I took a look at your loop. I might have made a mistake here, I'm old. Please correct me (anyone.) The governing physics equations are: 1) Kinetic energy (KE) = ½ mass times velocity squared or KE = ½ mv^2 2) Potential energy (PE) = mass times height times gravitational acceleration on the surface of earth (9.81m/sec/sec) or PE = mgh and 3) centripetal acceleration in a curve being velocity squared divided by radius or a = v^2/r If the speed going into the loop is 8.16 m/s, that equates to a kinetic energy (KE), assuming a mass of 105Kg (231 pounds for rider and cart), of KE = ½ mass times velocity squared = ½ mv^2, we have 3,496 Joules of energy. We'll assume for starters that your cart can handle a upper loop radius of 1.125 meters. We know we want it as small as possible. The centripetal acceleration equation acceleration is velocity squared divided by radius, a = v^2/r. Assuming, at the top of the loop, the centripetal acceleration in the upward direction equals the gravitational acceleration (9.81 m/sec/sec) in the downward direction, 9.81 = v^2/1.125. With some simple algebra, we solve for the velocity at the top of the loop required to offset gravity as being 3.32 m/s. That velocity would represent a KE of 579 Joules of energy. The remaining energy would be potential energy of the same mass raised by the total height of the loop, neglecting aerodynamic and mechanical friction drag. Potential energy (PE) = 3,496 Joules minus 579 Joules = 2,917 Joules. The height required would be determined by PE = mgh. Solving for h, we get h = PE/m/g. Height = 2,917/105/9.81 = 2.83 meters total loop height, which equates to 9.28 feet. We'll assume that your clothoid loop is simply two radii, the lower and upper. The total loop height is the sum of the upper loop radius and the lower loop radius. So the lower loop radius would be 2.83m - 1.125m = 1.705m, or 5.6 feet. Going into the loop, the centripetal acceleration would be a = v^2/r = 8.16^2/1.705 = 39m/sec/sec. This would be 39/9.81 = 3.98 g's. Added to 1g for gravity, and you'll be looking at 4.98 g's, for a short time, 2 meters divided by 8 m/s = 0.25 seconds. I think only one commercial roller coaster (South Africa) does more g's than 5. We can iterate these calculations. None of this accounts for drag, and you probably want a little higher speed at the top of the loop to avoid weightlessness, for safety. For a better summary of loop analysis, I suggest my 13 minute video on the subject “BYRCv09 Looping Roller Coaster Calculations.” More on drag and offset head accelerations to follow...
Nice analysis Paul, Great minds think alike! I ran similar numbers in a comment yesterday, but I totally forgot to add the 1G from gravity D'oh. So my theoretical min goes up to 4.17G. Could you check out mine to see if I missed anything else? Either way, I think a loop is not feasable unfortunately :( maybe a roll will work. Also, how would you calculate resistance? Between friction of 14 axels in the cart, slipping of the wheels, and deformation of the structure, I wouldn't even know where to begin! It seems like testing would be the best way to start?
On my BYRCs, I measured average energy loss simply by taking a known vertical drop over a known track distance. It came out to be about 4% on mine. Will could and should do the same experiment by letting his loaded cart go at the top of the hill, not breaking, and seeing how high it goes up the lift hill before stopping, then measuring the height drop, and by PE = mgh, you know the energy lost to drag, averaged over the whole track. In reality, the energy loss due to aero drag will increase dramatically with higher speed, and the mechanical friction loss will increase in turns, when side wheels engage, and in high g sections when wheels and rails deform.. But I did pretty well with averaging the total drag over a long section. I have a spreadsheet which calculates KE, PE, and velocity for a given table of points defined by height and distance down the track. At each new point, a little energy is subtracted based on the length of track traveled from the last point. I did this on all three tracks I have made, and tweaked the average drag coefficient until the spreadsheet predictions closely matched the test data. Like I said, it comes out to around 4%.
Hmm, Sorry about that. Will replied to it so it definitely exists... Here it is if you're interested: Hello Will! Here's some "Back of the Comment's Section Math": Assuming no Rolling resistance, mass cancels. Kinetic Energy entering = Kinetic energy at apex + Potential energy at apex 1/2 Vi^2 = 1/2 Vf^2 + gh For Radius of the loop: acceleration = Vf^2 / Rho (radius of loop at apex) =~ 5 m/s^2 For Vi = 8.16 m/s , Vf = 5mph = 2.235 m/s 1/2 (8.16)^2 - 1/2 (2.235)^2 = gh = 30.79 h = 30.79/g = 30.79 / 9.81 = 3.14 meters (Max theoretical height for the loop, actual should be lower, also it's almost pi that's awesome) The minimum possible apex radius is a constant circle at R= h/2 = 1.57 m The radius required for .5G at the apex is Rho = Vf^2/a = (2.235)^2 / 5 = .999 m (less than the minimum, that's good. we're not repeating flip flap railway.) THIS VALUE SHOULD BE AS SMALL AS THE CART CAN HANDLE! IT MAY REQUIRE A REDESIGN. The maximum radius you could start with would be such that the majority of the loop is as tight as possible, and only the exit tapers off. For simplicity, let's say that the loop is split 90 and 90 degrees where the top quarter is at 1m radius and the rest is at h-1 = 2.14m radius. Your acceleration at entry would be a = Vi^2/Rho = (8.16)^2/2.14 = 31.1m/s^2 = 3.17Gs ! ( that's about the minimum possible!) The point halfway up would be h = 2.14m, and V at that point is V= sqrt(2(1/2Vi^2 - gh)) = 4.96 m/s So the acceleration at that point is: a = V^2/Rho(1m) = 24.59m/s^2 = 2.5G (jumping from 1.17G at the large radius) It'll be real intense, and with some harder math you could smooth it out so that it starts around 3.5G and transients down to 0.5G and back. VERY IMPORTANT: THIS IGNORES ROLLING RESISTANCE AND IMPULSE LOSSES, you could test weighted carts to find a rough constant for energy loss per meter of track per kg of load (Car and passenger) dE/dS [J/kg*m]. You'll want to reduce the height if at all possible and fortify the hell out of the entry and exit ramps to minimize deformation loss. Thank's for being so specific with your values! It makes this job waaaaay easier. Best of Luck! Don't kill anybody. Sincerely, Your Friendly Neighborhood Dynamics Student.
