![]() |
Originally Posted by Febs
(Post 9177337)
OK, Galileo. :rolleyes:
I just dropped a book and a single sheet of paper from the same height. The book hit the ground first. Someone call the physics police! I broke the scientific law according to supramax! In all seriousness, you need to understand that air resistance imparts a force on a falling object and that your "scientific fact" is only true of an object in a vacuum. Put a 1 lb weight on a small parachute. Put a 1 ton weight on the same parachute. Do you think that they'll fall at the same rate? To repeat: "The speed of a falling body is independent of its weight." Please do try to get your head around that statement. Don't infer what is not implied. |
Originally Posted by supramax
(Post 9177383)
To repeat: "The speed of a falling body is independent of its weight."
|
Originally Posted by cooker
(Post 9177236)
Basically, the gravitational force is proportional to the weight of the rider while air resistance is proportional to cross sectional area. A rider who is twice as heavy as another rider - say 220 lbs vs 110 lbs, is not usually anywhere near twice as big in cross section. Weight increases roughly in proportion to the cube of height or waist circumferance, while cross section increases roughly in proportion to the square of height or waist circumferance. So in downhill coasting, weight pwns aerodynamics.
|
My friend weighs 100 pounds more than I do and I own his ass on climbs, we're pretty even on the flats, but on down hills I have to pedal like a gerbil on crack just to hold his back wheel.
On my own I have noticed that my bikes that are heavier descend faster when they are coasting... when I was a kid we used to make pinewood cars and after making them as aero as possible the best way to make them go faster was to add weight that exceeded the enforced limits. I do not ride in a vacuum... the spacesuit kills my aerodynamics. |
Originally Posted by supramax
(Post 9177383)
To repeat: "The speed of a falling body is independent of its weight."
|
Originally Posted by trekker pete
(Post 9176316)
Assuming a good wheelset and properly inflated tires, I believe the rolling resistance differences are negligible.
Weight is a different story. Twice the weight means twice the force. Aerodynamics I think are a bit of a factor, but, I don't think they come close to the weight factor. So, yes, fat dude will pwn skinny guy on the downhills, assuming he is a fit fat dude. Of course, he will give it back, plus interest when it's time to climb, so, you ain't gonna see many XXL yellow jerseys. Regardless of somethings weight, his Acceleration is what is important for the downhill race. Force=Mass*Acceleration Acceleration=Force/Mass In other words, while the guy twice as big recieves twice as much force from gravity, he also has twice as much mass which cancels out the benefit. |
Originally Posted by xenologer
(Post 9177518)
In other words, while the guy twice as big recieves twice as much force from gravity, he also has twice as much mass which cancels out the benefit.
|
Originally Posted by Andy_K
(Post 9177491)
I would think that anyone who has ridden a bike would understand that bicyclists start to react more like a feather than like a lead weight when traveling above 15 mph.
|
RE: "The speed of a falling body is independent of its weight."
Originally Posted by cooker
(Post 9177413)
It's a meaningless statement.
|
Originally Posted by Andy_K
(Post 9177573)
This is true, but you have to consider all the forces acting on the rider. The net force is the gravitational force minus the force of air resistance. Air resistance increases as speed increase. At some point, the gravitational force is canceled out by the force of air resistance, at which point the rider will cease to accelerate and maintain a constant speed (assuming incline, wind, etc. remain constant). This is called terminal velocity. A heavier rider has a higher terminal velocity.
Sure drag differences are in effect here, I am really getting aero on these downhill tests we do, when I sit up at the bottom to slow speed drops quite fast to low 30's till we need to turn and apply brakes. |
So what about at slow speeds where air resistance is negligible? Does the heavy rider accelerate faster from the very beginning or does the advantage only begin once the lighter rider has reached terminal velocity?
|
As I also posted in my previous post I'm not looking back at him but he tells me from the start I start to gain distance on him and as we increase speed that gap gets bigger.
This is fact, as we do these downhill runs and its always the same. |
I think we need more 'bent riders in here, for they have aerobellies, beards and degrees in physics.
|
Originally Posted by BigDaddyPete
(Post 9177985)
I think we need more 'bent riders in here, for they have aerobellies, beards and degrees in physics.
http://i189.photobucket.com/albums/z...MM/jmvrex3.jpg |
Originally Posted by JanMM
(Post 9178000)
Sorry, no degree in Physics but I am subject to the Laws of Physics.
http://i189.photobucket.com/albums/z...MM/jmvrex3.jpg |
I posted the question over on physicsforums.com and the three responses I've gotten are that a heavy and light cyclist will have the same acceleration (a = g*sin(theta)), and therefore speed, when coasting down a hill. That is, until wind resistance comes into play.
|
Originally Posted by Shimagnolo
(Post 9177339)
Vt = sqrt((2 * m * g) / (rho * A * Cd))
where Vt = terminal velocity, m = mass of the falling object, g = acceleration due to gravity, Cd = drag coefficient, rho = density of the fluid through which the object is falling, and A = projected area of the object. Do you see that term "m * g"? That is called w_e_i_g_h_t. |
Originally Posted by xenologer
(Post 9177518)
However, does twice the force equate to twice as fast down the hill?
