Gear Ratio in Bicycles using Rotational Motion

In summary: Similarly , if the bike is going too fast , the chain will start to 'skip' over the gears and the bike will eventually come to a halt .... This is because the ground pushes back on the bike tire and the bike can't keep up with the ground's force .... So the bike comes to a halt because the ground pushes back on the tire and the bike can't keep up with the ground's force ....When angular velocity of the back wheel increases , the wheel gains a tendency to slip on the ground . Friction opposes this tendency of relative motion by increasing - this causes acceleration of the wheel's center of mass , causing increase in velocity of the Com , thus making net velocity of point in contact with ground zero , thus preventing relative
  • #36
andyrk said:
Then what does changing gears even do?
Did you not read my post #28?

(Or, to answer a different way: exactly what AT just said and nothing more.)
 
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  • #37
I don't understand why the rear wheel accelerates more, by reducing the rpm of our legs/where we pedal.
 
  • #38
andyrk said:
I don't understand why the rear wheel accelerates more, by reducing the rpm of our legs/where we pedal.
You seem to be hazy on the concept of mechanical advantage.

Bikes use different gears to reduce the rider's effort at maintaining speed.

By using his feet on the pedals, the bike rider creates a torque which is transmitted by the chain to the rear wheel. This torque is what turns the wheel, which then moves the bike, and makes it go. If you want to go fast, it stands to reason that the rear wheel must turn fast, which in turn means the rider must pedal more quickly. This gets rather exhausting, even for competition riders. What if there were some way the rider could keep turning the pedals with his feet at the same rate all the time, but use a special device to make the rear wheel turn faster?

Fortunately, there is: the gear.

Study this article on gears:

https://en.wikipedia.org/wiki/Gear
 
  • #39
Velocity in the frame of the ground is actually zero - unless you have wheel spin.
andyrk said:
It has a velocity at the bottom-most point
 
  • #40
andyrk said:
I don't understand why the rear wheel accelerates more, by reducing the rpm of our legs/where we pedal.
Have you pedaled a bike before? What happens to your ability to apply force to the pedals as the rpm gets too high? Or: is it easier to apply a large force when your legs are moving fast or slow?
 
  • #41
andyrk said:
I don't understand why the rear wheel accelerates more, by reducing the rpm of our legs/where we pedal.
I don't understand why you think this must be the case.
 
  • #42
We really try to avoid spoonfeeding here because people learn better (and learn how to learn better) if they figure things out on their own after being given a gentle push in the right direction. But you're just not getting it, so I'm going to lay it out for you:

A bicycle begins to accelerate (from a stop or a constant speed) because more torque is applied to the back wheel. To get more torque to the back wheel, you push harder on the pedals.

The reason bikes are geared is because there is a limited range of speeds at which a human can push on a pedal. Above perhaps 120 rpm, our legs simply can't move fast enough to keep pushing on the pedals - you spend most of your effort basically "chasing" the pedal with your foot. So gearing-up allows you to push harder on the pedals, by lowering their rpms.
 
  • #43
andyrk said:
I don't understand why the rear wheel accelerates more, by reducing the rpm of our legs/where we pedal.
I get the feeling that you have stopped thinking about what your actual experience has been of riding a bike and latched yourself in a totally 'verbal' mode of approach. You expect too much if you insist on an answer that fits totally into your requirements for an 'explanation'.
The general topic of levers and gears is discussed all over the place (internet and books) and there are countless examples of Force Magnifiers and Velocity Magnifiers, where the terms Velocity Ratio and Mechanical Advantage are explained and discussed. You have to make a personal effort with this and not rely on other people to find the magic bullet to solve your problem.

It could be that you have read something, somewhere and you are trying to fit what you read or understood from it to the answers you have been getting on PF. It may be that your memory or interpretation has failed you and you should start again with another source (e.g. this thread).
 
  • #44
russ_watters said:
A bicycle begins to accelerate (from a stop or a constant speed) because more torque is applied to the back wheel
So what does more torque make the rear wheel do? Make it spin faster?
 
  • #45
andyrk said:
So what does more torque make the rear wheel do? Make it spin faster?
The increased torque at the pedals is delivered to the rear wheel by the chain, which exerts an increased torque on the rear gear. The rear gear is rigidly affixed to the wheel. So the increased torque on the rear gear is accompanied by an increased torque on the rear wheel. This causes it to exert a larger backward force on the ground, which, in turn, exerts the larger forward force on the wheel. This larger forward force is transmitted through the wheel to the axle, and from that, to the frame. This causes the body of the car to accelerate. The increasing linear velocity of the body is accompanied by an increasing rate of rotation of the wheel.

It is possible to model all this pretty easily. Such a model would include the rear wheel, the rear gear, the chain, the bar joining the rear wheel axle to the pedal axle, the pedal gear, and the pedal. Such a model would enable one to quantify the effect of different gear diameters on (a) the ratio of the pedal angular velocity to the rear wheel angular velocity and (b) the ratio of the imposed pedal torque to the forward force exerted by the ground on the rear tire.

Chet
 
  • #46
Amazing. I understood everything now.
 