If anyone doesn't know how much work Will does not just for these videos, his actual job, his family, friends, fans, all the coasters, etc.. you truly are amazing, brother, and I'm very grateful for all you do.. always be you , the world is better for it... and I don't know what you're doing now or what you're going through, we are here for you, Will, and I hope you and your family are doing great.. hope to see more soon. stay safe and take care.
Thank you so much! I'm actually working on a new thing. I'm calling it "ConnectiCOT!" Stay tuned for the first videos which are coming soon. And yes, I'm doing all the design, all the build, all the shooting, all the editing... and all the paying for it 😂 But seriously, I appreciate your kind words.
I honestly think you should not make the verical loop. Since the exit velocity of the loop would be significantly slower It would really take alot of the positive Gs away from the helix and the helix might be overbanked too much which could cause a lot of stress on the guide wheels. Maybe you could add an inline twisted at the top of the helix instead.
@@WillPemble Or you could just make a "twisted" drop. You have a pretty long straight section of track during the first drop where it would be, in my opinion, better to build a roll or maybe a corkscrew instead of a loop.
I've neglected drag to simplify these initial calculations. Aerodynamic drag goes up with higher speeds, by a non-linear function, It's a cube function (drag = K times v^3). Here's another potential mechanical drag problem. If you're really going to go through a 5 g lower loop, with 231 pounds of rider and cart, then you'll be looking at a 1155 pound equivalent static load. If you have 8 wheels, that's 144 pounds per wheel, if the weight is distributed equally, which it probably isn't. I don't know how hard your wheels are, but as the wheels (and steel tube track) distort under load, the friction between them goes up. So the drag in the first and last quarter of the loop may be high. I know this from two experiences: When I tested my strongest cart on my ~3g track, with 356 pounds of weight, it didn't make it over the overpass, due to extra friction between the deformed wheels and track rails. View the results on my video entitled “I Try to Destroy my Backyard Roller Coaster.” Also, I was helping some MIT students with a BYRC project last year, and they made a PVC flat track and 4-neoprene-wheel cart, and ran it back and forth more than 1000 times. The wheel deformation was extensive, and the wheels finally heated up and slipped off the hubs. Sidenote: The PVC track didn't break. They didn't build a PVC track nor steel cart. They did all wood, because they had done that before. Nothing stops innovation like perceived success. Before you build a loop, you should maybe perform the following experiments: Put your cart on a perfectly flat section of track, and load it with sandbags, gradually, up to 1000 pounds. With a fish scale, see how much force it takes to pull the cart with no weight, 200, 400, 600, 800, and 1000 pounds on it. There are two “coefficients of friction.” One static (how much load does it take to start it rolling?) and dynamic (how much force does it take to keep it rolling?) You will want to measure dynamic load on your fish scale, or how much force it takes to keep it rolling. IF you really want to see if your cart and track can take 5 g's with a 200 pound rider, and a 1.5 factor of safety, load it with 1600 pounds of sandbags on the flat track section, roll it back and forth a while, and see if any parts of the track or cart bends permanently or cracks. If any part or joint bent or cracked, it's not good enough. This doesn't address fatigue, but we'll assume you're not running more than a few hundred times, and that a fatigue crack would be noticeable. I think a loop on your track is feasible, but it might be a little violent, and make sure your cart is safe with the above testing. I'd have to do more analysis to see if the cart can make it round the remaining track.
Yes! Perfect this is exactly the kind of testing case I was thinking of! Will: You might want to make sure you want the high force maneuver before you do the 1600 lb test, cause if it fails you're out a cart :/
if you brought the drop over as close as you can to the helix then you could do a corkscrew ,(to the left as you look at it from the drop). then you change the entrance of the helix so from the corkscrew it enters more out side then it did before then it will work.i did the math ,it will work.
Good luck! Just remember to add rigidity to the track spine for the sudden increase in normal loading. Since loops have a small lateral component, a batter column (angled support) can account for varying three dimensional loads.
I doubt a vertical loop will work, it would have to be a small one if so or extend the lift hill to be able to make a decent sized loop which would feel comfortable. The loop would have to be more of a corkscrew if anything
I agree with the comments to simulate your loop proposal with FVD++ (free) or NoLimits2. I'm not sure if NL2 can simulate a one cart train with small track gauge, or if that would matter. I tried simulating two of my backyard coasters with NL2 (non-pro version), but I couldn't figure out how to make the train smaller, so it didn't seem right having this big train going around my small track. I also couldn't figure out how to change the drag. I can try to put your loop into a simplified BYRC loop spreadsheet I made up a while ago.... It might take me a few hours to get it right, so maybe later today sometime.
