# Bicycle gear ratios

1. Jun 8, 2009

### flycast

I am a cyclist and am starting to think about gear ratios. I have counted teeth on my bike gears (touring) and calculated ratios and the results do not make sense to me. I am left with the beliefe that ratio is not all at work here. It seems that it also has to do with the diameter of the gear in a kind of lever kind of way. In other words the bigger gear on the front the tougher it is, the smaller on the back the tougher it is.

Can someone please explane this to me in a way that makes sense to a non physicist?

2. Jun 8, 2009

### Staff: Mentor

Welcome to pf.

I'm not sure what the issue is - i don't see anything in the post that isn't true except where you express a doubt that it isn't!

Yes, the bigger the front gear the higher the torque.
Yes, the smaller the back gear the higher the torque.

3. Jun 8, 2009

### lividz

Are you talking of chain drive/gear drive?

Last edited by a moderator: Jun 9, 2009
4. Jun 8, 2009

### lividz

in both cases diameters matter....
for gears there a ratio in between no of teeth and diameter[oitch]=module..

5. Jun 8, 2009

### flatmaster

http://www.sheldonbrown.com/gain.html

Battery is almost dead, so I'll send you there.

Check out bikeforums.net

I'm flatmaster there as well.

6. Jun 8, 2009

### flatmaster

It's actually quite amazing. Your end gear ratio is determined not only by your gears, but also by your crank length and wheel diameter. This final ratio determines two things; the force required to push the pedals, and the cadence at which you must pedal. It's quite amazing that these two needs for gearing match up at at both a comfortable pedal force and cadence for human legs.

7. Jun 9, 2009

### lividz

Hmm...
well I'm a student ... and havnt yet covered "Chain drives" in academic syllabus...
but whats wrong with "Gear drive" mentioned above in my post...and I didnt specifically note you're a cyclist...so you aren't talking of gear drives [not bicycle gears]...but thats what I discovered after I posted...

And chain drives are considered as belt drives wrapped around rotating polygons..

Anyways sorry...coz I shouldnt have replied on chain drives[Actually I knew their similarity to belt drives mentioned above...and replied on belt drive basis]...

but still I've chosen to write the following regarding your post:

Thats has to do with moment...right?
does ur crank length mean the length of the rod connecting pedal and front wheel center......

Just trying to understand...dont give other 4m liks..I cant go there and register and then find out relevant things...its better u reply here...

Last edited by a moderator: Jun 9, 2009
8. Jun 9, 2009

### flycast

OK...

Sounds like gear teeth have very little to do with it once I change both front and back gears.

It sounds like the following make the most impact:

Length of the rod the pedal is attached to (crank)
Front gear diameter
rear gear diameter
rear wheel diameter

I am guessing that when people talk about gear ratios they are talking about mechanical systems where one gear diameter (and therefore teeth count) is fixed.

If I were to quantify the entire system into a number that I could use to compare gears and combination of gears how would I do that?

9. Jun 9, 2009

### uart

Wheel diameter and crank-arm length certainly do effect the overall mechanical advantage, but for a given bicycle those are normally fixed. So for a relative measure we usually just concentrate on the gear ratio itself, expressed as the number of teeth on the chain-ring divided by the number of teeth on the rear cog.

For example with a 52 tooth chain ring and a 13 tooth rear cog you have a ratio of 52:13 = 4:1. This means that for every one turn of the crank that the rear wheel turns four complete revolutions.

Now with a granny gear like 40 on the chain ring and 32 on the rear cluster the gear ratio is now 40:32 = 1.25:1. This means that for each turn of the crank that the rear wheel turns 1 and one quarter turns.

Now regarding the overall mechanical advantage of the system, from crank to road contact, this can be thought of in either of two equivalent ways. These are : 1. the ratio of distance moved at the road divided by the distance moved by the pedal (along it's arc around the bottom axel) or equivalently as 2. the ratio of force exerted at the pedal divided by the force delivered by the rear tyre to the road. Each of these equivalent ratios are equal to the gear ratio (as defined above) multiplied by the wheel radius divided by the crank-arm length.

So for an example, say you had a gear ratio of 4 (eg 52:13) and a wheel radius of 35cm and a crank length of 17.5cm; then the overall (speed) mechanical advantage would be 4 times 35 divide 17.5 which equals 8.

This means that the speed of the bike will be 8 times the speed that your pedals are moving along their arc, and equivalently that the force exerted by the tire onto the road is only one 8th of the net force (sum from both feet) being applied at the pedals.

10. Jun 9, 2009

### belliott4488

flycast:

As a cyclist, you've probably run into discussion of gears in terms of "gear-inches", which is the traditional way of measuring gears and has persisted to this day, despite its historical origins. Here's what that's all about:

If you had no chain drive, but could only drive your wheel by cranks and pedals rigidly connected to the wheel hub, then the only way to get a higher gear, i.e. to go farther with one turn of the cranks, would be to have a larger diameter wheel. This is exactly how the old "high wheelers" or velocipedes from the late 19th century worked, and it explains why they had such huge front wheels. Ultimately, the thing that limited their gearing (aside from the difficulty of riding a bike while perched some four or five feet in the air) was the length of the cyclists' legs, since they still had to be able to reach the pedals.

