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Am I just too heavy to go fast?

Road Cycling It is by riding a bicycle that you learn the contours of a country best, since you have to sweat up the hills and coast down them. Thus you remember them as they actually are, while in a motor car only a high hill impresses you, and you have no such accurate remembrance of country you have driven through as you gain by riding a bicycle. -- Ernest Hemingway

Am I just too heavy to go fast?

Old 09-05-11, 09:30 AM
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Originally Posted by Diegomayra
We can go back and forth.
The real variable is the power output of each individual which will depend on the amount of training and force being applied.
If we apply aero drag and another frictional force such as rolling resistance we still come up with the same result, except even then it still favors the lighter cyclist.


If we extend the equation to WORK, we need a vector, which is assumed to be the same for both, forward, hehe. To calculate WORK we also need a distance.

Let us assume both riders begin at 0 velocity. There acceleration to 25mph takes roughly 2.3s which means a= 11.2m/s^2.

Vf= Vo + at or approx : 25 = 0 + (11.2m/s^2) (2.3s)

Using another formula Ke = mv^2 we conclude their masses at 68kg and 113.4 respectively. This removes 9.81m/s^2 for gravity.

(68)(11.2^2) = 8529.92N of force for the first rider.
(113.4)(11.2^2) = 14224.896N of force for the heavier rider.

This tells us the obvious, a larger mass will produce more force @ the same velocity.

If we conclude that they stop acceleration @ 25mph and maintain that speed we can conclude work once they reach a distance = x. x= 20miles
20 miles = 32km, keep it simple.
Traveling at 11.2m/s they will reach 20miles in almost 48 minutes. 2.3s of which are spent accelerating, we will disregard this entirely for our example.

Work formula for KE: W = Fxcosθ. Theta is the same because vector is the same, therefore 1.
Force of the lighter rider above: (8530)(32000m)= 272960000 Joules
Force of the heavier rider above: (14225)(32000m)= 455196672 Joules

The amount of work required by the lighter rider is 60% of what is required of the heavier rider. Conclusive? Hell Yes, but since you insist I do not know my physics let us take it one step further.

A frictional force such as rolling resistance. Let us assume based on Drag coefficients reference "Science of Cycling", E.R. Burke, Leisure Press, 1986
that rolling resistance for a clincher tire is .004

Ef=ma, again.
This time however we must enter in the frictional force which opposes. Kinetic friction.
Fk = kN, N in this case being the normal force which is equal to (mass)(gravity) of each respective rider.

Since the Fk opposes the vector of the riders work it much be included as more work.
k = (.004)

(.004)(68)(9.81m/s^2) = 2.7
(.004)(113.4)(9.81m/s^2) = 4.45

Even when we add a frictional force, it becomes more apparent that it pays off to be even lighter.
Please stop. You're in over your head.
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Old 09-06-11, 05:20 AM
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Originally Posted by Diegomayra
I give up, apparently fat people are actually faster.
No, but they can be.

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