Without heavily modifying the simulation parameters, I doubt NL2 would be very helpful for simulating resistance over the whole run, since full coaster are much more rigid than this one. However I see it being beneficial for profiling as a guide for manufacturing, which could be scaled from a more reasonable NL2 scale.
yes thank you that's what I was referring to. The aerodynamics will be similar but the deformation in the track, supports, and ground will we relatively significant on the goal boss. Maybe a Wooden roller coaster will have slightly higher impulse loss and would be a closer approximation?
I think the cart design will need to change also, it may be to long. But what do I know. Love the idea tho and looking forward to see what you come up with.
I personally think a dive loop despite being more complicated will be the best option going forward. 1: u wont need to bend the drop. 2: the helix entry can stay the same. 3 You can use the terrain to your advantage and use smaller supports
Another concern with a tight loop will be acceleration/deceleration of the riders head. The riders head is offset from the track by some distance, so when the track goes from straight to curved, as in going into the loop, the riders head will experience a very abrupt change in velocity, and the reverse coming out of the loop. You can see in the blue flash videos how the riders head thrusts forward bit going into the loop, and is thrust back coming out of the loop. I address this in a short video “BYRCv 06 Offset Accelerations and Jerk.” If you knew what was coming, you could brace yourself for it. It seems manageable on the blue flash. The blue flash has something to hold onto in front of the seat, and that helps.
Good point, with radii as low as 2 meters, and the head 1 meter up from the track, then it has to decelerate to half its speed fairly quickly. The same issue is present but differently in a zero g roll, where you might want to err the hearline higher as to not introduce angular jerk on the head, whipping it around. ...How fast would this twist be happening?
Hello Will! Here's some "Back of the Comment's Section Math": Assuming no Rolling resistance, mass cancels. Kinetic Energy entering = Kinetic energy at apex + Potential energy at apex 1/2 Vi^2 = 1/2 Vf^2 + gh For Radius of the loop: acceleration = Vf^2 / Rho (radius of loop at apex) =~ 5 m/s^2 For Vi = 8.16 m/s , Vf = 5mph = 2.235 m/s 1/2 (8.16)^2 - 1/2 (2.235)^2 = gh = 30.79 h = 30.79/g = 30.79 / 9.81 = 3.14 meters (Max theoretical height for the loop, actual should be lower, also it's almost pi that's awesome) The minimum possible apex radius is a constant circle at R= h/2 = 1.57 m The radius required for .5G at the apex is Rho = Vf^2/a = (2.235)^2 / 5 = .999 m (less than the minimum, that's good. we're not repeating flip flap railway.) THIS VALUE SHOULD BE AS SMALL AS THE CART CAN HANDLE! IT MAY REQUIRE A REDESIGN. The maximum radius you could start with would be such that the majority of the loop is as tight as possible, and only the exit tapers off. For simplicity, let's say that the loop is split 90 and 90 degrees where the top quarter is at 1m radius and the rest is at h-1 = 2.14m radius. Your acceleration at entry would be a = Vi^2/Rho = (8.16)^2/2.14 = 31.1m/s^2 = 3.17Gs ! ( that's about the minimum possible!) The point halfway up would be h = 2.14m, and V at that point is V= sqrt(2(1/2Vi^2 - gh)) = 4.96 m/s So the acceleration at that point is: a = V^2/Rho(1m) = 24.59m/s^2 = 2.5G (jumping from 1.17G at the large radius) It'll be real intense, and with some harder math you could smooth it out so that it starts around 3.5G and transients down to 0.5G and back. VERY IMPORTANT: THIS IGNORES ROLLING RESISTANCE AND IMPULSE LOSSES, you could test weighted carts to find a rough constant for energy loss per meter of track per kg of load (Car and passenger) dE/dS [J/kg*m]. You'll want to reduce the height if at all possible and fortify the hell out of the entry and exit ramps to minimize deformation loss. Thank's for being so specific with your values! It makes this job waaaaay easier. Best of Luck! Don't kill anybody. Sincerely, Your Friendly Neighborhood Dynamics Student.
Hey Charlie, Thanks for the awesome analysis! In my monkey calcs. I gotta say, >3G makes me a little nervous. Cages Coaster Reviews commented that maybe twisted inline inversion might be better. What if we make a counter-clockwise heart line roll that transitions into the helix turn? We'd carry much higher speed into the helix, which would keep that part of the ride awesome, and we'd still get our inversion. Thoughts?
Honestly I love that Idea a lot more. I didn't want to reject the loop idea without running the numbers but loops are really hard. Cage is totally right that a roll will be much less lossy and if done well will be way more fun. I can foresee two key points to it: It should follow a free-fall path, so you might have to reprofile a lot of the first drop by dropping, pulling up, then twisting while dropping and pulling up again into the helix. Luckily it looks like you have enough rail as long as nothing goes wrong in manufacturing (HAHAHAH) The other tricky bit is the actual bending of the track as what you are trying to create is a circular-ish bend that is twisted on it's own axis, my immediate thought for pulling it off would be to run a little bit (6-12") through the bender, then loosen it and twist the stock slightly and bend the next bit. Given multiple passes, offsetting where the twists are, you might get a pretty smooth result. There is probably a better way to do it. I'm not sure what you've used, if anything, to CAD it out, but I recommend having someone model it on NL2 Pro, then use the lightbox to STL trick to make a spline in any CAD software. It'll really help as the maneuver is much more delicate on the rider than most of the coaster. This is an awesome improvement to a great coaster! Best of luck to the team there! I'll be on the lookout for numbers to crunch in future videos.
