Ash
Warming-Up
- Apr 23, 2006
- 686
- 1
Hmmm...I found the following on Velonews this morning from Leonard Zinn's column. I have his 'Zinn and the Art of Road Bike Maintenance', which as far as I am concerned is the best maintenance book ever written, so I usually look at his columns.
I am not a mathematician, but I think I could have figured out to pedal fast over the top of the hill before descending without doing a Phd in the subject. Thought you might like this Thomas, you are a technical geezer!
Here it is:
When cycling, you must consider the forces of rolling resistance, air resistance and gravity, which when added together, give the power the rider puts out to maintain his or her speed, "s." From DiPrampero et al., 1979 (sorry, I only found it in Portuguese), the equation looks like this:
P = krMs + kaAsv2 + giMs
where P is power, kr is the rolling resistance coefficient, M is the combined mass of cyclist and bicycle, s is the bicycle speed on the road, ka is the air resistance coefficient, A is the combined frontal area of cyclist and bicycle, v is the bicycle speed through the air (i.e. road speed plus head wind speed), g is the gravitational acceleration constant, and i is the road incline (grade; however, this is only an approximation, as the sine of the road angle to the horizontal should technically be used).
When you coast downhill, you are adding no power, so you reach terminal velocity when the gravitational force pulling you down the hill equals the drag forces resisting your motion. So, terminal velocity is when:
krMs + kaAsv2 = giMs
When you are moving fast downhill, the aerodynamic drag forces are huge compared to the rolling resistance forces, so you can ignore them. And you clearly want to know the answer in general, meaning on a calm day, rather than how fast you go into a headwind, so s = v, making the equation become:
kaAs3 = giMs
or
sterminal = (giM/kaA)1/2
Obviously, the heavier the rider and the steeper the hill, the bigger the giM term becomes, and the bigger sterminal becomes. Similarly, the lower the frontal area, A, of the bike and rider, the higher the terminal velocity becomes. This, of course, is the reason for tucking on your bike when you descend.
An interesting thing that you can see from this terminal-velocity equation is that there is a terminal velocity beyond which you will not go unless you pedal or the hill becomes steeper. This means that you will get down the hill the fastest the sooner you get to terminal velocity. So it always pays to sprint over the top of the hill and the beginning of the descent before you go into your tuck. You will arrive at the bottom sooner than if you had simply gradually coasted up to sterminal.
Lennard
I am not a mathematician, but I think I could have figured out to pedal fast over the top of the hill before descending without doing a Phd in the subject. Thought you might like this Thomas, you are a technical geezer!
Here it is:
When cycling, you must consider the forces of rolling resistance, air resistance and gravity, which when added together, give the power the rider puts out to maintain his or her speed, "s." From DiPrampero et al., 1979 (sorry, I only found it in Portuguese), the equation looks like this:
P = krMs + kaAsv2 + giMs
where P is power, kr is the rolling resistance coefficient, M is the combined mass of cyclist and bicycle, s is the bicycle speed on the road, ka is the air resistance coefficient, A is the combined frontal area of cyclist and bicycle, v is the bicycle speed through the air (i.e. road speed plus head wind speed), g is the gravitational acceleration constant, and i is the road incline (grade; however, this is only an approximation, as the sine of the road angle to the horizontal should technically be used).
When you coast downhill, you are adding no power, so you reach terminal velocity when the gravitational force pulling you down the hill equals the drag forces resisting your motion. So, terminal velocity is when:
krMs + kaAsv2 = giMs
When you are moving fast downhill, the aerodynamic drag forces are huge compared to the rolling resistance forces, so you can ignore them. And you clearly want to know the answer in general, meaning on a calm day, rather than how fast you go into a headwind, so s = v, making the equation become:
kaAs3 = giMs
or
sterminal = (giM/kaA)1/2
Obviously, the heavier the rider and the steeper the hill, the bigger the giM term becomes, and the bigger sterminal becomes. Similarly, the lower the frontal area, A, of the bike and rider, the higher the terminal velocity becomes. This, of course, is the reason for tucking on your bike when you descend.
An interesting thing that you can see from this terminal-velocity equation is that there is a terminal velocity beyond which you will not go unless you pedal or the hill becomes steeper. This means that you will get down the hill the fastest the sooner you get to terminal velocity. So it always pays to sprint over the top of the hill and the beginning of the descent before you go into your tuck. You will arrive at the bottom sooner than if you had simply gradually coasted up to sterminal.
Lennard