Recent content by martonhorvath

Thank you, it is clear.

I agree with this, as I said earlier, I think we have been talking about the same thing, just you brought kinetic energy into the conversation and I was trying to explain my thoughts through constant velocity examples after a point.

The speed does not stay the same, I am sorry if I have been unclear about this before. That is correct, indeed. If we want to look at an equation, we could write something like ##P_{downhill} = P + P_{gravity} - (P_{friction} + P_{air})##, where P is the same 200 W as in the uphill example.

The 200 W is a capacity in the example, it is spent regardless of incline. That would be the point of my question. To have the same work rate from the engine both uphill and downhill, so from 200 = 100 + 50 + 50, it would be e.g., 200 = 120 + 80, I want to keep this 200W which is produced by the...

Thank you for your detailed reply, I think your answer reassures me that I do not think wrongly about this. If I understand you correctly, you described ##P_{gravity}## from the perspective of the gravitational force, thus in your thread of thoughts if gravity does work on the car, ##P_{gravity}...

Okay, you are right concerning acceleration but can you see what I mean under 'net power' now? If we take a look at the power balance equation and want to calculate the car's contribution to the 'net power' (i.e., power corresponding to the speed of travel) at an incline less than zero, it...

Maybe my question formulation was wrong here because I am starting to get a feeling that both of us are right here. So I will try to illustrate what I am trying to say here through another example, I think this one enlightens more about the problem I want to know: let's assume that the car...

I see, but that power is not produced by the car but by the hill. My point would be that in this case, the car has to overcome only three forces and there is an additional one which "helps" it. E.g., if the car would produce 1 kW under both conditions then I think that in the case of the...

By applying some simplifications, the power ##P## of a body moving at a speed of ##v## at an incline of ##\alpha > 0^\circ## can be expressed as: ##P = P_{acceleration} + P_{friction} + P_{gravity} + P_{air} = (ma + mgsin(\alpha) + C_{rr}mgcos(\alpha) + 0.5C_dA\rho v^2)\cdot v## If we would...

I realized that I left out multiplication by ##v## on the right side in the beginning when typing my post. Therefore the correct expression for ##\alpha## should look like as: $$\alpha = arcsin(\frac {P} {mgv\sqrt {4+\mu^2}}) - arctan(\frac {\mu} {2})$$

Related to the part highlighted by you, these are supposed to be metrics describing performance. Power output normalized to body weight (W/kg) and power normalized to the coefficient of air drag (##W/C_{d}A##).

##C_{d}A## - coefficient of air drag, ##C_{rr}## - rolling resistance, ##P## - overall power output, ##\alpha## - incline, ##\rho## - density of air, ##m## - weight of the cyclists. I think everything else should be trivial or explained in the original post.

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...

Calculating the resulting velocity: ##\sqrt{(v+v_{w}×sin(\beta))^2+(v_{w}×cos(\beta))^2}=\sqrt{v^2+2vv_{w}×sin(\beta)+(v_{w}×sin(\beta))^2+(v_{w}×cos(\beta))^2}=\sqrt{v^2+v_{w}^2+2vv_{w}×sin(\beta)}## I can see it until here, but how does the multiplication by ##(v+v_{w}×sin(\beta))## come?

Thank you for the formula, I have also thought that using only ##v_{wind}×sin(\beta)## is an oversimplification. Yes I take ##\beta=0## for corss-wind. It has turned out that my calculations are not as wrong as I thought, just made a stupid and fatal mistake when compared to the other method...