Also I made a mistake! As Paul Gregg did correctly in his analysis, I forgot to add back 1G of gravity into my calculations, that makes the initial acceleration 4.17G and the top now -0.5G! Sorry about that, it would need to be shorter and/or tighter to get the right apex acceleration, and that would drive up the max a even more (why Paul got ~5G) Now I really think a roll is a better idea.
You can always make a barrel-roll as it requires almost no speed to go through. If you’re really creative I bet you can make a snake dive after the barrel-role if you really tried to too!
To me, it seems that constructing the loop is going to be a lot easier than a zero-g roll. Bending and rotating the track to ride and feel comfortable for a zero-g roll would probably be really difficult. Although, I do think an RMC like barrel roll drop coming down that hill would be really cool.
If it's OK, I will address your questions in several comments, so they each are not so long, and it will be more organized as far as replies. First off, nice job on measuring your speeds. You might show your readers more details on how you calculated the speed from camera frame rate, number of frames, and a known distance on the track, if that's how you did it.
Hey Paul! Comments? Yes, please! We shot the cart at 240 frames per second. It takes 15 frames (0.0625 seconds) for the cart to travel .51 meters. A little arithmetic like this... 0.51/0.0625=8.16m/s. Slomo is fantastic!
I think you should make the slope steeper when you offset the track for the loop so it has more speed. Then you should be clear for the car to travel the rest of the track to the end!
When you change the seats it will alter the weight and weight distribution. You need to account for that change. So rebuild your seat FIRST and then do your math equations.
The lowest part of the track is the middle of the helix, put the loop there and then a small hill up onto the ‘bridge’ section. By doing away with the helix altogether you will have more speed into the loop, it will be easier to place as the track will be on a straight and you won’t have to worry about it making it through the helix.
Just remember not to make it round, but like a Spiral so the radius gets smaller, the Higher the cart proceeds and going back down, radius getting bigger. I know youre familliar with this, but just wanted to mention it 🤪
@@BikerFail16 A normal magnet firstly pulls you to it, but slows you down afterwards. So that would be pointless. An LIM or LSM motor on the other hand would work, but both of these require precision, lots of electricity, a cooling system and a computer who controls it. Not something you can achieve on a backyard coaster in other words.
Blunder Loop. I still think an oblique loop is the way to go. Goal Boss is already banking at drop so seems like less work and less g force. Hey, great work and keep moving forward. Blunder On.
I don't know the equations, but the set up now seems to go fast at the top of the helix, so I think there's a good amount of energy that could go to making an epic loop...
I think you should use some sort of Force Vector Design (FVD) to design the loop. I’m sure you could find someone who’s pretty good with it to help you out.
Will your wheel assemblies need modifying on the count that the up stops are now effectively also going to be the main road wheels whilst the cart is upside down?
Great question, Dan. The wheel assemblies are crazy strong. No worries there. Also, the cart as built now can roll freely around a 1 meter radius loop without getting jammed up. The big question is, would a rider want to be on the cart?
Will Pemble I think you should definitely consider a 1-meter-radius "headline" loop. It sounds insane. ;) On the other hand, I do worry about the G's in this project overall. Love the discussion that's going on here!
if you are going to be remaking the cart, does that also mean you are going to enhance the harness/restraint system? Do you plan to add some sort of lap bar? or keep it as just a seat belt? Because with it only being a seat belt, wouldn't that limit riders to a height restriction?
For a loop, we'll have to change out the seat with a high-back seat. Not sure whether we'll use 2 point or 4 point restraints yet. Blue Flash (best BYRC ever) has a handle for passengers to hold onto. Might give that some thought.
On second thought, I'm not completely certain you need a high back seat. Look closely at the blue flash riders, and see if you think you could handle that. Try scaling the blue flash from the videos. I think you could make about the same loop, but smoother and more efficient.
I took a look at your loop.
I might have made a mistake here, I'm old. Please correct me (anyone.)
The governing physics equations are:
1) Kinetic energy (KE) = ½ mass times velocity squared or KE = ½ mv^2
2) Potential energy (PE) = mass times height times gravitational acceleration on the surface of earth (9.81m/sec/sec) or PE = mgh
and 3) centripetal acceleration in a curve being velocity squared divided by radius or a = v^2/r
If the speed going into the loop is 8.16 m/s, that equates to a kinetic energy (KE), assuming a mass of 105Kg (231 pounds for rider and cart), of KE = ½ mass times velocity squared = ½ mv^2, we have 3,496 Joules of energy.
We'll assume for starters that your cart can handle a upper loop radius of 1.125 meters. We know we want it as small as possible.
The centripetal acceleration equation acceleration is velocity squared divided by radius, a = v^2/r. Assuming, at the top of the loop, the centripetal acceleration in the upward direction equals the gravitational acceleration (9.81 m/sec/sec) in the downward direction, 9.81 = v^2/1.125. With some simple algebra, we solve for the velocity at the top of the loop required to offset gravity as being 3.32 m/s. That velocity would represent a KE of 579 Joules of energy.
The remaining energy would be potential energy of the same mass raised by the total height of the loop, neglecting aerodynamic and mechanical friction drag.
