank you for your helpful posts.
I like your reason for sharing. It is worthwhile, in terms of human life to save.
Thanks for posting!
Wheels Magazine did an experiment similar to this, they found that brake quality was more important than speed.
> brake quality was more important than speed.
When the tires lose their grip on the pavement, brake quality is no longer an issue. Ultimately, the tire friction limit is the deciding factor.
#1. I've never heard anyone (at least anyone of authority) claim braking distance increased proportionally with speed. It has always been square proportional (2x speed = 4x distance, 3x speed = 9x distance, etc...). That is basic middle/high school physics (Ek = mv^2/2).
#2 is splitting hairs. While it may be theoretically possible to engineer a truck that stops as fast as a small car (there's a video out there of a kid dashing across the road and a well engineered truck stopping on a dime avoiding tragedy), most trucks in the US at least are not engineered to stop as quickly. In practice, a larger (more mass) vehicle will take longer to stop. The difference is glaring if you've ever driven a truck.
#4 should be obvious, though from personal experience having lived in an area with wildfires and hearing about pileups because people insisted on driving through smoke so thick you can't see 10 feet in front of you...maybe its not as obvious to some.
@LC-uh8if • 19 minutes ago
> #1. I've never heard anyone (at least anyone of authority) claim braking distance increased proportionally with speed.
That's why I posted a few sources verbatim for those kinds of claims. They are legion.
> ... The difference is glaring if you've ever driven a truck.
Also covered in the video, where I say that trucks really do take more distance to stop, and the reason is economics.
> ... maybe its not as obvious to some.
We regularly hear pileup stories involving hundreds of vehicles, so clearly someone isn't getting the message.
Thanks for writing!
Stopping distance is doubled for a mass that is doubled when equal amount of force is used. Larger mass must use higher force to stop in same distance. With Road Vehicles, slip must also be considered since even if brakes had infinite force, it means nothing when wheel slides. Drivers will then opt to use less brake force so there's less wheel slip, but doing so will increase stopping distance. So yes mass isn't the only factor in stopping distance for road vehicles, but it isn't irrelevant.
> Stopping distance is doubled for a mass that is doubled when equal amount of force is used ...
Yes, but that's not the claim. The erroneous claim is to say that a heavier vehicle requires more stopping distance. That's wrong -- such a vehicle requires better equipment to deal with more energy. There is no physical principle that mandates a greater stopping distance for heavier cars, when other factors are properly controlled for.
7:50 but at least the formula in Q1 works out right? The brakeforce is given and by no means dependant on the mass.
Sorry, I'm on the road and can't see the specific equation you're referring to. If it includes the friction coefficient μ, then yes, that factor is dependent on mass -- more mass, more friction. The conversion isn't perfect in practice, but it's the reason there is nearly equal stopping distances for light and heavy vehicles, but only if they're engineered to manage the different energies.
Thanks for writing!
I talk about F=m*a leading to the car slowing down at 2m/s²
So five cars with five sets of tires break the same as one vehicle with the same mass and number of cars. Twenty tires have more traction than four tires. The principle may be sound but there is a traction limit. Four tires simply do not have the same tractive effort as twenty tires. Longer stopping distances are required to avoid the limitations of the equipment. Maybe if we put more sets of tires on a big rig we could get it to stop at the same distance as a car, but that's just not practical.
> Twenty tires have more traction than four tires.
Traction depends on a combination of surface area and downward force, so ... no. This doesn't mean a tiny tire surface will bear the load, but friction force increases as area decreases. Smaller area, greater frictional force per unit area.
Self-driving cars will still crash, but hopefully less often than we do as they can understand the physics and engineering issues you teach here, if not as a concept as a practical implementation.
Yes, and as people are discovering, self-driving cars are a much tougher problem that we thought a few years ago. Two companies have recently stopped their self-driving pilot programs after injuries and deaths, and Tesla's "self-driving" claims are being challenged by regulators.
But eventually we will solve this one. Thanks for writing!
