Cycling calculations

Is this up the hill or coasting down? I can't think of a way of calculating either without a bunch of other independant variables.

Check with your old Physics teacher 😄

Going up.

All the other variables will be constants as I'm just looking for a general idea.

Thanks.

Pharkas said:

Going up.

All the other variables will be constants as I'm just looking for a general idea.

Thanks.

It all depends on how much power the cyclist is able to generate. See:

http://www.kreuzotter.de/english/espeed.htm
http://www.analyticcycling.com/

Berend

Pharkas said:

Hi,

I'm looking for formulas or info on how to approximate speeds with basic info.

Something like :

Rider Weight : 160 lbs
Hill Grade : 6 %
Hill Distance : 5 KM

Speed = ?

Any info & links would help me out.

Thanks !

None of that info really feeds into the equation. You need gearing and cadence to figure speed (in theory).

Jon Packard said:

You need gearing and cadence to figure speed (in theory).

No you don't. Power is power no mater how fast you're going. Unless you're considering efficiency changes due to chain cross, gearing is best left out of it.
So, the power input of the rider is constant (it's not, but anything here can be a function of anything else if you want it to be) We'll call it Pr.
Power in equals power out, so Pr = Po.
The right hand side breaks down into vertical and tangential components (screwy coordinates, but it makes this easier). Po = Pv + Pt
In the vertical direction, it's all gravity. Since P = W/t, Pv = Wv/t
Wv = Fv*Dv (work=force*distance)
Dv/t = Vv (vertical distance/time=vertical velocity)

Pv = M*g*Vv (mass*gravitational accelaration*vertical velocity

Vv comes from the grade, the tangential velocity, and some geometry that I'm too lazy to work out right now. g=9.8 m/s^2 or 32.2 ft/s^2.

Tangential power is all drag. Again, Pt = Wt/t = Fd*Vt (drag force*tangential velocity)
From basic fluids, Fd = Cd*(1/2)*rho*A*Vt^2

From Alexander and Smits, Cd = about .88 and A = about 3.9ft^2
At sea level and 70F, 2.329e-3 slugs/ft^3

So you only need the rider's mass, power output, and the grade of the hill to back out the average speed, Vt

Too much for me...

Speed = Distance divided by time

That's the only thing i know :p

Artmichalek -> Thanks for all the info !

However, I'm a little lost with all the formulas.

What is rho in Fd = Cd*(1/2)*rho*A*Vt^2 ?

Would you be able to provide a sample detailled calculation with the following data ?

Power : 200 W
Weight : 180 lbs.
Grade : 5%
Distance : 2 KM

This will help me to understand the formulas a bit more.

Thanks again for your help and time !

Pharkas said:

Artmichalek -> Thanks for all the info !

However, I'm a little lost with all the formulas.

What is rho in Fd = Cd*(1/2)*rho*A*Vt^2 ?

Would you be able to provide a sample detailled calculation with the following data ?

Power : 200 W
Weight : 180 lbs.
Grade : 5%
Distance : 2 KM

This will help me to understand the formulas a bit more.

I'll admit that the proof is a little bit dense. To start with, rho is the fluid density and distance doesn't matter. If you want to run through the calculations, you have to work in one set of units. I recomend converting everything to metric. The attached equations are all you really need to calculate this. The alpha in the first equation is just a constant to represent my extreme resistance to doing geometry when I don't have to. It relates velocity in the vertical direction to measured velocity relative to the road. For the second equation just input your constants and the ones from my previous post. All units should be kg, m, s. Once everything is substituted, you can solve for V.

Artmichalek,

Thanks again but my math & physics are definitely not as advanced as yours and I can't figured this out.

Could you run me through a sample calculations with all the data ?

Thanks,

Pharkas said:

Could you run me through a sample calculations with all the data ?

Not much to run through:
P=200 W
g=9.8 m/s^2
rho=1.2 kg/m^3
A=.3623 m^2
Cd=.88
M=81.65 kg
alpha = 0.05 (not really, but for this grade it's close enough)

Plug all that in, solve for V, and you get about 4.5 m/s or 10mph. If alpha drops to 0, you're doing about 22mph

The info you give is too vague. This would be like a problem in applied mathematics, where there is an inclined plane of x degrees, a coefficient of friction of y, a contact surface area of z, and a mass of a, and so on.

Niblox said:

The info you give is too vague. This would be like a problem in applied mathematics, where there is an inclined plane of x degrees, a coefficient of friction of y, a contact surface area of z, and a mass of a, and so on.

Of course it's vague. If you want to include every detail, it would be faster and easier to just go out and ride the hill. It's a good rough estimate though. If you have good wind tunnel data to narrow down Cd and A, you can even discretize those equations over a time trial course and optimize it relative to your peak power/duration. But that takes some serious programing time.