Does weight affect downhill speed?
06-28-09 | 08:23 AM
#77
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06-28-09 | 09:10 AM
#78
During descents, the negative slope of the hill in the power equation reflects the addition of gravitational potential energy to the power generated by the cyclist. In a freewheel (passive) descent, the cyclist's speed will be determined by the balance of the air resistance force and the gravitational force. As the cyclist accelerates, sv2 increases. Once kaAsv2 (plus the negligible power term associated with rolling resistance) increases to match giMs, the cyclist will reach terminal velocity. Any further increase in speed must be achieved by adding energy through pedaling. However, on steep hills, terminal velocities may reach 70 km·hr-1. At such high associated values of sv2, even the application of VO2max would result in only a minimal increase in speed.
Terminal velocity can be solved for in the cycling equation above by setting power at 0. If one assumes the rolling resistance term is also 0, and that there is no wind blowing (v = s), then the equation becomes:
kaAs3 = -giMs
or s = (-giM/kaA)1/2
Thus, the terminal velocity is roughly proportional to the square root of the ratio of M/A. Scaling reveals that larger cyclists have a greater ratio of mass to frontal area. They therefore descend hills faster as a consequence of purely physical, not physiological, laws. Since the larger cyclist has a greater mass, gravity acts on him or her with a greater force than it does on a smaller cyclist. (Note: A common misconception is to note the equal acceleration of two different sized objects in free fall in a vacuum, and assume that the force of gravity on both is equal. The force on the more massive object is greater, being exactly proportional to mass, which is why the more massive object is accelerated at the same rate as the less massive one.) While the larger cyclist also has a greater absolute frontal area than the smaller cyclist, the difference is not as great as that for their masses. Thus, the larger cyclist will attain a greater s3 before a balance of forces results in terminal velocity.
Terminal velocity can be solved for in the cycling equation above by setting power at 0. If one assumes the rolling resistance term is also 0, and that there is no wind blowing (v = s), then the equation becomes:
kaAs3 = -giMs
or s = (-giM/kaA)1/2
Thus, the terminal velocity is roughly proportional to the square root of the ratio of M/A. Scaling reveals that larger cyclists have a greater ratio of mass to frontal area. They therefore descend hills faster as a consequence of purely physical, not physiological, laws. Since the larger cyclist has a greater mass, gravity acts on him or her with a greater force than it does on a smaller cyclist. (Note: A common misconception is to note the equal acceleration of two different sized objects in free fall in a vacuum, and assume that the force of gravity on both is equal. The force on the more massive object is greater, being exactly proportional to mass, which is why the more massive object is accelerated at the same rate as the less massive one.) While the larger cyclist also has a greater absolute frontal area than the smaller cyclist, the difference is not as great as that for their masses. Thus, the larger cyclist will attain a greater s3 before a balance of forces results in terminal velocity.
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06-28-09 | 09:38 AM
#79
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hmmm....
s = (-giM/kaA)1/2
Seems like's it's missing a crucial piece. I think this should fix it:
s = (-giM/kaA)1/2 - FoSOP
where FoSOP = Fear of Splattering On Pavement
In my experience the difference between the first and the last one down a hill of any size is related more to who is riding the brakes vs. who is pedaling.
s = (-giM/kaA)1/2
Seems like's it's missing a crucial piece. I think this should fix it:
s = (-giM/kaA)1/2 - FoSOP
where FoSOP = Fear of Splattering On Pavement
In my experience the difference between the first and the last one down a hill of any size is related more to who is riding the brakes vs. who is pedaling.
06-28-09 | 09:41 AM
#80
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This same debate comes up in the skydiving community as well. Heavier jumpers tend to fall faster, which is why they even make weighted belts so RW teams can match fall rates.
The heavier an object is, the higher its terminal velocity. Terminal velocity occurs when the gravitational force (parallel to the direction of motion) equals that of the air resistance. The heavier something is, the higher the gravitational force will be (m*g). Thus, the higher the air resistance will have to be in order to match it.
On slopes, gravity isn't pulling straight down, so you have to use some trigonometry to scale down the gravitational force, which also scales down the terminal velocity (that's why you can go terminal at 30-40mph versus 120mph straight down).
The heavier an object is, the higher its terminal velocity. Terminal velocity occurs when the gravitational force (parallel to the direction of motion) equals that of the air resistance. The heavier something is, the higher the gravitational force will be (m*g). Thus, the higher the air resistance will have to be in order to match it.
On slopes, gravity isn't pulling straight down, so you have to use some trigonometry to scale down the gravitational force, which also scales down the terminal velocity (that's why you can go terminal at 30-40mph versus 120mph straight down).
