- #1
martonhorvath
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- TL;DR Summary
- I would need some help verifying my approach to a problem in Newtonian mechanics with a practical approach. Any additional insights are welcome as well!
Hi!
As written in the summary, I would like to get feedback about my approach to an issue I wanted to explore. In the example I am using cycling, but principally the same theory would apply to similar sports as well.
So, the problem was that I wanted to investigate whether the athletes' W/kg ratio or W/CdA ratio matters more in different scenarios. Of course, the answer should be highly incline- and speed-dependent.
To solve the question, I wanted to express a threshold incline (##\alpha##) at which the ratio of ##P_{gravitational} = P_{air}##, accepting the preconception that W/kg will 'matter' more in terms of performance if power against gravity is greater than power against air resistance.
So when expressing power as ##P = P_{gravity} + P_{rolling res.} + P_{air}##, (not accounting for minor losses such as loss due to the friction of the drivetrain), my initial equation looked like this:
$$\frac {P-P_{rolling}} {2} = P_{grav} = P_{air}$$
as follows:
$$\frac {P - C_{rr} mg cos(\alpha)} {2} = mg sin(\alpha) = 0.5 × C_{d}A×\rho×v^3$$
where m is the weight of the cyclist, ##C_{rr}## is the rolling resistance coefficient, etc.
When expressing P, it follows that:
$$P = mg(2sin(\alpha) + C_{rr}cos(\alpha))$$
$$\frac {P} {mg} = 2sin(\alpha) + C_{rr}cos(\alpha))$$
let's define ##k= \frac {P} {mg}##, so the equation can be expressed as:
$$k = \sqrt {4 + C_{rr}^2} (\frac {2} {\sqrt {4+C_{rr}^2}} × sin(\alpha) + \frac {C_{rr}} {\sqrt {4+C_{rr}^2}} × cos(\alpha))$$.
If ##R = \sqrt {4 + C_{rr}^2}##, ##cos(\beta) = \frac {2} {R}## and ##sin(\beta) = \frac {\mu} {R}##,
$$k = R(sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)$$.
After sine addition:
$$k = R × sin(\alpha+\beta) $$
$$\alpha = arcsin(\frac {k} {R} - \beta)$$
where $$\beta = arctan (\frac {C_{rr}} {2})$$
therefore
$$\alpha = arcsin (\frac {\frac {P} {mg}} {\sqrt {4+C_{rr}}}) - arctan (\frac {C_{rr}} {2})$$
Can someone validate my initial thoughts and assure me that my calculations are correct?
As written in the summary, I would like to get feedback about my approach to an issue I wanted to explore. In the example I am using cycling, but principally the same theory would apply to similar sports as well.
So, the problem was that I wanted to investigate whether the athletes' W/kg ratio or W/CdA ratio matters more in different scenarios. Of course, the answer should be highly incline- and speed-dependent.
To solve the question, I wanted to express a threshold incline (##\alpha##) at which the ratio of ##P_{gravitational} = P_{air}##, accepting the preconception that W/kg will 'matter' more in terms of performance if power against gravity is greater than power against air resistance.
So when expressing power as ##P = P_{gravity} + P_{rolling res.} + P_{air}##, (not accounting for minor losses such as loss due to the friction of the drivetrain), my initial equation looked like this:
$$\frac {P-P_{rolling}} {2} = P_{grav} = P_{air}$$
as follows:
$$\frac {P - C_{rr} mg cos(\alpha)} {2} = mg sin(\alpha) = 0.5 × C_{d}A×\rho×v^3$$
where m is the weight of the cyclist, ##C_{rr}## is the rolling resistance coefficient, etc.
When expressing P, it follows that:
$$P = mg(2sin(\alpha) + C_{rr}cos(\alpha))$$
$$\frac {P} {mg} = 2sin(\alpha) + C_{rr}cos(\alpha))$$
let's define ##k= \frac {P} {mg}##, so the equation can be expressed as:
$$k = \sqrt {4 + C_{rr}^2} (\frac {2} {\sqrt {4+C_{rr}^2}} × sin(\alpha) + \frac {C_{rr}} {\sqrt {4+C_{rr}^2}} × cos(\alpha))$$.
If ##R = \sqrt {4 + C_{rr}^2}##, ##cos(\beta) = \frac {2} {R}## and ##sin(\beta) = \frac {\mu} {R}##,
$$k = R(sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)$$.
After sine addition:
$$k = R × sin(\alpha+\beta) $$
$$\alpha = arcsin(\frac {k} {R} - \beta)$$
where $$\beta = arctan (\frac {C_{rr}} {2})$$
therefore
$$\alpha = arcsin (\frac {\frac {P} {mg}} {\sqrt {4+C_{rr}}}) - arctan (\frac {C_{rr}} {2})$$
Can someone validate my initial thoughts and assure me that my calculations are correct?