That's why gears were originally measured in gear-inches - it was literally the diameter of the wheel in question. Nowadays, we all use bikes that change the ratio of pedal/crank rotations to distance traveled by changing the ratio of pedal rotations to wheel rotations (by some mechanical means), so our wheels all have the same diameter (for a given style of bike).

In order to compare our gears with the traditional definition, you have to ask "what would be the diameter of a fixed wheel that had the same mechanical advantage as my selected gear on my bike?" To answer that, you have to multiply the ratio of wheel rotations to pedal rotations - which is equal to the the ratio of chain ring teeth to cog teeth - by 27 inches, which is the approximate diameter of a standard road bike wheel in inches.

For example, if you have a 50-tooth chain ring and a 25-tooth rear cog, then your rear wheel turn twice for every rotation of the pedal cranks. Thus you will move a distance equal to twice the circumference of your wheel. That's the same as one rotation of a wheel with twice the diameter, i.e. 54 inches. So you move as far for each rotation of your pedals as you would on a velocipede with a 54 inch wheel, hence you have a gear of 54 gear-inches.

11. Jul 4, 2009

### wilsonxb

Do you know the sizes of the front sprocket and the sizes of the rear cog

12. Jul 4, 2009

### Naty1

If you recognize the gear teeth size is equal on two meshing gears you are half way there!
The gear rato is determined by the number of relative teeth...one gear has a larger diameter than it's mate because there needs to be more room, more diameter hence larger circumference in order to fit in more same sized teeth. Try reading wikipedia
http://en.wikipedia.org/wiki/Gear_ratio

13. Jul 15, 2009

How much torque you end up getting on the rear wheel is directly dependent on solely the gear ratio and the crank arm length. The smaller the gear ratio(ex: 0.5:1), the less torque you have on the rear wheel. the larger the gear ratio(ex: 2:1), the greater torque you have on the rear wheel. Greater torque means faster acceleration and greater hill-climbing ability. The torque that you generate on the pedals is directly dependent on the length of the crank arm. Torque=Force*Radius. The Radius is the crank arm itself. Therefore, take 2 crankarm lengths: 175mm, and 220mm. Both are given the same force input on the rider's behalf of 200N's. The 175mm crankset develops: 0.175*200= 35Newton-meters of torque, the 220mm crankset on the other hand develops: 0.22*200=44Newton-Meters of torque.

14. Jul 15, 2009

As far as determing how much torque ends up on the rear wheels, just take that torque developed at the crankset and multiply it by the gear ratio. For example, let's say that you're pedaling up a steep hill out of the saddle: you're exerting your body weight, let's say 600N's on the 175mm pedals. this equates to 105Newton-meters. Let's say your in mid-gear on the front: 42 tooth. And you're on a 18 tooth sprocket in the rear. That's a gear ratio of 0.428:1. Just take your 105newton-meters that you're generating and multiply by this gear ratio and you get 45newton-meters on the rear wheel. Which by the way, assuming you weigh 132 pounds, means that you could ascend a 28% slope, you'd just be out of the saddle the entire time.

15. Jun 14, 2010

### bill9999

We all know that there are several options of front sprocket/rear sprocket combinations that achieve the same gear ratio. i.e. A large front sprocket with medium rear or medium front with larger rear will produce the exact cadence at a given speed.
What I've always wondered, is there any advantage (in torque, power or energy) in using a medium front versus a large front sprocket assuming equivalent cadence-speed?

16. Jun 14, 2010

### sophiecentaur

I think the optimum choice involves considering the angle of the chain. It isn't going to be as good if you connect an outside sprocket on the crank with an inside sprocket on the wheel as there will be more rubbing of the cheeks of the links coming on and off the sprocket.

17. Jun 14, 2010

### uart

18. Sep 12, 2010

### Blazerwolf

I am getting a bicycle that had 24 gears as opposed to a single speed. Based on information I have gathered up the following data and formatted a chart based on that information that I guess one could call a Penny Farthing Chart.

First assumption is that the ideal pedaling gear ration will be 55 on our chart. As the ratio goes down the slower you go and the higher the ration the faster you go.
This chart is developed in an Excel worksheet
Front Sprocket Tire Size Freewheel Ratio
24 29 16 43.50
25 29 16 45.31
26 29 16 47.13
27 29 16 48.94
28 29 15 54.13
29 29 16 52.56
30 29 16 54.38
31 29 16 56.19
32 29 16 58.00
33 29 16 59.81
34 29 16 61.63
35 29 16 63.44
36 29 16 65.25
37 29 16 67.06
38 29 20 55.10
39 29 16 70.69
40 29 16 72.50
41 29 16 74.31
42 29 16 76.13
43 29 16 77.94
44 29 16 79.75
45 29 16 81.56
46 29 16 83.38
47 29 16 85.19
48 29 26 53.54
30t sprocket X 26" ÷ 15t freewheel = 52

As you can see when the sprocket size is 38t and the cog size is 20 the ratio is 55.10 The ideal gear ration using 55 as the nominal ratio. You will also see that with a 48t and a 26t cog that ratio is 53.54 being close to ideal.

Please analyze and tell me where I am with this concept.