Potential energy (PE) = 3,496 Joules minus 579 Joules = 2,917 Joules.
The height required would be determined by PE = mgh. Solving for h, we get h = PE/m/g.
Height = 2,917/105/9.81 = 2.83 meters total loop height, which equates to 9.28 feet.
We'll assume that your clothoid loop is simply two radii, the lower and upper.
The total loop height is the sum of the upper loop radius and the lower loop radius. So the lower loop radius would be 2.83m - 1.125m = 1.705m, or 5.6 feet.
Going into the loop, the centripetal acceleration would be a = v^2/r = 8.16^2/1.705 = 39m/sec/sec. This would be 39/9.81 = 3.98 g's. Added to 1g for gravity, and you'll be looking at 4.98 g's, for a short time, 2 meters divided by 8 m/s = 0.25 seconds. I think only one commercial roller coaster (South Africa) does more g's than 5.
We can iterate these calculations.
None of this accounts for drag, and you probably want a little higher speed at the top of the loop to avoid weightlessness, for safety.
For a better summary of loop analysis, I suggest my 13 minute video on the subject “BYRCv09 Looping Roller Coaster Calculations.”
More on drag and offset head accelerations to follow...
Nice analysis Paul, Great minds think alike! I ran similar numbers in a comment yesterday, but I totally forgot to add the 1G from gravity D'oh. So my theoretical min goes up to 4.17G.
Could you check out mine to see if I missed anything else?
Either way, I think a loop is not feasable unfortunately :( maybe a roll will work.
Also, how would you calculate resistance? Between friction of 14 axels in the cart, slipping of the wheels, and deformation of the structure, I wouldn't even know where to begin! It seems like testing would be the best way to start?
On my BYRCs, I measured average energy loss simply by taking a known vertical drop over a known track distance. It came out to be about 4% on mine. Will could and should do the same experiment by letting his loaded cart go at the top of the hill, not breaking, and seeing how high it goes up the lift hill before stopping, then measuring the height drop, and by PE = mgh, you know the energy lost to drag, averaged over the whole track. In reality, the energy loss due to aero drag will increase dramatically with higher speed, and the mechanical friction loss will increase in turns, when side wheels engage, and in high g sections when wheels and rails deform.. But I did pretty well with averaging the total drag over a long section. I have a spreadsheet which calculates KE, PE, and velocity for a given table of points defined by height and distance down the track. At each new point, a little energy is subtracted based on the length of track traveled from the last point. I did this on all three tracks I have made, and tweaked the average drag coefficient until the spreadsheet predictions closely matched the test data. Like I said, it comes out to around 4%.
I don't see your comment on running numbers. If your comment contained a link, it won't show up on RUclips for anybody but you.
Hmm, Sorry about that. Will replied to it so it definitely exists...
Here it is if you're interested:
Hello Will! Here's some "Back of the Comment's Section Math":
Assuming no Rolling resistance, mass cancels.
Kinetic Energy entering = Kinetic energy at apex + Potential energy at apex
1/2 Vi^2 = 1/2 Vf^2 + gh
For Radius of the loop:
acceleration = Vf^2 / Rho (radius of loop at apex) =~ 5 m/s^2
For Vi = 8.16 m/s , Vf = 5mph = 2.235 m/s
1/2 (8.16)^2 - 1/2 (2.235)^2 = gh = 30.79
h = 30.79/g = 30.79 / 9.81 = 3.14 meters (Max theoretical height for the loop, actual should be lower, also it's almost pi that's awesome)
The minimum possible apex radius is a constant circle at R= h/2 = 1.57 m
The radius required for .5G at the apex is Rho = Vf^2/a = (2.235)^2 / 5 = .999 m (less than the minimum, that's good. we're not repeating flip flap railway.) THIS VALUE SHOULD BE AS SMALL AS THE CART CAN HANDLE! IT MAY REQUIRE A REDESIGN.
The maximum radius you could start with would be such that the majority of the loop is as tight as possible, and only the exit tapers off.
For simplicity, let's say that the loop is split 90 and 90 degrees where the top quarter is at 1m radius and the rest is at h-1 = 2.14m radius.
Your acceleration at entry would be a = Vi^2/Rho = (8.16)^2/2.14 = 31.1m/s^2 = 3.17Gs ! ( that's about the minimum possible!)
The point halfway up would be h = 2.14m, and V at that point is V= sqrt(2(1/2Vi^2 - gh)) = 4.96 m/s
So the acceleration at that point is: a = V^2/Rho(1m) = 24.59m/s^2 = 2.5G (jumping from 1.17G at the large radius)
It'll be real intense, and with some harder math you could smooth it out so that it starts around 3.5G and transients down to 0.5G and back.
VERY IMPORTANT: THIS IGNORES ROLLING RESISTANCE AND IMPULSE LOSSES, you could test weighted carts to find a rough constant for energy loss per meter of track per kg of load (Car and passenger) dE/dS [J/kg*m]. You'll want to reduce the height if at all possible and fortify the hell out of the entry and exit ramps to minimize deformation loss.
Thank's for being so specific with your values! It makes this job waaaaay easier.
Best of Luck! Don't kill anybody.
Sincerely,
Your Friendly Neighborhood Dynamics Student.