  • #47
andyrk said:
So what does more torque make the rear wheel do? Make it spin faster?
Yes. And because of friction with the ground, instead of just spinning free and slipping, it accelerates the bike forward as it spins faster.
 
  • #48
Also, when we change gears, why does it become easier to pedal(less effort, pedals feel lighter) or difficult to pedal (more effort, pedals feel heavier), depending on the gear we shifted to? Does this have any physics related to it?
 
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  • #49
andyrk said:
Also, when we change gears, why does it become easier to pedal(less effort, pedals feel lighter) or difficult to pedal (more effort, pedals feel heavier), depending on the gear we shifted to? Does this have any physics related to it?
Have you looked into the physics of simple levers? Gears are levers working continuously.
 
  • #50
A.T. said:
Have you looked into the physics of simple levers? Gears are levers working continuously.
What does it have to do with physics?
 
  • #51
andyrk said:
What does it have to do with physics?
Physics describes levers quite well.
 
  • #52
I don't understand how the fact that it becomes harder or easier to pedal, can be explained by physics.
 
  • #53
andyrk said:
I don't understand how the fact that it becomes harder or easier to pedal, can be explained by physics.
Do you understand it for levers? Pushing down at different points of a playground seesaw?
 
  • #54
A.T. said:
Do you understand it for levers? Pushing down at different points of a playground seesaw?
Yes I know it becomes tougher or easier. But how do you relate it to circular motion and what part of circular motion would it use? Torque? The relation between angular velocity and tangential velocity? What?
 
  • #55
andyrk said:
But how do you relate it to circular motion and what part of circular motion would it use?
I would start with a static case, and try to understand the relation between applied forces (that's what you "feel").
 
  • #56
andyrk said:
Yes I know it becomes tougher or easier. But how do you relate it to circular motion and what part of circular motion would it use? Torque? The relation between angular velocity and tangential velocity? What?
Do you see a connection between gears and levers (and all other machines, in fact)? Have you actually read any basics of this or are you trying to get your education just by asking questions here on PF. That is a very inefficient method of learning.
 
  • #57
andyrk said:
Also, when we change gears, why does it become easier to pedal(less effort, pedals feel lighter) or difficult to pedal (more effort, pedals feel heavier), depending on the gear we shifted to? Does this have any physics related to it?
This:
A.T. said:
It changes the ratios of angular velocities and torques, between back wheel and pedals.
Also: https://en.wikipedia.org/wiki/Mechanical_advantage
That discusses both levers and gears...and pulleys. Pick whichever one makes the most intuitive sense to you to start off with -- they all do the same thing from an "easier/difficult" perspective.

Or, another way: changing gears changes the length of the "lever arms" that transmit the torque.
 
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  • #58
It boils down to two moment balances, one on the pedal axle (and attached front gear) and the other on the rear wheel axle (and attached rear gear). The chain attaches to both gears and transmits torque between them. It is able to do this because the tension in the upper part of the chain becomes higher than the tension in the bottom part of the chain. Does this contain enough physics in it for you?

Draw a free body diagram of the rear wheel with the chain split top and bottom at some location between the rear axle and the pedal axle. Then draw a free body diagram of the pedal axle with the chain split at the same location

Chet
 
  • #59
Why should two gears connected by the same chain rotate with the same torque?
 
  • #60
andyrk said:
Why should two gears connected by the same chain rotate with the same torque?
They don't. Again, read this:
A.T. said:
It changes the ratios of angular velocities and torques, between back wheel and pedals.
 
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  • #61
andyrk said:
Why should two gears connected by the same chain rotate with the same torque?
As Russ said, the torques are not the same. You would already know this if had followed my suggestion in post #58 and drawn the free body diagrams that I indicated. This is a very straightforward system to analyze if you use the free body diagrams. Show us what you can do. On the other hand, if you feel that you have advanced beyond the point where you need to use free body diagrams, I think you will now agree that you haven't.

Chet
 
  • #62
Chestermiller said:
As Russ said, the torques are not the same. You would already know this if had followed my suggestion in post #58 and drawn the free body diagrams that I indicated. This is a very straightforward system to analyze if you use the free body diagrams. Show us what you can do. On the other hand, if you feel that you have advanced beyond the point where you need to use free body diagrams, I think you will now agree that you haven't.

Chet
No I can understand it by drawing the FBD. But I am studying this just for fun and I am lazy, so I don't want to bother with FBD and instead understand it just by discussion. But if that won't make me understand this completely, then I would have no other option than to draw the FBD. But I am just waiting till then.
 
  • #63
andyrk said:
No I can understand it by drawing the FBD. But I am studying this just for fun and I am lazy, so I don't want to bother with FBD and instead understand it just by discussion. But if that won't make me understand this completely, then I would have no other option than to draw the FBD. But I am just waiting till then.
You've already waited too long. Your admitted laziness has cost those of us who are trying to help you lots of our valuable time. Doesn't that matter to you?

Chet
 
  • #64
Semi-rhetorical question: it matters to us.

We've been humoring you based on the possibility that you might be young or slow, but lazy is something we can't abide.

This thread has gone around in circles long enough and is therefore locked. Tomorrow, when you have another question on the subject, reread the thread; I suspect you will find that it has already been answered.
 

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