Calcuting forces on a slope (cycling) -online calculators wrong?

AI Thread Summary
The discussion centers on discrepancies in calculating the forces acting on a cyclist going uphill, particularly between various online calculators. The user notes that their calculations for a 5º slope yield a parallel force of 66.6 N, consistent with one calculator but differing significantly from others that report 38.2 N. They seek to understand the additional forces at play, such as air resistance and friction, by conducting a calibration run downhill. The conversation highlights the importance of distinguishing between slope angle and percent grade in calculations. Accurate force measurement is crucial for determining the work done against gravity and other resistive forces during uphill cycling.
Festina
Messages
2
Reaction score
0
I'm trying to deduce the forces acting on a cyclist going up a hill and I need help. There's 2 questions:

Q.1

I've looked at online calculator already in existence:

http://www.analyticcycling.com/ForcesPower_Page.html
http://www.gribble.org/cycling/power_v_speed.html


What puzzles me about these calculators is when I caculate the parallel force I get numbers that are different to these calculator. but are the same as http://thecraftycanvas.com/library/...pers/incline-force-calculator-problem-solver/


e.g. for a 5º slope and a 78 kg rider+equipment: -9.81 * sin(rad5) * 78 = 66.6 N "thecraftycanvas.com" get the same result, yet Gribble and Analytic cycling get 38.2 N

Q.2

Following on from this, what I am trying to do is calculate what the force are acting on the rider by doing a calibration run down the hill, the other forces (air resistance (wind, pressure), friction) besides gravity being the difference between the force on the calibration run and a hypothetical run just with the force of gravity.

So that when the rider goes up hill in X amount of time we can calculate the how much work against gravity by itself and all the other forces rolled together (deduced from the calibration run).

Am I missing something fundamental?
 
Physics news on Phys.org
Festina said:
yet Gribble and Analytic cycling get 38.2 N
You confuse slope angle and percent grade.

tan(slope angle) * 100 = percent grade

Am I missing something fundamental?
Air drag is velocity dependent. You can let him roll downhill, and measure the acceleration at different velocities. From acceleration you get the net force. Substract gravity to get air drag + rolling resistance. Transmission loss is not included.
 
Last edited:
I did indeed A. T., thanks!
 
Festina said:
I did indeed A. T., thanks!
I added an answer to the 2nd question.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Thread 'Gauss' law seems to imply instantaneous electric field (version 2)'
This argument is another version of my previous post. Imagine that we have two long vertical wires connected either side of a charged sphere. We connect the two wires to the charged sphere simultaneously so that it is discharged by equal and opposite currents. Using the Lorenz gauge ##\nabla\cdot\mathbf{A}+(1/c^2)\partial \phi/\partial t=0##, Maxwell's equations have the following retarded wave solutions in the scalar and vector potentials. Starting from Gauss's law $$\nabla \cdot...

Similar threads

Back
Top