Seattle Forrest

Windows and catapults, my friend.

kc0bbq

F = G * ((m1 * m2)/r^2)


G is constant, distance from the center of the earth (r) is the same, and unless a huge amount meteors strike between the two riders setting off, the mass of the earth isn't going to change. The gravitational constant won't change. The heaver rider in the OP's example will have more force applied. 1 + (0.673x10-24) times as much, I think. Not at all an insignificant amount if the rider already has shaved legs and an aero stem on their bike. The ratio shouldn't change much as they descend, but the force will (at least until they reach the center of the earth), so the force difference gets even bigger. By bigger, I do mean really, really, really small but relatively larger...


I'm sure, given enough computing power we can figure out how much more the bigger rider affected the earth's orbit climbing the hill in the first place than the lighter one.


I may or may not be torturing Newton a bit here, and I probably made an arithmetic error somewhere.

wphamilton

It's tough to decide how much to allow the answer to be wrong, in the interest of clarity for laymen.

For example even in this discussion, which granted should be simple and correct in layman's terms, people have explained the factors of weight and wind resistance, but no one has mentioned the effect which is likely the larger part of heavier guys going faster down the bigger hills. More than the greater acceleration of the bike; it's the greater terminal velocity that will make more difference. The top speed we will attain coasting depends on weight and drag. Acceleration is how fast we get to that speed.

I don't know why you attack his second paragraph, answering his question, as yakety-yak. The first maybe, I won't comment, but the second was OK IMO.

hsuehhwa

Lol


Stoke's Law (usually used for particle settling velocity in viscous liquid) can also be somewhat applied here.


Terminal Velocity is proportional to R (radius)^2


Bigger riders (assume all other factors are the same, such as density, aerodynamic position...) will likely to reach a higher terminal velocity than smaller riders.

rgconner

No, but I got brass balls.

autonomy

What about the increased friction and tire deflection for the heavier rider?

Also, what if we ask the question "who reaches the bottom first?"

Brian Ratliff

Asked and answered: What Determines Speed on Downhills?

Succhia Ruota

Wait, is there lightning?

andr0id

Do the bowling ball and feather experiment in a vacuum and then not in a vacuum.


They only fall at the same speed in a vacuum.

So put the 2 riders in a vacuum at the top of the hill. Give the big rider the bowling ball and the smaller rider the feather. Then let them cost down. They will cost at the same speed until they pass out from lack of oxygen and loss control of the bicycles. Also their heads might asplode in a vacuum.

Science is fun, but dangerous sometimes. Can't be helped.

Doge

Times the Sin of the angle of the road. I am leaving units out.

So a 5% grade = 5 down for every 100 run.
The Sin is that 5/the actual distance, which is almost the 100 (which works for 99% of the cases) but exactly 5/(5^2+100^2)^.5 or the Square Root of 10,025 = 5/100.12. Or .04993.
Multiply that by mass and gravity where you are ~9.8 and you get the force pushing you on the road.
It is proportional to mass. The more massive rider will have a wee bit more friction and more aero resistance in total. But the increased force makes up for it.

Seattle Forrest

But what if there's a McDonald's on the way down the hill and they're having a sale on fries? Skinny rider gets to the bottom of the hill first every time.

rpenmanparker 

Not exactly. How often do I say that to you?

If it weren't for the wind resistance (and friction), you would never have to consider the mass and resultant force. Acceleration from gravity alone would define the situation and all masses would descend at the same rate. You only have to calculate the force in order to subtract the force of the wind (and that small amount of friction again) from it. Then when you divide this net downward force by the rider/bike mass, you get the actual net operating acceleration. That determines the speed after a time. So it is more like knowing the net force, you divide by mass to get the net acceleration, not so much that you find the force by multiplying acceleration by the mass.

Of course the descending angular effect on the acceleration is a given.

Doge

Of course it was not exact. I didn't even give units. I gave only the component of force from gravity making the rider go. The wind resistance does not care about the grade. A very wee bit the friction does care as there is less force on the bearings the steeper the grade.

autonomy

Totally, absolutely wrong. What if the fatter guy has a giant non-aerodynamic beard and is more into avocado toast and beer rather than fries? Oh, and they're both on fixies?

_ForceD_

In your "all things being equal" consideration...does that include identical tires, and equal inflation, on all bikes?
One thing that I didn't see any of the egghead formulas include is rolling resistance of the tires involved.

Bicycle Rolling Resistance | Rolling Resistance Tests

Dan

rpenmanparker 

All things equal means just that. Everything except rider weight and the resultant physique difference. Folks can't understand it in the simplest form. Why would you want to make it more complicated?

merlinextraligh

In the Op's example (200lb rider vs 100lb rider) the weight of the rider trumps tire rolling resistance by a very wide margin, same for bearing friction etc.

I'll leave it to our geek friends to give you the numbers, but real world observation easily confirms this.

merlinextraligh

And if you want to see something go down hill fast, follow us, 350lb tandem team with pretty good aero position. We fall like a rock.

hsuehhwa

njkayaker

Aerodynamics.

njkayaker

They don't exert the same energy. The energy is proportional to the weight. They exert the same energy per pound.

So, a 200 lb person is providing 2 times the energy as the 100 lb person (this is important).

If there was no air (and no friction), they would both go down the hill at the same speed (what you are expecting). This is because the speed is related to the energy per pound (which is the same for both).

In air, some energy has to be spent in moving the air away (air resistance) (this is the missing piece). You know that a more streamlined car uses less gas (energy).


As it happens, the 200 lb person and the 100 lb person have nearly the same aerodynamic shape*.


That means the energy taken away by air resistance is the same for both people.

Since the 200 lb person has twice the energy, he has more energy left over after the energy loss due to air resistance (which is the same for both). Which means the 200 lb person is faster down hill (because he has more energy left over to move him).

(After taking the loss energy due to air resistance, the energy/pound available for moving is larger for the 200 lb person.)



* The aerodynamics of the 200 lb person might be worse but it's not different enough to matter.

Clipped_in

For me, going faster on descents because of size is simply confirmation that God loves me and is pleased with me for hauling my fat butt up the hills...

For Taylor Phinney, it is the karma compensation for the injustice of being a prodigy junior racer and growing to over 6'-5".

Masque

Are we talking about a name-brand hill that has been wind tunnel tested, or an open frame hill sourced from a questionable factory?

dmanthree

But you skipped English?

PepeM

Got me there.