View Full Version : What is the physics behind cars and bicycles needing bigger gear ratios at low speeds

Wesley Clark

03-14-2005, 10:35 AM

A bicycle going from 0-10 mph or 5-15mph in 15th gear requires more effort to make this acceleration than it does to go from 25-35mph in 15th gear. A car is the same way, the gear ratio in 1st gear when you go from 0-20mph is about 3:1, but when you get to 50-60mph you are in overdrive where the ratio is 0.8:1.

I don't know alot about physics, i'm only halfway through my first semester of a real physics course so i do not understand this. I understand that force = acceleration multiplied by mass. A bicycle going from 2-12mph over 20 seconds has the same acceleration as a bicycle going from 20-30mph in 20 seconds and the same mass, but the energy needed to increase velocity by 10mph over 20 seconds in a high gear ratio is alot less when you are going 20mph than at 2mph. It is the same with cars, you need a 3:1 gear ratio to go from 0-20 but a 0.8:1 to go from 55-75mph even if it takes 20 seconds in both situations.

Why is that, why do you need more energy to accelerate at low speeds than at high speeds? Is decceleration at lower speeds stronger than at higher speeds? Is it related to torque.

David Simmons

03-14-2005, 10:47 AM

A bicycle going from 0-10 mph or 5-15mph in 15th gear requires more effort to make this acceleration than it does to go from 25-35mph in 15th gear. A car is the same way, the gear ratio in 1st gear when you go from 0-20mph is about 3:1, but when you get to 50-60mph you are in overdrive where the ratio is 0.8:1.

I don't know alot about physics, i'm only halfway through my first semester of a real physics course so i do not understand this. I understand that force = acceleration multiplied by mass. A bicycle going from 2-12mph over 20 seconds has the same acceleration as a bicycle going from 20-30mph in 20 seconds and the same mass, but the energy needed to increase velocity by 10mph over 20 seconds in a high gear ratio is alot less when you are going 20mph than at 2mph. It is the same with cars, you need a 3:1 gear ratio to go from 0-20 but a 0.8:1 to go from 55-75mph even if it takes 20 seconds in both situations.

Why is that, why do you need more energy to accelerate at low speeds than at high speeds? Is decceleration at lower speeds stronger than at higher speeds? Is it related to torque.

A gear set that reduces output speed for a given input speed multiplies the input torque by the square of the speed reduction ratio. This allows the engine to run at its RPM for best output and multplies that engine torque at the rear wheels.

And, by the way, there is no reduction in bicycle gearing. The sprocket on the pedal end is bigger than that on the rear wheel end.

Gorsnak

03-14-2005, 10:48 AM

It's all about leverage. The energy requirements are identical, but it's easier put the required amount of energy into the system with a leverage advantage. It's exactly the same situation as contrasting trying to loosen an overtightened lug nut with a 6" wrench versus a 3' breaker bar. The same amount of force at the nut is required with either, it's just a lot easier to apply that force with the breaker bar.

In the case of the bicycle, it's an issue of what pedalling rpm results in the cyclist's legs producing the most power. At low rpm, the cyclist doesn't produce nearly as much power as he does at higher rpms (until he can't spin any faster without losing efficiency). So accelerating at low speeds with a high gear is difficult for him, since he can't produce power at low rpm, and hence has difficulty putting the energy required for acceleration into the system.

Gorsnak

03-14-2005, 10:53 AM

And, by the way, there is no reduction in bicycle gearing. The sprocket on the pedal end is bigger than that on the rear wheel end.

Some mountain bikes have a gear or two that result in the drive wheel spinning slower than the pedals. 24-tooth chainring to 28-tooth cog isn't unheard of. Nearly all of them drop down to 1:1 at the highest.

Hoodoo Ulove

03-14-2005, 11:05 AM

And, by the way, there is no reduction in bicycle gearing. The sprocket on the pedal end is bigger than that on the rear wheel end.

True of most road bikes - not true of mountain bikes and bikes set up for loaded touring. My mountain bike's small front chainwheel has 22 teeth, the largest sprocket on the rear cluster has 32.