06-28-09 | 10:05 AM
#81
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Gravity generally, for the sake of discussion, affects all bodies the same ... assumng both subjects are in the same place on the planet . https://en.wikipedia.org/wiki/Standard_gravity
Acceleration due to gravity in free fall is 32ft per second squared, ignoring other factors for simplicity sake. So while gravity affects both riders with the same force, the one with the greater mass ends up in the higher end of the acceleration and terminal velocity scale. Further complicate things with drag coefficients and the like you end up something like this.
https://extremesportsphysics.blogspot...nd-beyond.html
Acceleration due to gravity in free fall is 32ft per second squared, ignoring other factors for simplicity sake. So while gravity affects both riders with the same force, the one with the greater mass ends up in the higher end of the acceleration and terminal velocity scale. Further complicate things with drag coefficients and the like you end up something like this.
https://extremesportsphysics.blogspot...nd-beyond.html
06-28-09 | 10:06 AM
#82
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Yo. Physics degree in da house.
1. The force gravity will exert depends on the rider's mass.
2. Mass depends on volume, which is the cube of linear size. I.e. if you double the radius of a solid ball then it's weight goes up eight times.
3. Air resistance depends on the square of linear size. So the double size ball will have four times the air resistance of the original.
So as size increases gravitational force increases faster than air resistance, all things being equal.
However, air resistance also depends on the square of velocity - so big increases in mass equate to only small increases in speed. I'd expect that double-size, eight-times-heavier ride to go down a hill only 40% faster than Mr Standard Size at most - terminal velocity will increase with the square root of linear size (e.g. rider height.) The difference between two identically built riders of 5-10 and 6-4 will be about 4%.
The case is different for a tandem - it has double weight, but air resistance isn't doubled - the stoker is drafting behind the steerer.
1. The force gravity will exert depends on the rider's mass.
2. Mass depends on volume, which is the cube of linear size. I.e. if you double the radius of a solid ball then it's weight goes up eight times.
3. Air resistance depends on the square of linear size. So the double size ball will have four times the air resistance of the original.
So as size increases gravitational force increases faster than air resistance, all things being equal.
However, air resistance also depends on the square of velocity - so big increases in mass equate to only small increases in speed. I'd expect that double-size, eight-times-heavier ride to go down a hill only 40% faster than Mr Standard Size at most - terminal velocity will increase with the square root of linear size (e.g. rider height.) The difference between two identically built riders of 5-10 and 6-4 will be about 4%.
The case is different for a tandem - it has double weight, but air resistance isn't doubled - the stoker is drafting behind the steerer.
06-28-09 | 10:11 AM
#83
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Oh - and in a simple analysis like the one above, which neglects rolling resistance and the rider's own propulsion and what have you, the slope doesn't matter. It will affect the absolute terminal velocities, but not their ratio.
Talking of propulsion, if we figure that strength is proportional to weight and that the rider is pedaling all out, then the heavier rider gets a bit faster again. But you can't say how much so until you get down to real numbers for slope, surface area, and rider power output.
Talking of propulsion, if we figure that strength is proportional to weight and that the rider is pedaling all out, then the heavier rider gets a bit faster again. But you can't say how much so until you get down to real numbers for slope, surface area, and rider power output.
06-28-09 | 10:42 AM
#85
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Assuming all potential energy is converted to kinetic energy (oversimplified), then PE = KE so m*g*h = 1/2*m*(velocity)^2. So you have mass on both sides of the equation, they cancel, and the velocity is then independent of the mass of the rider.
That said, if you remove all those simplifications (vacuum, 100% conversion efficiency), the final velocities won't be the same... but it has absolutely nothing to do with the extra potential energy being stored. The equations become more complex, as detailed by some posters above.
06-28-09 | 10:42 AM
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we're not talking about free falling, we're talking about lbs of force exerted on a machine, don't forget that a wheel is a machine
06-28-09 | 11:18 AM
#87
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The extra energy stored by a big person does not give them a higher velocity over a small person. That extra energy results in them achieving the same velocity (in a vacuum), since they are heavier they need more energy to achieve that velocity.
Assuming all potential energy is converted to kinetic energy (oversimplified), then PE = KE so m*g*h = 1/2*m*(velocity)^2. So you have mass on both sides of the equation, they cancel, and the velocity is then independent of the mass of the rider.
That said, if you remove all those simplifications (vacuum, 100% conversion efficiency), the final velocities won't be the same... but it has absolutely nothing to do with the extra potential energy being stored. The equations become more complex, as detailed by some posters above.
Assuming all potential energy is converted to kinetic energy (oversimplified), then PE = KE so m*g*h = 1/2*m*(velocity)^2. So you have mass on both sides of the equation, they cancel, and the velocity is then independent of the mass of the rider.
That said, if you remove all those simplifications (vacuum, 100% conversion efficiency), the final velocities won't be the same... but it has absolutely nothing to do with the extra potential energy being stored. The equations become more complex, as detailed by some posters above.
I was thinking in terms of the energy, rather than the velocity.
I have deleted the erroneous section of that post.
06-28-09 | 11:28 AM
#88
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06-28-09 | 12:16 PM
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Yesterday as I was on my ride, I came across a Farmer's Market with various vendors. One of them had a machine for making mini-donuts. Mmmm, were those good. Need to keep up the aerobelly somehow.
06-28-09 | 02:16 PM
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06-28-09 | 04:15 PM
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06-28-09 | 05:46 PM
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06-28-09 | 05:55 PM
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06-28-09 | 06:54 PM
#99
This is getting old real fast.
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06-28-09 | 07:04 PM
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