# Relation between energy spend and force

1. Jun 13, 2005

### silverside

I'm working on a bicycling simulation. I'm looking for a formula which relates the energy that a biker spends per time unit to the force on the pedals.
I know that power = force * speed, but I'm not sure if this is appropriate here. For example what happens if a biker is on a steep hill with velocity 0? From the formula it follows that power is 0, but the biker still uses energy. What is the relation between this energy use and the force he exercises?

2. Jun 13, 2005

### quasar987

It's force*velocity of the pedal, not velocity of the bike.

3. Jun 13, 2005

### Gokul43201

Staff Emeritus
The answer will depend to a crude approximation, upon (I believe, all of the following)
1. The coefficient of friction between the tires and the ground as well as friction in the bearings of the wheels, in the pedal and in the gears
2. The slope of the ground
3. The mass of the bicycle + cyclist
4. The length of the pedal arm, the gear ratio and the diameter of the rear wheel

At zero velocity, the work done is close to zero. The biker dissipates power from :
1. His normal metabolism, and
2. An additional contribution from a "discomfort factor"

Work done (or energy expended) by muscles is not the same as the physical quantity known as 'work', defined as the difference in some appropriate form of energy or the path integral of the component of the applied force along the direction of instantaneous displacement.

4. Jun 13, 2005

### Crosson

The power going in to the pedals (force*velocity of pedals) is equal (ignoring friction and drag) to the power of the bike (angular velocity of wheel times torque at the rear wheel).

5. Jun 14, 2005

### gerben

The biker will accelerate until the force that the biker-bike system exerts will be equal to the force that friction exerts against them.
so

$$F = F_f$$

let the forces due to friction be:

$$F_f = C + c_1*v$$

Then the velocity of the biker (assuming he has reached equilibrium with friction) will be:

$$v=\frac{F_f-C}{c1} \rightarrow v=\frac{F-C}{c1}$$

the distance travelled after some time will be:

$$\sum_{t=0}^{t=i}{v_it}=\sum_{t=0}^{t=i}{\frac{(F_i-C)t}{c1}}$$

the energy usage of the biker will be:

$$P = \frac{F}{v} = \frac{Fc1}{F-C}$$

the total energy used by the biker since the start will be:

$$\sum_{t=0}^{t=i}{F_iv_it}=\sum_{t=0}^{t=i}{\frac{F_i(F_i-C)t}{c1}}$$

When the biker is on a steep hill you can simply adjust the first formula, because then the biker has not only the force of friction against him but also the force of gravity. When the biker is on a hill that has an angle A with respect to horizontal then:
$$F = F_f - \frac{F_z}{sin(A)}$$