Bicycle Transmission, Schwartz, Cal Poly Physics
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- Опубликовано: 5 дек 2014
- This video is part of a free online mechanics course with all videos and a textbook.
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How does the power go from your feet to the wheels? 
Хобби
wow...super clear explanation...many thanks!
After finding all the videos, reading the blogs and articles in the end God finaaly I find something worthy man. You make my day (actually it's night rn😂). But thank you can. U please explain a little detail or any way to contact as I'm working on the bike to travel at speed of 100km/h and I am not sure about the calculations and gear ratio
Thank you.
2:50 - It's not really helpful to describe the larger diameter rear gear (it's actually the number of teeth, and not technically the diameter of the rear sprocket) as delivering twice the power. Given half the pedaling force and at double the velocity (cadence) we'd have the same power input, which is exactly how pro cyclists conserve their muscles. With the changing of the gear we also experience a lower velocity of the bike for a constant velocity of the crank. So no more power, just a different force needed to move the chain.
It's true that it allows the mechanical advantage of same crank velocity x less crank force, to move the bike less distance. In reality the power that a rider is able to produce may be the same, but the various gears provide a variable to allow the rider to perform more WORK with the same power. W = P * T. When we start a bike, inertia of your body + bike weight is great, so we require the mechanical advantage of the low gearing to help our fixed power ability to overcome that inertia. In a vacuum (no aero drag) and with no hysteresis of the tires (no rolling resistance) and with no mechanical friction in the drivetrain, we would be able to continue to increase gearing as we picked up speed, and with the same power, do more and more work with the same pedaling force, and incidentally stop pedaling at 1000mph and coast infinitely because of our inertia and no opposing force. But in the real world we encounter increased opposing forces as we increase speed. We use transmissions to change the crank speed needed to maintain a desired speed or force required to accelerate, and as you said, because our velocity of pedaling has limits. Technically, we stop pedaling to move the bike, but instead pedal to oppose wind drag and hysteresis to maintain speed.
But changing to the lower gear (bigger rear sprocket) doesn't increase power instantaneously, because we tend to go slower and cover less ground meaning less velocity. It actually decreases the power needed to propel the bike forward from a stop or slower speed, thus sacrificing speed and allowing lower force because we lack excessive instantaneous power. Thus the real world use of a transmission is about acceleration and changing velocity, and not changing power output from a fixed power input, which your formula showed, does not happen.
Hello Dr. Schwartz.
Amazing video, it really clarified a lot for me and I truly appreciate it. I have one question if you have a free second. In the beginning of the video you basically said that the power is the same, but forces/torques/radii change. That makes sense to me, then right at the end you said power doubles. That got me a little lost. I really appreciate your help and again, very informative!
Kyle, Thanks for the question. The power from your legs is the same as the power to the wheel (minus the tiny amount lost to friction in the drive system). However, then you switch gears and you're now pushing with the same force (torque) at twice the rotational velocity, so now you are putting out twice the power (as you were before), and the wheels are receiving twice the power. Because you've switched gears, your legs produce twice the power through spinning twice as fast, and the rear wheel (which is still spinning at the same speed because you haven't sped up yet) experiences twice the power by receiving twice the force. I hope that helps.
If you lower your gear by a factor of two, and you are peddling just as hard, then you will be putting out twice the power. The power in is still the same as the power out.
Thanks for the video! I got my BSMET at Cal Poly SLO about 20 years ago. Wonderful campus! Now I'm designing an ebike I want to build. The power limit here in MI is 750 watts to be classified as a bicycle (and not an electric motorcycle, requiring registration, insurance, etc). I'd like to figure out what gear ratio I'd have to use to have the bike to have sufficient torque (assume 50 pedal rpm) to carry two adults weighing a total of 375 lbs up a 5% grade at a constant speed, with a tire of 29" outside diameter. I plan to build it single speed, so I need to get benchmarks on what speed it can climb the hill, and what will be the top speed when the crank is at 100 rpm. Do any of your videos address this? Thanks!
I think you need a physicist of ME collaborator!!! It's a pretty straightforward calculation
Mr. Schwartz, thanks for making this film. I have a question for you. How does physics explain mathematically why it takes more power to peddle a bike up at 15% grade than it does to peddle a bike on a 0% grade? What is the best way to quantify the feeling that gravity is pulling harder when climbing a hill in contrast to riding on flat ground?