Paul Gregg I just did me exam on energy’s and wasted energy’s
If anyone doesn't know how much work Will does not just for these videos, his actual job, his family, friends, fans, all the coasters, etc.. you truly are amazing, brother, and I'm very grateful for all you do.. always be you , the world is better for it... and I don't know what you're doing now or what you're going through, we are here for you, Will, and I hope you and your family are doing great.. hope to see more soon. stay safe and take care.
Thank you so much! I'm actually working on a new thing. I'm calling it "ConnectiCOT!" Stay tuned for the first videos which are coming soon. And yes, I'm doing all the design, all the build, all the shooting, all the editing... and all the paying for it 😂
But seriously, I appreciate your kind words.
I honestly think you should not make the verical loop. Since the exit velocity of the loop would be significantly slower It would really take alot of the positive Gs away from the helix and the helix might be overbanked too much which could cause a lot of stress on the guide wheels. Maybe you could add an inline twisted at the top of the helix instead.
I think you're onto something, Cage! I'm gonna think about that. A LOT!
@@WillPemble sounds good, im really looking forward to your next video
@@WillPemble Or you could just make a "twisted" drop. You have a pretty long straight section of track during the first drop where it would be, in my opinion, better to build a roll or maybe a corkscrew instead of a loop.
I've neglected drag to simplify these initial calculations.
Aerodynamic drag goes up with higher speeds, by a non-linear function, It's a cube function (drag = K times v^3).
Here's another potential mechanical drag problem. If you're really going to go through a 5 g lower loop, with 231 pounds of rider and cart, then you'll be looking at a 1155 pound equivalent static load. If you have 8 wheels, that's 144 pounds per wheel, if the weight is distributed equally, which it probably isn't. I don't know how hard your wheels are, but as the wheels (and steel tube track) distort under load, the friction between them goes up. So the drag in the first and last quarter of the loop may be high.
I know this from two experiences: When I tested my strongest cart on my ~3g track, with 356 pounds of weight, it didn't make it over the overpass, due to extra friction between the deformed wheels and track rails. View the results on my video entitled “I Try to Destroy my Backyard Roller Coaster.”
Also, I was helping some MIT students with a BYRC project last year, and they made a PVC flat track and 4-neoprene-wheel cart, and ran it back and forth more than 1000 times. The wheel deformation was extensive, and the wheels finally heated up and slipped off the hubs. Sidenote: The PVC track didn't break. They didn't build a PVC track nor steel cart. They did all wood, because they had done that before. Nothing stops innovation like perceived success.
Before you build a loop, you should maybe perform the following experiments: Put your cart on a perfectly flat section of track, and load it with sandbags, gradually, up to 1000 pounds. With a fish scale, see how much force it takes to pull the cart with no weight, 200, 400, 600, 800, and 1000 pounds on it. There are two “coefficients of friction.” One static (how much load does it take to start it rolling?) and dynamic (how much force does it take to keep it rolling?) You will want to measure dynamic load on your fish scale, or how much force it takes to keep it rolling.
IF you really want to see if your cart and track can take 5 g's with a 200 pound rider, and a 1.5 factor of safety, load it with 1600 pounds of sandbags on the flat track section, roll it back and forth a while, and see if any parts of the track or cart bends permanently or cracks. If any part or joint bent or cracked, it's not good enough. This doesn't address fatigue, but we'll assume you're not running more than a few hundred times, and that a fatigue crack would be noticeable.
I think a loop on your track is feasible, but it might be a little violent, and make sure your cart is safe with the above testing. I'd have to do more analysis to see if the cart can make it round the remaining track.
Yes! Perfect this is exactly the kind of testing case I was thinking of!
Will:
You might want to make sure you want the high force maneuver before you do the 1600 lb test, cause if it fails you're out a cart :/
if you brought the drop over as close as you can to the helix then you could do a corkscrew ,(to the left as you look at it from the drop). then you change the entrance of the helix so from the corkscrew it enters more out side then it did before then it will work.i did the math ,it will work.
Good luck! Just remember to add rigidity to the track spine for the sudden increase in normal loading. Since loops have a small lateral component, a batter column (angled support) can account for varying three dimensional loads.
I doubt a vertical loop will work, it would have to be a small one if so or extend the lift hill to be able to make a decent sized loop which would feel comfortable. The loop would have to be more of a corkscrew if anything
Zachery Rankin anything is possible
I agree with the comments to simulate your loop proposal with FVD++ (free) or NoLimits2. I'm not sure if NL2 can simulate a one cart train with small track gauge, or if that would matter. I tried simulating two of my backyard coasters with NL2 (non-pro version), but I couldn't figure out how to make the train smaller, so it didn't seem right having this big train going around my small track. I also couldn't figure out how to change the drag.
I can try to put your loop into a simplified BYRC loop spreadsheet I made up a while ago.... It might take me a few hours to get it right, so maybe later today sometime.
Without heavily modifying the simulation parameters, I doubt NL2 would be very helpful for simulating resistance over the whole run, since full coaster are much more rigid than this one.
However I see it being beneficial for profiling as a guide for manufacturing, which could be scaled from a more reasonable NL2 scale.
But NL2 must make some assumptions for aero and mechanical drag, based on experience. I think NL2 assumes perfectly rigid track.
yes thank you that's what I was referring to. The aerodynamics will be similar but the deformation in the track, supports, and ground will we relatively significant on the goal boss. Maybe a Wooden roller coaster will have slightly higher impulse loss and would be a closer approximation?
I think the cart design will need to change also, it may be to long. But what do I know. Love the idea tho and looking forward to see what you come up with.