While I do agree there is nothing in physics that says heavy vehicles should stop slower (or faster), the real world is slightly more complicated than the simple dry friction model taught at school, F=mu*N. In practice, as the friction force gets so large that it causes stresses in the materials involved approaching the strength of the material it self, the coefficient of friction starts to taper off, as the material starts to give. In the case of rubber, it is also a viscoelastic material, which causes the onset of degradation of grip to set in even sooner.
The result is that vehicles with lots of weight compared to the contact patch of the tires do have worse stopping distances. Such vehicles of course also have better rolling resistance, and are thus more economical to drive. So yes economics are indeed the main driving factor of elongated stopping distances of utility vehicles.
We can take the energy approach and see that during braking _all_ the kinetic energy has to be converted to something else for the vehicle to become stopped. Most of the energy transfers into heat; in the brakes, the tyres and I suppose the road as well. The original equation for KE at 9:04 is the correct one, it just doesn't directly give the distance. The "corrected" one relates to distance, but is _NOT_ the energy (rather closer to a momentum). So the amount of heat generated is directly proportional to the mass, and a 40t vehicle would need to dissipate 20 times as much heat as a 2t one with the same pre-brake speed. For the distance to remain the same, the time must remain the same, which means that in practice the brakes might have to become so hot that they might break any reasonable material and design limitation. Therefore they must be designed to extend the time instead, and consequently will have a longer stopping distance. (Of course it also helps to have larger and more wheels&brakes to distribute the heat dissipation among.)
> The result is that vehicles with lots of weight compared to the contact patch of the tires do have worse stopping distances.
Yes, but that is an engineering problem, not a physics problem. If you control for extraneous factors, the result agrees with the common-sense expectation.
I did say that an expensive make-over would allow the truck to stop in the same distance as a small car. And it's true -- too expensive for everyday reality, but I was addressing a principle, not a budget.
Thanks for writing!
Whilst I agree from a physics point if view on stopping distance being not dependent on mass - You are assuming ideal brakes working efficiently , and this being an engineering issue means in the real world stopping distance can depend to some degree on mass, and how slowly some vehicles are allowed be designed to stop
> ... and this being an engineering issue means in the real world stopping distance can depend to some degree on mass
Yes, and I made that point. Real trucks really do take (much) more distance to stop than a small car, and this results from how they're designed.
Your clinical values of pavement friction are ignoring tire slip on the rim. At such great frictional loads that fewer tires would have you're more likely to increase rim slip. These are fractional increases in stopping distance . . . that is until you blow out a tire. This is the reason the 18 wheeler needs so many wheels, it needs the reduction of frictional loading. Observe a cement truck that has a drop down extra pair of tires in the back. These tires are for traveling when the truck is full and can stay stowed out of the way when the truck is empty. Your point that a mega vehicle stops at the same distance 5 individual vehicles would stop is sound except you'd need specialized tires.
> Your clinical values of pavement friction are ignoring tire slip on the rim.
If the tires slip along the wheel rim, something went wrong in the tire shop and the company is responsible. But this isn't a realistic scenario if the wheels and tires are appropriate to the vehicle.
Thanks for writing!
@@lutusp >appropriate to the vehicle.
So just math then and not real world physics. Have some special tires been invented that would be appropriate on both a 20,000 ton car and a 2 ton car?
Your math checks out, but something is going to give because material science isn't keeping up.
@@djdickey > Have some special tires been invented that would be appropriate on both a 20,000 ton car and a 2 ton car?
No one made that claim. The right tires and brakes can be designed for each example vehicle, not some universal solution for any vehicle.
How drunk or decrepit do you have to be to have a 1.5 second reaction time?
> How drunk or decrepit do you have to be to have a 1.5 second reaction time?
Cell phones. Car radios. Children in the car. ... Children in the car will cell phones. Dogs and cats.
In science, the idea is to control for all extraneous factors in order to be able to study just one.
@@lutusp If you use a cell phone while driving, if you don't have the presence of mind to focus on the road while adjusting the radio, if you did not adequately prepare to accomodate children or pets OR are one of the people who "Can't walk and chew gun at the same time" or "can't talk and walk at the same time", any accident as a result of your failure to hit the brakes in time is 100% your fault and should not even be considered here.
