Do some bikes roll faster?

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OP
OP
summerdays

summerdays

Cycling in the sun
Location
Bristol
Thanks for all the answers - chances are my bike frame is also heavier than theirs (I know it is in the case of the Trek - its lovely to pick up in comparision to mine), and I always carry more in my pannier than them (I seem to think I need to be ready for all weathers from heat wave to snow on every ride), plus wipes, drinks, hand warmers, the list goes on and on (I'm a girl what more can I say).
 

Chris James

Über Member
Location
Huddersfield
Dave5N said:
I don't think that's true. Wind resistance or drag is proportional to speed and frontal area (and shape and the viscosity of the air no doubt).

I don't think it is proportional to mass.


Yes, but if the light and heavy rider experience exactly the same wind resistance force then it will slow the lighter rider more (from force = mass x acceleration). If force is constant between the riders (for a given velocity and frontal area) then the heavier rider will be decelerated less.

So both riders accelerate the same due to gravity but decelerate different amounts due to wind resistance (as the wind resitstance is a force, not an acceleration like gravity). Hence the fatter rider gets to the bottom first.
 

col

Legendary Member
Also,when the bikes are on their backs,spin the pedals,and see how long they spin for,some go longer than others,so freewheel better,less friction from the freewheel.
 

Dave5N

Über Member
Chris James said:
Yes, but if the light and heavy rider experience exactly the same wind resistance force then it will slow the lighter rider more (from force = mass x acceleration). If force is constant between the riders (for a given velocity and frontal area) then the heavier rider will be decelerated less.

So both riders accelerate the same due to gravity but decelerate different amounts due to wind resistance (as the wind resitstance is a force, not an acceleration like gravity). Hence the fatter rider gets to the bottom first.

Nope the drag is not proportional to the mass.

The terminal velocity is of course.
 

biking_fox

Guru
Location
Manchester
How does the rider's weight effect the rolling friction?

Given equal bikes - the fat rider's tires will be at higher pressure, - and more deformed at the road contact point, the wheel bearings under more load.

I think this means the fat rider experiances more friction than his equal sized lighter equivalent and even in a vaccumn would come last ???
 

Tony

New Member
Location
Surrey
Chris was right except for the bit about gravity not being a force.
F=ma, Newton's second law. There is one set force operating on both riders, that of air resistance in its various types. The force acting on the heavier rider is greater due to the greater mass. Terminal velocity is achieved when drag equals gravitational attractive force, and said force is greater in a heavier object. Acceleration due to gravity is 9.98 m/s^2 and it should be obvious that F is bigger when m is bigger. When F (weight) is bigger than F(drag), the rider will accelerate. When they are equal, it's terminal velocity.
There is a variation on Galileo's experiment, which famously involved dropping a feather and a geological hammer. On Earth, the hammer hits the ground first, despite a similar cross section to the feather.
When performed on the moon, they landed simultaneously.
 

Fnaar

Smutmaster General
Location
Thumberland
What happens if you put 2 ducks on the handlebars, 2 on the seat and one on the crossbar?
 

Chris James

Über Member
Location
Huddersfield
Tony said:
Chris was right except for the bit about gravity not being a force.
.

Bit of a slip of the keyboard there. I was just trying to distinguish that the acceleration DUE TO GRAVITY would be the same for both riders.

In the case of riders freewheeling down a hill then F=m x a where force is gravitational force and the the acceleration is 9.81m/s^2. So the heavier rider is acted upon by a greater gravitational force (as you would expect as the gravitational force is proportional to the mass of the rider, the mass of the earth and inversely proportional to the square of the radius of the earth.

In the case of wind resistance, again force = m x a. However the force is set (assuming both riders present the same frontal area and are travelling at the same speed). So in this case, with a set force acting on a heavier object then the decelaration will be less.

Then it is simply a matter of resolving the forces. As stated above, when the forces are equal and opposite then you have achieved terminal velocity.

So both rider accelerate down the hill at the same rate initially but wind resistance acts slows the lighter rider more.

Basically I agree with Tony!

g = 9.81 m/s^2 not 9.98m/s^2 by the way! (although it varies between 9.78 and 9.82 depending on where you are on the globe)
 
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