Why do mountain bikers choose smaller front sprockets when climbing hills?

[R Power]

Hi friends

I have a little doubt on the gear ratios of a bike like a mountain bike which have front sprockets also.
It's clear that we get maximum RPM and min torque out of a bike when the chain is on smallest sprocket at the rear and largest at the front(highest gear ratio).
But i read that when climbing hills bikers choose smallest sprocket at the front and largest on the rear in order to ride with constant effort. This is what i can't understand. Why would they choose smallest sprocket on front?
Basically, my question is what is the need for small sprockets at front?
Because while climbing hill we need more torque and so we should choose largest sprocket at rear that's ok but why smallest or smaller at front. Because choosing small sprocket at front will just reduce the rear wheel rpm but will have no effect on output torque. So it will give no help while climbing.
If you say that the advantage is more torque at lower rpm then rethink that torque remains
same only rpm of rear wheel reduces and if we just want to lower the rpm we can pedal slowly.
Torque at rear wheel depends only upon rear sprockets. Then what is need for small sprockets at front while climbing hills. Plz help me clear out this.

 

[brewnog]

The torque is indeed increased using a smaller crank gear, I don't know where you've got the idea that torque is only a function of the sprocket diameter. It's the overall gear ratio which makes the difference.

(Output speed x output torque) = (input speed x input torque), ignoring losses. This means that, for a constant bike speed (and constant power input), an increase in input torque is accompanied by a decrease in input speed. This is why it's easier when climbing a hill to drop a gear (either smaller crank gear or larger sprocket) and pedal faster!

 

[DaveC426913]

Hi friends

I have a little doubt on the gear ratios of a bike like a mountain bike which have front sprockets also.
It's clear that we get maximum RPM and min torque out of a bike when the chain is on smallest sprocket at the rear and largest at the front(highest gear ratio).
But i read that when climbing hills bikers choose smallest sprocket at the front and largest on the rear in order to ride with constant effort. This is what i can't understand. Why would they choose smallest sprocket on front?
Basically, my question is what is the need for small sprockets at front?
Because while climbing hill we need more torque and so we should choose largest sprocket at rear that's ok but why smallest or smaller at front. Because choosing small sprocket at front will just reduce the rear wheel rpm but will have no effect on output torque. So it will give no help while climbing.
If you say that the advantage is more torque at lower rpm then rethink that torque remains
same only rpm of rear wheel reduces and if we just want to lower the rpm we can pedal slowly.
Torque at rear wheel depends only upon rear sprockets. Then what is need for small sprockets at front while climbing hills. Plz help me clear out this.

Let's say the largest rear gear has 60 teeth, and the front 2 gears have (A and B) 30 and 20 teeth respectively (not possible but I'm using easy numbers).

So, on gear A (which has 30 teeth), every full rotation of the pedals turns the rear wheel 1/2 turn (30/60).

Now the rider switches to gear B (20 teeth). For each full pedal, the rear wheel will now only turn 1/3 turn (20/60).

Because he has to do much less work per full pedal cycle (only turning the rear wheel 1/5 of a turn), his pedaling is significantly easier.

This is the whole basis of gear ratios.

If the front gear has 10 teeth and the rear geear had 10000 teeth, you could pedal a truck uphill.

 

[R Power]

(Output speed x output torque) = (input speed x input torque), ignoring losses. This means that, for a constant bike speed (and constant power input), an increase in input torque is accompanied by a decrease in input speed. This is why it's easier when climbing a hill to drop a gear (either smaller crank gear or larger sprocket) and pedal faster!

As we choose small sprocket at front an increase in front sprocket RPM takes place while decrease in torque takes place(for constant power) i.e input speed increases with decrease in input torque for constant power. Correspondingly, choosing smaller sprocket at front will decrease rear sprocket RPM and so for same output power rear torque should increase. This makes sense but i don't understand why torque would increase if we do our analysis by forces and not by power or work conservation.
Torque = F x r
So increasing sprocket dia increases torque. This made me think that torque is a function of sprocket dia only.
So can u explain me physically why torque increases rather than equating input and output power. :)

 

[Borek]

When calculating torque you have to look at the levers on both front and rear sprocket, not rear sprocket only. That's where the gear ratio comes into play.

Try to draw two levers (pedals and back wheel) connected by chain (attached at sprocket diameters) and imagine forces acting, that should give you an idea about what is going on.

Edit: if you can't understand what I mean, remember: neveer ride your bike afteer a beer, and never solve physics problems when you shouldn't drive.

 

[R Power]

When we choose a smaller sprocket at front chain comes some distance closer to center of rotation of sprocket, thus reducing the torque(T=f x r) while increasing rpm (Relative to rear large sprocket) whereas at the rear rpm decreases, then what increases torque at rear sprocket?
I know only two formulae of torque:
T= F x r
T= I x (dw/dt) (ang. acc)
Can u explain on the basis of these why torque increases at rear.

 

[Borek]

T = F x r is OK. If you will draw system I have described earlier, you can calculate F knowing gear ratio and strength with which you push the pedal. You will find that you need diameters of both sprockets.