1:10 It is *true* that the braking distance goes up by the square of the starting speed. But your explanation is *also* a myth.
You assumed that the rate at which we can dissipate energy will be constant. Even though in the case of dissipating energy as heat via friction, the rate increases linearly with speed. Which actually cancels out the squaring of the energy. And so it will only take _twice_ as long to get rid of all the energy when our starting speed is _twice_ as fast.
(If the dissipation rate was constant, then the deceleration would increase as speed decreases. Which is not what we observe in reality, nor in your own simulation.)
BUT! That is time, not distance! The total distance the car will travel is the same as the average speed multiplied by time. And when we double the starting speed, (assuming liniar deceleration) we also double the average speed. And that is why the braking distance goes up by the square of the starting speed!
We spend *twice* as long getting rid of that speed. _And_ we go *twice* as fast (on average) while doing it! (2time * 2speed) = 4distance!
If you want a visual example. Draw a graph with the vertical axis as speed, and the horizontal axis as time. (0, 0 preferably where the two axis's intersect.)
Draw a line that starts at some amount of speed and has some downwards slope. Now draw a second line with the same slope, but have it start at double the speed. You now have two triangles, and the area of these triangles will be (speed * time). Also known as distance.
Visually, 4 of the first will fit in the second. And the two edges that form the right angle at (0, 0) have independantly doubled from the first to the second, just from changing the starting point.
By doubling one value (or starting speed), we have actually doubled two seperate values (Average speed and time), which are then multiplied together for our result (braking distance), turning our liniar increase in starting speed, into the exponential increase in braking distance. And that is *actually* why the braking distance goes up by the square of the starting speed!
Your fault was to look at the quadratic increase in energy with speed, and the quadratic increase in braking distance, and assume they were related. Even though they are not.
> Your fault was to look at the quadratic increase in energy with speed ...
The values I showed are based on physics, then verified in field trials, which agree. Thanks for writing!
I don't think it would be completely inaccurate to say that mass increases the braking distance as there is a noticeable difference to my carry truck when I have a 170KG generator in the back. Though I get that bigger cars could probably stop quicker with the hefty brakes they have on them, especially high performance spec cars.
That should not be the case. The car's on board computer would have detected more weight and adjusted the braking system accordingly so there is no noticeable difference when braking and accelerating. If the added load means the car is taking noticeably longer to stop then there is either something wrong with the electronics/braking system or you have dangerously overloaded the vehicle, and in either case you must not be operating it on the road in such unsafe state.
@@bethhughes559 It is a little compact pickup tuck from 1990, it has no electronics other than the lights, gauges, fans, ignition, starter, and the bed tilt pump. I am also not overloading it as the SWL for the bed is 500KG.
@@arianaashworth3091 From 1990? A 33 year old vehicle shouldn't really be on the road anyway, they are unsafe and give off dangerous emissions. Perhaps consider replacing it with an EV.
@@bethhughes559 I am not getting rid of it just because it is older, there's nothing even wrong with it! It passed the MOT with no issues and it has had a catalytic converter retrofitted. A few people I know who own electric cars have been having problems with the charging network, which is actually one of the reasons the generator ends up being transported around on it in the first place.
@arianaashworth3091 • 8 days ago (edited)
> I don't think it would be completely inaccurate to say that mass increases the braking distance as there is a noticeable difference to my carry truck when I have a 170KG generator in the back.
Yes, but my video addresses the basic physics, not specific examples that have extraneous, uncontrolled factors. In science one must eliminate all factors but one -- the factor being studied.
Thanks for writing!
Of course, assuming humans will be humans, whatever you do may be completely irrelevant: you may get into an accident and it could still be someone else's fault. Happened to my mom twice.
You should show some practical demonstrations with toy cars.
I assume you mean actual physical cars, not computer graphic cars. It's not easy to get the desired forces and effects without full-size cars, or computer models of full-size cars, which is what I chose.
Thanks for writing!
@@lutusp Yea, I meant full size real cars, obviously it is very easy to lie and cheat with computers, you could give them any stopping distance you want.
