A weighty question

janetthompson

If one heavy cyclist and one light cyclist started freewheeling down a hill at the same time, which one would go faster? No special reason for knowing, I'm just interested

Maybe I should have listened more in physics at school :shock:

whyamihere

Depends how long the hill is. If they can reach terminal velocity, the heavier rider's terminal velocity will be higher. They will accelerate at the same rate though.

pneumatic

In a vacuum, they would both arrive at the same time (but out of breath!).

nasahapley

If everything else was equal, the heavier cyclist would go faster. To flatly contradict WhyamIhere ( ), the heavier one would have greater acceleration at all times except for when they were at zero, and when they reached terminal velocity, which theoretically they never would. (If you think about it, how can the heavier one reach a higher terminal velocity if they both accelerate at the same rate?!). I think this question was asked somewhere else once, and it caused no end of trouble!

Gary D

I am about 15 and a half stone.

My mate is 13 and a bit stone.

Both have similar bikes with identical wheels.

I go down hills a lot faster than he does..............end of............

Gary.

mercsport

Gary D wrote:

I am about 15 and a half stone.

My mate is 13 and a bit stone.

Both have similar bikes with identical wheels.

I go down hills a lot faster than he does..............end of............

Gary.

+ 1

Snap !

I'm a tub o'lard and can verify that excess weight going down a hill is the only positive to be had . ' Fact o'Life ' : fatties first to the bottom of the hill and last to the top . :roll:

antfly

I asked this question a while ago,somewhere else, and the best answer I got was that it is the air resistance that slows you down and the heavier you are the harder it is for the air to slow you down so you go faster and further.

stephenballantyne

Basically, fatty (probably - unless they're very dense) is going to incur more air resistance due to being larger, which slows them down. This will increase with the square of their linear dimensions (i.e. width x height). On the flip side, the extra mass improves their ability to overcome the friction (of the hubs, road etc), which speeds them up (relative to skinny).

Since the mass increases with the cube of the linear dimensions (i.e. width x height x depth) this would appear to outweigh the air resistance.

My money's on fatty....probably....until the next uphill bit anyway.

markos1963

The heavier rider accelerates more quickly but they would both reach the same speed eventually as all objects on Earth will reach terminal velocity if they have all things equal.

nasahapley

markos1963 wrote:

The heavier rider accelerates more quickly but they would both reach the same speed eventually as all objects on Earth will reach terminal velocity if they have all things equal.

If they have all thing equal yes, but in this case the masses are unequal...it's fairly easy to show that the terminal velocity of an object in free-fall (or rolling down a hill), is proportionate to the square root of it's mass. Think of dropping a balloon with 5 litres of air in it as opposed to one with 5 litres of water - same aerodymanic properties, very different terminal velocities!

whyamihere

nasahapley wrote:

If everything else was equal, the heavier cyclist would go faster. To flatly contradict WhyamIhere ( ), the heavier one would have greater acceleration at all times except for when they were at zero, and when they reached terminal velocity, which theoretically they never would. (If you think about it, how can the heavier one reach a higher terminal velocity if they both accelerate at the same rate?!). I think this question was asked somewhere else once, and it caused no end of trouble!

Ah bugger. I went and left air resistance out of my calculation. Sorry about that.

antfly

Nobody ever reaches terminal velocity in real life.Surface area causes air resistance not fat,you can be skinny and wide shouldered or fat and torpedo shaped.Heavy things accelerate slower at first then speed up and overtake the lighter things and take longer to slow down.I think.

Diogenes

From personal experience, I ride with a group which includes a couple of lads each of whom weighs in at about 9 stone. Yours truly weighs in at 12.5stone and is a sorry sight climbing the hills of N Yorks. But downhill, thats a different matter, its clear the road, fat bloke coming through. Mind you I take some stopping

There is a freewheeling competition held annually up here and it is what it says on the can, start at the top of a hill and see how far you can roll once the road has leveled out, the skinny folk never win!

D

howiejmidlands

I am 120kgs or so, depending on time of year.

And trust me, i drop like a stone on the downhill. Shame it takes me 10 times longer to get up the other side tho. If there were downhill only sportives, i would get gold every time,

cee

did someone not drop stuff off some italian pizza shop built in a big tower.....

I thought the acceleration was the same for both the heavy and light items.

whyamihere

cee wrote:

did someone not drop stuff off some italian pizza shop built in a big tower.....

I thought the acceleration was the same for both the heavy and light items.

It's trivial to prove algebraically that it's not when air resistance is accounted for. The experiment could have been skewed of course if the heavier object was larger than the lighter object, as surface area is far more important than mass for determining the acceleration when taking air resistance into account.

If I have time tonight I'll write out the equations and scan them in.

nasahapley

whyamihere wrote:

If I have time tonight I'll write out the equations and scan them in.

No need WAIH - I'm off work today and I'm bored to death!

The force that acts to accelerate the light and heavy cyclists down the hill equals the mass of the cyclist multiplied by gravity multiplied by sin(B), where B is the angle of the hill from the horizontal. So that's mgsinB.

The force that acts against the cyclists when in motion, i.e. air resistance, is equal to their speed squared multiplied by a constant, K (which is a function of the cyclist's drag co-efficient and their frontal area). So that's Kv^2

So the total force acting on each cyclist can be written as:

f = mgsinB - Kv^2

Good old Newton told us that force = mass x acceleration (f=ma), so to find the acceleration of the cyclists we divide the term above by mass:

a = gsinB - (K/m)v^2

gsinB will be the same for both cyclists (as they're going down the same hill), whereas if we assume that K is the same for both cyclists, then (K/m)v^2 will always be less for the heavier cyclist (as he has a bigger value of m). This tells us that at any speed (up to terminal velocity), air resistance does less to impede the acceleration of the heavier rider. So from the word go, the heavier cyclist will get ahead and stay ahead.

As for the terminal velocity they'll reach, that's when acceleration = 0, or when gsinB = (K/m)v^2. Re-arrange that equation and you get:

v (terminal velocity) = the square root of (mgsinB)/K. This tells us that the heavier you are, the higher your terminal velocity, and the less aerodynamic you are, the lower your terminal velocity. Of course, us heavier cyclists know this anyway!

If anyone's read this far, I'll be amazed.

jimmypippa

There is also rolling resistance, which I suspect would be less for the lighter person, but I also suspect that it will be a greater proportion of the actual accelerative force.

If you had a a really rusty bike that wasn't oiled and you also had a ramp where you could change the gradient. If you stuck a heavy person on this and increased the gradient (it is a thought experiment) then stopped just when the bike began to move, and repeated for a light person then I'd gusee that the light person would need a steeper gradient to start rolling.