You do work to increase your gravitational potential energy
thanks
How do you calculate the force static friction applies on the rear wheel based on the 1) force of the chain on the rear gear and 2)radius of the rear gear and radius of the rear wheel?
You could use an energy consideration: the power to the wheel = power to the bicycle, so Tension in chain * speed of chain = force of friction on road * speed of bike. You could also just multiply the tension in the chain by the radius of the rear cog , divided by the length of the wheel radius.
Sorry to take two years to reply to this. If the coefficient of friction is high enough for the wheel not to slide out, then you can just conserve power. P = F*v = Torque * rotational velocity. So, you can calculate the power at any point in the drive train (because it's the same), and set it equal to F*v, where v is the speed and F is the force of friction. Divid by speed and you have force.
Mi Mr. Schwartz,
Love to video. I'm currently working on a physics project and I am trying to find the difference in efficiency between different gear ratios while using the same amount of force to push the pedals. But because I don't have a power meter on my bike, I am trying to find a way to calculate the necessary rpm for each gear ratio in order to know I will be exerting about the same force during my real-world trials. Do you have any ideas on how I would go about finding this?
you dude, im doing a very similar experiment and I'm facing the same issue as you are. if you sound a solution to this, can you please reply to me telling me how
@808mas nope i changed the experiment all together lol
Mr Schwartz,
Hypothetically, if a third gear (half the diameter of the your gear #1) was introduced in the beginning where foot applied force after doubling the radius of the last gear, could speed and force of foot remain the same while doubling output force?
That conservation of energy is going to prevail. If you gain in speed, it will cost you in force.
Understood, thank you!
Mr Schwartz,
If the conservation of energy prevails, why did the tour de france outlaw incumbent bicycles because they have a "mechanical advantage"?
Isnt the fact that gears are used at all a mechanical advantage?
@@jamescongleton6146 Good question. You could interpret "mechanical advantage" a number of ways. The Tour de France is seeking to standardize the bicycle so that it is the person that is being tested against each other. The recumbent bike has the advantage of a lower wind resistance, so they won't allow them. However, this is something very different from finding a way to deliver more power to the wheels than you produce with your legs... which is impossible, unless you use an energy storage mechanism such as a battery/electric motor. The gears provide a "mechanical advantage" by allowing one to spin their legs at their optimum frequency (about 90 rpm) so that the rider generates the maximum power from their legs under all biking speeds (for all levels if incline). having different gears allows the legs to spin at the same frequency for different bike speeds. The Tour de France allows this kind of "mechanical advantage". I hope this helps.
So question.. In a perfect world the pedal force would be constant around all 360 degrees of the pedal stroke. But in reality it's not. We apply more force from 2:00 to 4:00, than from 11:00 to 1:00 and so on. Would this be why I ALWAYS prefer the big crank ring over small even at exactly the same gear ratios? For example (crank-to-cassette, 2.00 ratio): 54t/27t is preferred over 42t/21t. Why is this?
I think it's purely imaginary on your part. Should have no effect
with the possible exception that you may have slightly better connection with larger rings, but that should have little effect.
With this formula, is having a longer crank an advantage?
Marvin, Thanks for the question. Having a longer crank allows you to increase your torque and therefore deliver more power if you can keep your rotation rate up. The downside of that is you have to move your feet further, so it is a trade off. In particular, you don't want the crank to be longer than comfortable for you leg size... but now we're getting into bicycle fitting.!
@@pvs242ful totally agree, thanks for the reply. useful for those climbs.
Also can u please explain how can we maintain speed on a bicycle as we are giving it an acceleration by putting a continuous force through the peddles
You do work to increase your kinetic energy... is that OK?
@@pvs242ful actually sir I want to say that if I apply a continuous force on some circular disc(like peddling ) then as force=mass*acceleration then technically it has to be accelerating may be to an infinite value but in actual practice when we ride bicycle it does not happen the only thing that comes to my mind is our force or work is used to get rid out of friction..... I'm I right ??.. I want someone to clear this doubt ?? Please help sir !!!
Good point... but there is not a NET force on the pedals because there is an equal and opposite force caused by the gravitational gradient... So there is a net force of zero as you struggle to climb a hill at constant speed. And you will see if you take your feet off the pedals you will accelerate downward.
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