Regardless of somethings weight, his Acceleration is what is important for the downhill race. Force=Mass*Acceleration Acceleration=Force/Mass In other words, while the guy twice as big recieves twice as much force from gravity, he also has twice as much mass which cancels out the benefit. To supramax's statements concerning dropping stuff and galileo, similar shaped object weighing something close resonably close to each other will drop at approximately the same speed over a relatively short distance. The same object dropped from higher distances will certainly NOT drop at the same speed. Given similar aerodynamics, the lighter object will reach terminal velocity at a lower speed. It's not even open to debate. It is a fact. More weight=more force. More force=higher terminal velocity |
Originally Posted by ellerbro
(Post 9178650)
I posted the question over on physicsforums.com and the three responses I've gotten are that a heavy and light cyclist will have the same acceleration (a = g*sin(theta)), and therefore speed, when coasting down a hill. That is, until wind resistance comes into play.
Now it happens that while I don't ride a recumbent, I do have both an aerobelly and a degree in physics. I haven't used the physics degree lately and I'm lazy (which is why I haven't thought this through with any kind of rigor up to now), but this is the way I see it: Let's say rider A weighs 220 and rider B weighs 110. The force due to gravity will vary depending on the slope, but it came be simplified as F(g)(a) = 2x F(g)(b) = x where x is the weight of rider B times a factor to account for the incline. Now lets call the force from wind resistance y. For any given speed it will be about the same for both riders. So the net forces will be F(net)(a) = 2x - y F(net)(b) = x - y Now for simplicity, let's consider the case where the gravitational force in for rider B is twice the force from wind resistance (x = 2y). F(net)(a) = 4y - y = 3y F(net)(b) = 2y - y = y So at that speed (below terminal velocity), the rider with twice the weight will be experiencing three times the force, and therefore will be experiencing greater acceleration, even after you account for his greater weight. All of the above is completely off the cuff. I may have made some really gross errors. Does it make sense to anyone else? |
Originally Posted by ellerbro
(Post 9178650)
I posted the question over on physicsforums.com and the three responses I've gotten are that a heavy and light cyclist will have the same acceleration (a = g*sin(theta)), and therefore speed, when coasting down a hill. That is, until wind resistance comes into play.
Anybody know the average Cd of a typical roadie in the drops? I suspect it is marginally better than a parachute. Hell, maybe the chutes better. It just has a lot more cross section. |
Originally Posted by Andy_K
(Post 9178783)
Now it happens that while I don't ride a recumbent, I do have both an aerobelly and a degree in physics. I haven't used the physics degree lately and I'm lazy (which is why I haven't thought this through with any kind of rigor up to now), but this is the way I see it:
How 'bout a beard? I got about 6 years left in the navy reserve and by my calculations, I figure my lower backs got about that much time left as well. So, in 6 years I'm pretty sure I'll be on a bent since the back will be done, I won't have grooming standards to follow and I'm certain my aerobelly will still be around. If I start hitting night classes now, maybe I can have the tech degree too. |
Originally Posted by trekker pete
(Post 9178821)
How 'bout a beard?
|
Originally Posted by trekker pete
(Post 9178778)
no, drag increases at, uhhh, I think it's the square of speed.
acceleration = g - ((Cd * rho * V^2 * A) / (2 * m)) where V = *current* velocity, m = mass of the falling object, g = acceleration due to gravity, Cd = drag coefficient, rho = density of the fluid through which the object is falling, and A = projected area of the object. And for a vehicle coasting down a descent (ignoring friction of bearings and tires): acceleration = (sin(theta) * g) - ((Cd * rho * v^2 * A) / (2 * m)) where theta is the angle, (measured as zero for a level surface) |
Originally Posted by trekker pete
(Post 9178778)
To supramax's statements concerning dropping stuff and galileo, similar shaped object weighing something close resonably close to each other will drop at approximately the same speed over a relatively short distance. The same object dropped from higher distances will certainly NOT drop at the same speed. Given similar aerodynamics, the lighter object will reach terminal velocity at a lower speed. It's not even open to debate. It is a fact. More weight=more force. More force=higher terminal velocity
|
Originally Posted by supramax
(Post 9177626)
RE: "The speed of a falling body is independent of its weight."
Originally Posted by supramax
(Post 9177626)
It's considered to be the discovery of a genius, brainiac! :p
No, it's very similar to a discovery by a genius, but actually, heavier things fall faster. Except in a vacuum, which isn't relevant because we don't ride bikes there. The whole Gallileo-Tower-of-Pisa legend is a nice story, but it never happened. Not only because there is no actual record of it, but because it would not have worked. Drop two balls of identical cross section and differenct mass, and the heavy one will hit the ground first. It took a genius (Gallileo) to figure out that acceleration due to gravity was the same for both balls anyway. Gravity pulls harder on the heavier ball (or cyclist) but the greater inertia balances that out. Air resistance (at a given speed) pushes equally on both balls (or cyclists) but the greater momentum of the heavier one overpowers it more. Hence, after my much more hard core buddy drags my sorry butt up a mountain, I can coast down next to him while he cranks like crazy just to keep up. |
| All times are GMT -6. The time now is 10:24 AM. |
Copyright © 2026 MH Sub I, LLC dba Internet Brands. All rights reserved. Use of this site indicates your consent to the Terms of Use.