I personally think a dive loop despite being more complicated will be the best option going forward.
1: u wont need to bend the drop.
2: the helix entry can stay the same.
3 You can use the terrain to your advantage and use smaller supports
Nice!!
Another concern with a tight loop will be acceleration/deceleration of the riders head. The riders head is offset from the track by some distance, so when the track goes from straight to curved, as in going into the loop, the riders head will experience a very abrupt change in velocity, and the reverse coming out of the loop. You can see in the blue flash videos how the riders head thrusts forward bit going into the loop, and is thrust back coming out of the loop. I address this in a short video “BYRCv 06 Offset Accelerations and Jerk.” If you knew what was coming, you could brace yourself for it. It seems manageable on the blue flash. The blue flash has something to hold onto in front of the seat, and that helps.
Good point, with radii as low as 2 meters, and the head 1 meter up from the track, then it has to decelerate to half its speed fairly quickly.
The same issue is present but differently in a zero g roll, where you might want to err the hearline higher as to not introduce angular jerk on the head, whipping it around.
...How fast would this twist be happening?
Hello Will! Here's some "Back of the Comment's Section Math":
Assuming no Rolling resistance, mass cancels.
Kinetic Energy entering = Kinetic energy at apex + Potential energy at apex
1/2 Vi^2 = 1/2 Vf^2 + gh
For Radius of the loop:
acceleration = Vf^2 / Rho (radius of loop at apex) =~ 5 m/s^2
For Vi = 8.16 m/s , Vf = 5mph = 2.235 m/s
1/2 (8.16)^2 - 1/2 (2.235)^2 = gh = 30.79
h = 30.79/g = 30.79 / 9.81 = 3.14 meters (Max theoretical height for the loop, actual should be lower, also it's almost pi that's awesome)
The minimum possible apex radius is a constant circle at R= h/2 = 1.57 m
The radius required for .5G at the apex is Rho = Vf^2/a = (2.235)^2 / 5 = .999 m (less than the minimum, that's good. we're not repeating flip flap railway.) THIS VALUE SHOULD BE AS SMALL AS THE CART CAN HANDLE! IT MAY REQUIRE A REDESIGN.
The maximum radius you could start with would be such that the majority of the loop is as tight as possible, and only the exit tapers off.
For simplicity, let's say that the loop is split 90 and 90 degrees where the top quarter is at 1m radius and the rest is at h-1 = 2.14m radius.
Your acceleration at entry would be a = Vi^2/Rho = (8.16)^2/2.14 = 31.1m/s^2 = 3.17Gs ! ( that's about the minimum possible!)
The point halfway up would be h = 2.14m, and V at that point is V= sqrt(2(1/2Vi^2 - gh)) = 4.96 m/s
So the acceleration at that point is: a = V^2/Rho(1m) = 24.59m/s^2 = 2.5G (jumping from 1.17G at the large radius)
It'll be real intense, and with some harder math you could smooth it out so that it starts around 3.5G and transients down to 0.5G and back.
VERY IMPORTANT: THIS IGNORES ROLLING RESISTANCE AND IMPULSE LOSSES, you could test weighted carts to find a rough constant for energy loss per meter of track per kg of load (Car and passenger) dE/dS [J/kg*m]. You'll want to reduce the height if at all possible and fortify the hell out of the entry and exit ramps to minimize deformation loss.
Thank's for being so specific with your values! It makes this job waaaaay easier.
Best of Luck! Don't kill anybody.
Sincerely,
Your Friendly Neighborhood Dynamics Student.
Hey Charlie,
Thanks for the awesome analysis! In my monkey calcs. I gotta say, >3G makes me a little nervous. Cages Coaster Reviews commented that maybe twisted inline inversion might be better. What if we make a counter-clockwise heart line roll that transitions into the helix turn? We'd carry much higher speed into the helix, which would keep that part of the ride awesome, and we'd still get our inversion.
Thoughts?
Honestly I love that Idea a lot more. I didn't want to reject the loop idea without running the numbers but loops are really hard. Cage is totally right that a roll will be much less lossy and if done well will be way more fun.
I can foresee two key points to it: It should follow a free-fall path, so you might have to reprofile a lot of the first drop by dropping, pulling up, then twisting while dropping and pulling up again into the helix. Luckily it looks like you have enough rail as long as nothing goes wrong in manufacturing (HAHAHAH)
The other tricky bit is the actual bending of the track as what you are trying to create is a circular-ish bend that is twisted on it's own axis, my immediate thought for pulling it off would be to run a little bit (6-12") through the bender, then loosen it and twist the stock slightly and bend the next bit. Given multiple passes, offsetting where the twists are, you might get a pretty smooth result. There is probably a better way to do it.
I'm not sure what you've used, if anything, to CAD it out, but I recommend having someone model it on NL2 Pro, then use the lightbox to STL trick to make a spline in any CAD software. It'll really help as the maneuver is much more delicate on the rider than most of the coaster.
This is an awesome improvement to a great coaster! Best of luck to the team there! I'll be on the lookout for numbers to crunch in future videos.
Also I made a mistake! As Paul Gregg did correctly in his analysis, I forgot to add back 1G of gravity into my calculations, that makes the initial acceleration 4.17G and the top now -0.5G!
Sorry about that, it would need to be shorter and/or tighter to get the right apex acceleration, and that would drive up the max a even more (why Paul got ~5G)
Now I really think a roll is a better idea.