And I don't mean to accuse you of anything, my point is only that even by accident it is easy to mislead with computers.
@@MegaLokopo I should explain something -- the numbers in the computer simulation result from a combination of theory and simple confirmation through field tests using real cars. The field tests came first, then the theoretical explanation. Put more simply, it says, "You have the field measurements with real cars, now here's the explanation."
Also, in science, a theory than cannot be confirmed in nature must be discarded. This one is amply confirmed by experiment.
I hope this helps.
@@lutusp so why not show the ample confirmation made by the physical experiment. In the video you make a claim but you don't prove it or show real world evidence, if you have the evidence you should show it.
@@MegaLokopo That corpus of evidence is large and readily available. My video is meant to summarize that evidence.
Consider -- if someone performed a test that contradicted modern theory -- the theory my video explains -- that person would become famous overnight.
But hey -- knock yourself out:
www.google.com/search?q=car+stopping+distances
Braking 60-0 in the dry is more like 130 ft in sedans. Many can do it in even less depending on ABS and driver skill.
> "Braking 60-0 in the dry is more like 130 ft in sedans."
In field tests, and consistent with theory, and assuming adequate brakes, tires and equal axle loading, all vehicles stop in the same distance for a given speed. This is why safety agencies list vehicle speed and correlate that with stopping distance, but omit vehicle mass.
But if a vehicle doesn't have brakes and tires consistent with its mass, or is loaded improperly, this relationship goes out the window.
> Many can do it in even less depending on ABS and driver skill.
No, not so. Once ABS kicks in, the stopping distance is fixed, and beyond a driver's ability to change it -- except, of course, to make it longer by releasing the brakes.
@@lutusp 130 feet is not unrealistic. Motor trend reports 2013 Audi A5 coupé to have stopping distance 60-0 of 101 feet in the dry. 3500 lb is typical sedan weight. Therefore by your logic it should be more like 100 ft 60-0 right? Not 180?
ABS gives up braking distance to allow steering. If the tire was using literally 100% of its grip to stop, you wouldn't be able to turn. ABS pulses the brake, when the brake is released, turning capability is restored. Therefore ABS is technically not the greatest stopping distance possible. However I will say this is a minor caveat I wouldn't hold against your analysis. ABS is good enough and is often super close to the theoretical limit. Just a quirk of how it works if it's using PWM
@@lutusp just wanted to add, I don't dispute your conclusions at all. Just the actual realistic braking distance. It's fine to present a worst case scenario but you should at least mention that 180 ft is just about a worst case 60-0.
I have watched so many dashcam videos where people act like they've never even heard of a brake. Probably at least half of drivers lack the necessary understanding of car handling enough to be safe at modern highway speeds. So, it's genuinely not that bad to assume worst case distances. But, attentive and skilled drivers will do far better than the numbers you are presenting in this video.
@@HarryTheOwlcat > But, attentive and skilled drivers will do far better than the numbers you are presenting in this video.
Yes, that's true, mostly having to do with reaction time, which varies greatly from driver to driver.
Other factors are road surface conditions, brake design, and differential axle loading, all of which can undermine the effectiveness of a stop.
In many ways my numbers are worst-case, for example the three-second rule and an imagined 1.5 second reaction time. But safety agencies agree with these conservative limits and precautions.
Thanks for writing!
@@HarryTheOwlcat > It's fine to present a worst case scenario but you should at least mention that 180 ft is just about a worst case 60-0.
Yes, true, but in a publicly released video that might change a young driver behavior, it's best to look at published field-test stopping distances and choose the worst case.
Consider the alternative -- a distraught mother decides to hold me to account for her son's avoidable accident. Weirder things have happened. :)
Thanks for posting!
I'm confused where the line is between physics and engineering.
If a car had the weight of the moon, of course its braking distance would increase above that of a regular car. But I guess that is because the load has too little contact (only 4 tyres). Braking force isn't increasing proportionally to its mass. So I guess the solution is engineering more wheels?
In realistic circumstances, increasing vehicle mass doesn't increase braking distance -- that was the point of my five-car monster vehicle example. As you reduce the tire contact area, the relationship begins to break down, but not in a linear way, until the brakes cannot handle the energy or the tires melt under load.
