# Need help with my bicycle computer project

1. Aug 20, 2009

### wmazz

I am trying to work out the physics used in a micro-controller project. I am having problems with even simple relationships, like F = ma. My answers end up with much larger numbers than, a man on a bicycle should be capable of. This is just a hobby for me, and I haven't taken a physics class before.

My Goals are:

(1) Calculate rolling resistance over an uneven surface area. My project does have a 2-axis accelerometer to determine the current angle.

(2) Calculate Watts/sec and Newton meters used for any given gear ratio, speed or angle.

(3) Calculate the rate Calories are being used up.

(4) (the easy part) Displaying and recording the data on a thumb drive.

I have used many dynamometers in the past, so I am trying to stick with what's familiar. I am basing my calculations off a SAE Paper #942478 by Kee and G.P. Blair, about Inertia Dynos and they use a go-kart as an example.

(no character map in Win 7)

The formulae for Engine torque is: Torque eng = (flywheel Inertia )(Engine speed in rad/sec)(ratio of flywheel speed to engine^2) + torque due to dyno bearing friction.

They use what looks similar Kinetic Energy formulae for linear speed to determine flywheel inertia required, to simulate a given load: The formulae for Engine torque is: Torque eng = (flywheel Inertia )(Engine speed in rad/sec)(ratio of flywheel speed to engine^2) = (Kart Mass)(Engine speed in rad/sec)(ratio of engine to flywheel speed ^2)(radius, wheels^2) = Torque in Nm or N?

It's this last formulae I am trying to use : Torque in Nm or N = (Bike and rider mass)(Pedal Crank speed in rad/sec)(gear ratio^2)(radius, wheels^2) + rolling resistance

Mass = 145kg
Pedal Speed = 37pi/45 rad/sec (1 revolution of the tire is = to 37pi/45 rad/sec of the pedals)
Gear Ratio = 44 : 18 (2.444 to 1) (This is gear ratio I use the most)
Rolling Resistance = ??
Pedal Crank Length = .175m

Calcs to (145kg) (37pi/45 rad/sec) (2.44^2) ( .350^2) = 274 kg-m or Nm or N, If Newton’s than 274N * .175m = 47.97Nm or Watt/seconds or 11.45 Calories

Seems to large and doesn't account for rolling resistance, or changes in acceleration or surface angle.

I would like to do this in baby steps. Also is there a link on this site that provides a character map and how to make a fine looking post?

Thank You

Bill M.

Last edited: Aug 20, 2009
2. Aug 21, 2009

### wmazz

I have never taken a physics class, but I will start Pre-Calculus this fall.

I was able to quantify my formulae by changing it to be similar to F=ma. Then comparing my results by converting them to HorsePower http://users.frii.com/katana/biketext.html" [Broken].

Mass = 145kg
Gear Ratio = 44 : 18 (2.444 to 1) (This is gear ratio I use the most)
Pedal Crank Length = .175m
Rolling Resistance (Rr) = ??

I was using, Watts = ma*gear-ratio*crank length = Kg-m^2/sec^2

And I would like to compute the extra watts required for a hill. Would this work?

s = 22pi/9 rad/sec*2.444*2*r*pi = distance traveled

So, s*sin(theta) = h

Then add h to my formulae, Watts = ma*gear-ratio*crank length*h

After testing I could add Rolling Resistance (Rr) to the formulae:

Watts = ma*gear-ratio*crank length*h + Rr

So is my addition of " s*sin(theta) = h" correct, and the idea that I can add Rolling resistance gatherd by testing to the equation ??

Thanks

this may be helpfull http://users.frii.com/katana/biketext.html" [Broken]

Last edited by a moderator: May 4, 2017
3. Aug 22, 2009

### sganesh88

The bike+rider mass will have a say in the rolling resistance expression only.