A car is the same way, the gear ratio in 1st gear when you go from 0-20mph is about 3:1, but when you get to 50-60mph you are in overdrive where the ratio is 0.8:1.

The overall gear reduction is typically about three times that. You neglected the final reduction at the ring and pinion.

A bicycle going from 0-10 mph or 5-15mph in 15th gear requires more effort to make this acceleration than it does to go from 25-35mph in 15th gear.

First of all, it takes more energy to go from 5->15mph than 0->10mph. Kinetic energy goes as square of speed, so the the faster your speed, the more energy it takes to add 1 mph to your speed.

Second, gear ratio has very little to do with power output. The function of a gear system is to match the RPM of the engine to the RPM of the wheel. Some types of engines (including human legs) work best at a certain RPM, but the output power falls off at higher or lower RPM. So you'd use higher gear ratios at higher speeds. The gear you're in tells you nothing about the power output.

Finagle

03-14-2005, 11:43 AM

First of all, it takes more energy to go from 5->15mph than 0->10mph. Kinetic energy goes as square of speed, so the the faster your speed, the more energy it takes to add 1 mph to your speed.

Not to mention wind resistance which increases at least by the square of the velocity. I don't know if I can accelerate my bicycle from 20 to 30 mph, but I know I can't do it in 20 seconds.

Stranger On A Train

03-14-2005, 11:44 AM

Second, gear ratio has very little to do with power output.

Amend that to "...has nothing to do..." and you've got it right.

Backintheday when I was TAing a Basic Engineering class, one of the projects was to use a petal driven generator to power a lightbulb. The students had to figure out how to connect the petal system to the generator (using chain and sprocket) and wire up the bulb; a simple task, even for college freshmen who are reveling in their new-found freedom to drink and **** (or more likely, spend all night playing computer games and Magic: The Gathering) and manage to show up to an 08.30 class bleary-eyed and only sporatically attentive. Most students managed to grasp the basic concepts, but I remember a leader of one team who kept trying to figure out how to use gear ratios and longer crank arms to multiply the 1/3hp that a typical human being can produce into something greater. Nope, can't do it...power is power. You can change the forces, but Pout = Pin - L. Also, many students couldn't believe that a mere lightbulb could make the petals impossible to push. We'd hook up ten bulbs in a row and they'd stand on the cranks barely able to move them. It was funny as hell.

My job now pays much better, but isn't nearly as fun, even when the people are just as clueless as freshman engineering students. :rolleyes:

Stranger

David Simmons

03-14-2005, 01:15 PM

Also, many students couldn't believe that a mere lightbulb could make the petals impossible to push. We'd hook up ten bulbs in a row and they'd stand on the cranks barely able to move them. It was funny as hell.

My job now pays much better, but isn't nearly as fun, even when the people are just as clueless as freshman engineering students. :rolleyes:

Stranger

During the State High School Basketball Championships at Iowa City when I was there, the College of Engineering put on an open house with exhibits of all kinds. One that we EE's used was a hand-cranked generator hooked up to a light bulb. In parallel with the bulb was a circuit that ran down to the far end of the lab to a shorting switch.

We would wait until a teeny little girl and a great big athlete type would come in and have the girl generate some electricty for us. She would spin the crank and the bulb would light up nicely. Then we'd give the athlete a chance and surrepticiously close the shorting switch. Some of them would almost have a stroke trying to light the light.

TheLoadedDog

03-14-2005, 02:14 PM

We would wait until a teeny little girl and a great big athlete type would come in and have the girl generate some electricty for us. She would spin the crank and the bulb would light up nicely. Then we'd give the athlete a chance and surrepticiously close the shorting switch. Some of them would almost have a stroke trying to light the light.

I'm a sucker for stuff like that. One of my favourite Candid Camera stunts involved a frail old lady with a suitcase. She'd ask passing burly young men to carry the case for her. Of course, the suitcase was generally light, but it did have a sheet of iron in the bottom, and there just happened to be an electro-magnet in the sidewalk.