@@lutusp You keep referencing "until...the tires melt under load" - this is not the only limit, mechanical grip of the tire can easily be exceeded without them 'melting'. Braking on ice or loose surfaces shows this. This limit of mechanical grip is certainly (in the real world) a significant factor in the reality of heavier vehicles requiriing more stopping distance (eg large SUV or semi vs a compact). Tires can not be manufactured to have 'infiinite grip' especially on poor surfaces like dirt and ice.
While we like to simplify problems to get at the 'base' understanding in physics, we can't lose sight of the 'real-world'. In reality, heavier vehicles take more distance to stop - that is the fact that exists and people need to know, most do not need to know the 'ideals' underpinning forces and motion. (Yes, I do have a physics degree - currently working on my PhD but a ways to go yet).
All your points are correct and important, bur in a video meant for non-specialists I try to avoid too much technical detail. If the point is to alert people to driving dangers, there's no point in creating a perfectly accurate video that no one watches.
Thanks for posting!
@@lutusp Alert drivers to what? The fact of the matter is, your model of a constant deceleration is flawed enough it could be considered misleading. Big trucks simply do take longer to stop due to material constraints, and limited contact area between disk and brake pad. Drivers should therefore take more caution around them. Saying that we shouldn't model with mass in respect to stopping distance is flawed.
@@BrainiacManiac142 > Saying that we shouldn't model with mass in respect to stopping distance is flawed.
Yes, it would be better to instead provide a graphic, easily understood example that falsifies the presumption. But then, I did that in the video. Everything after that is about engineering, not theory. Falsifying a commonly held wrong belief was the primary aim.
Thanks for writing!
So in theory I could fit the brakes, wheels and tyres from a classic 690kg Mini to my 1525kg Jaguar XJ and it would stop just as well as a classic Mini?
I don't believe that for 1 second and clearly neither did Jaguar.
> So in theory I could fit the brakes, wheels and tyres from a classic 690kg Mini to my 1525kg Jaguar XJ and it would stop just as well as a classic Mini?
Nope. But then, I never made that claim. More massive vehicles require more massive braking systems.
Thanks for writing!
@@lutusp Well you did make the claim the weight makes little to n difference though.
@@matthewjenkins1161 > Well you did make the claim the weight makes little to n difference though.
No, I proved graphically that heavy vehicles can be engineered to stop in the same distance as light vehicles. The five-car monster vehicle example demonstrates this point in a way that's impossible to dismiss.
In everyday reality, if appropriate tires and brakes are provided, the same principle applies.
@@lutusp Of course they can be engineered to stop in the same distance.
They need to be because weight DOES make a difference.
@@matthewjenkins1161 > They need to be because weight DOES make a difference.
Yes, as I explain in the video. That difference can be engineered away, but only if we are willing to do it.
Uh.. Kinetic energy of a car certainly is E=mv^2 and not whatever formula you showed at 9:05
> Uh.. Kinetic energy of a car certainly is E=mv^2 and not whatever formula you showed at 9:05 ...
Yes, that's true, but the discussion at that point is stopping distance, which requires us to take friction into account.
Thanks for writing!
@@lutuspYes, absolutely.
The formula he showed is actually similar to the acceleration of the force of gravity on a falling object. Braking is the reverse, much like the Deceleration of an object, that has been tossed straight up. The formula for kinetic energy applies to the total heat dissipated by the brakes, bringing the car to a stop from the various speeds. Yes, the car doing 80 takes 4 X the time to stop as the car doing 20, but, since the velocity is 4 X faster, it indeed covers 16 times more distance, given the higher velocity, multiplied by the longer time it takes to bring it to a stop.
Great, informative video. Thanks!
I thank you for this welcome feedback! More videos on the way!
Car physics may be well understood... But not by any stupid millenials and zoomers who drive them, apparently.
I often ask people why we wear a seatbelt, and most people never think about the glass cutting you open from the windshield. With simple propagandized safety messages, people never learn WHY we do what we do.