Here we go again! Looking forward to this 🤗
You can always make a barrel-roll as it requires almost no speed to go through. If you’re really creative I bet you can make a snake dive after the barrel-role if you really tried to too!
To me, it seems that constructing the loop is going to be a lot easier than a zero-g roll. Bending and rotating the track to ride and feel comfortable for a zero-g roll would probably be really difficult. Although, I do think an RMC like barrel roll drop coming down that hill would be really cool.
If it's OK, I will address your questions in several comments, so they each are not so long, and it will be more organized as far as replies.
First off, nice job on measuring your speeds. You might show your readers more details on how you calculated the speed from camera frame rate, number of frames, and a known distance on the track, if that's how you did it.
Hey Paul! Comments? Yes, please!
We shot the cart at 240 frames per second. It takes 15 frames (0.0625 seconds) for the cart to travel .51 meters. A little arithmetic like this... 0.51/0.0625=8.16m/s. Slomo is fantastic!
You're a madman. Amazing nonetheless.
True but awesome
I think you should make the slope steeper when you offset the track for the loop so it has more speed. Then you should be clear for the car to travel the rest of the track to the end!
I think you should consider a Zero-G roll instead. Would be easier incorporate with the speed you have to work with, and they are super fun
I'll go looking for part 2, but I think a loop may have to be a teardrop (shallower in, tightening at the top and out) to handle relatively low speed
And You Shold But The New Cart Handles
Love your vids
Did you say you made an analysis for energy loss? I would love to see it! Where could it be found?
When you change the seats it will alter the weight and weight distribution. You need to account for that change. So rebuild your seat FIRST and then do your math equations.
When you do the loop the thing that brings you up has to be high yours is low if you make it higher it might work!
Can I just say that you're a freaking hero, Will!
Stay awesome!
The lowest part of the track is the middle of the helix, put the loop there and then a small hill up onto the ‘bridge’ section.
By doing away with the helix altogether you will have more speed into the loop, it will be easier to place as the track will be on a straight and you won’t have to worry about it making it through the helix.
Well what is the update!?????? I want to see what you came up with!
Excited to see this progress!
Just remember not to make it round, but like a Spiral so the radius gets smaller, the Higher the cart proceeds and going back down, radius getting bigger.
I know youre familliar with this, but just wanted to mention it 🤪
Vane | Let's Play in the last video at the end he showed a drawing on his driveway of what you are referring to.
@@BikerFail16 totally missed that! Thanks!
Pembleland would be a cool place to visit. Maybe I make some concepts for home-made flat rides for you!
I think it might be able to do a loop then a helix. It might not have enough speed to do the helix.
You should make a launch coaster one day
What if you used magnets on the loop (like a launch coaster) to make it exit the loop faster?
rct3isepic very expensive
@@BikerFail16 A normal magnet firstly pulls you to it, but slows you down afterwards. So that would be pointless. An LIM or LSM motor on the other hand would work, but both of these require precision, lots of electricity, a cooling system and a computer who controls it. Not something you can achieve on a backyard coaster in other words.
Filipolis oh it could be done, but as I said it would be very expensive.
Id like to see the making of the loop
Make a Zero G Roll in the flinsh of the Helix
Blunder Loop. I still think an oblique loop is the way to go. Goal Boss is already banking at drop so seems like less work and less g force. Hey, great work and keep moving forward. Blunder On.
I don't know the equations, but the set up now seems to go fast at the top of the helix, so I think there's a good amount of energy that could go to making an epic loop...
I think you should use some sort of Force Vector Design (FVD) to design the loop. I’m sure you could find someone who’s pretty good with it to help you out.
what did you name your roller coaster?
How do u attach the supports to the ground
Will your wheel assemblies need modifying on the count that the up stops are now effectively also going to be the main road wheels whilst the cart is upside down?
Great question, Dan. The wheel assemblies are crazy strong. No worries there. Also, the cart as built now can roll freely around a 1 meter radius loop without getting jammed up. The big question is, would a rider want to be on the cart?
Will Pemble I think you should definitely consider a 1-meter-radius "headline" loop. It sounds insane. ;) On the other hand, I do worry about the G's in this project overall. Love the discussion that's going on here!
How do you bend the tubes in 3D ?
Maybe get jhon iverse he knows how to put a loop on a backyard coaster .
if you are going to be remaking the cart, does that also mean you are going to enhance the harness/restraint system? Do you plan to add some sort of lap bar? or keep it as just a seat belt? Because with it only being a seat belt, wouldn't that limit riders to a height restriction?
A seat belt is more accommodating than a lap bar or OTSR.
For a loop, we'll have to change out the seat with a high-back seat. Not sure whether we'll use 2 point or 4 point restraints yet. Blue Flash (best BYRC ever) has a handle for passengers to hold onto. Might give that some thought.
On second thought, I'm not completely certain you need a high back seat. Look closely at the blue flash riders, and see if you think you could handle that. Try scaling the blue flash from the videos. I think you could make about the same loop, but smoother and more efficient.
I think the helix will be over backed now. Don't forget shoulder restrain... 0.5 G is not a lot to be kept in your seat.
A seat belt should be enough. At least you wouldn't be trapped if something went wrong
How much did it cost to build this?
Also, use a program like Newton 2 or FVD for the loop design.
Next project: station
Haha! I didn't even think about that!
Yesss, when's the next video on this??
👍🏻✅
Firsttt
Ur obviously not... If there is comments before u.
_Urban-one cuz there are to be fair