Magnet ON

"Certainly Ma'am...... ugh.. um... I don't understand this... You got rocks in this thing?"

Magnet OFF

"Oh come now young man, it's easy. Look!"

Wesley Clark

03-14-2005, 03:48 PM

A gear set that reduces output speed for a given input speed multiplies the input torque by the square of the speed reduction ratio. This allows the engine to run at its RPM for best output and multplies that engine torque at the rear wheels.

And, by the way, there is no reduction in bicycle gearing. The sprocket on the pedal end is bigger than that on the rear wheel end.

Do you mean a 3:1 ratio has 9x the torque of a 1:1 ratio?

A car will not drive at 10mph in 4th gear, it will just shut down. A bicyclist cannot go from 0-5mph in 18th gear because his legs are not strong enough. Why does this happen if force is just mass times acceleration and acceleration is the same in both situations?

Gorsnak - can you cover that again, or someone else who understood it? Is it just that the RPMs of the power supply (whether that power supply is the engine or the pedals) has to be a certain number irrelevant of speed to produce quality power, and that at lower speeds a higher gear ratio on a bike will result in RPMs 5-10x too slow for a bike to function?

Stranger On A Train

03-14-2005, 04:17 PM

A car will not drive at 10mph in 4th gear, it will just shut down. A bicyclist cannot go from 0-5mph in 18th gear because his legs are not strong enough. Why does this happen if force is just mass times acceleration and acceleration is the same in both situations?

F=Ma in all situations, including these, but it isn't the only relevent equation. There is also T=Fr; that is, the smaller your "r" (radius of rotation or moment arm) the more force is required. Torque through a transmission, neglecting mechanical losses, will be the same in and out. throwing the parameters into the equation and setting the torques equal gives you:

Fout=Fin * (rin/rout)

For a 3:1 ratio, you get Fout = 3 Fin. Cool, huh? For lower ratios, it takes more force to get the same torque, and there is a limit to how much force a mechanical engine or a human being can produce. The tradeoff for the mechanical advantage of a high ratio is that you have to go take more revolutions to get the same output, which is why your car moves okay in low gear at low speeds, but at highway speeds (where it doesn't need a lot of torque) the engine would be screaming if you left it in first or second gear.

For any engine, be it Otto-cycle, diesel, or human powered, there is an optimum range of engine speed for torque output. Too little speed and the engine can't maintain enough momentum to keep running efficiently. Too much and it can't keep up and becomes unbalanced, or can't burn fuel fast enough, or starts tearing itself apart due to inertial loads.

Stranger

A car will not drive at 10mph in 4th gear, it will just shut down. A bicyclist cannot go from 0-5mph in 18th gear because his legs are not strong enough. Why does this happen if force is just mass times acceleration and acceleration is the same in both situations?

The answer is that it doesn't really happen the way you say. A biker in 18th gear can get the bike moving to 5mph, he just has to put steady force on the pedals (not even necessarily a huge amount of force), and the acceleration happens very slowly. The only reason a car can't do the same thing is that an internal combustion engine has a lower boundary for how slow it can crank before it dies. Little explosions don't happen in slow motion.

Is it just that the RPMs of the power supply (whether that power supply is the engine or the pedals) has to be a certain number irrelevant of speed to produce quality power, and that at lower speeds a higher gear ratio on a bike will result in RPMs 5-10x too slow for a bike to function?

I think you got it.

If you choose your gears properly so that you can pedal at 90 rpm, you need about 28 lb of force on the pedals to achieve 200 W of power output*. If you choose the wrong gear and try to pedal at 5 rpm, you need 505 lb of force to achieve 200 W power output - which is highly inadvisable, if not impossible.

*Assuming 170mm crank length, the pedal moves at a speed of 2*pi*170mm*90/min = 1.6 m/s. Work is force times distance, so power is force times speed. 1.6 m/s * 28 lb = 200 W (with appropriate unit conversions).

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