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May i introduce to you a method to estimate the power output of the 2014 PUs.

At first i calculated a generic F1 car with:

Power = 750 hp = 552000 W (at every single rpm, the 2014 cars should come close to this)
rho/2*cd*A = 0,655 (= power / (340km/h / 3,6)^3 )
rho/2*cl*A = 3*0,655 = 1,965 ( this car has an efficiency of 3=cl/cd )
mass = 695 kg
friction_coefficient = 0,015 (it better should read rolling_resistance_coefficient)

From that i could calculate P_drag(v), P_roll(v) and P_acc = P - P_drag - P_roll. P_acc is the power that can be used to accelerate the car, in reality this is only valid above the traction limit.

The following plots were the result, the first showing P [W] as a function of v [km/h], the second one P as a function of v^3.
As the drag is a function of v^3, the power that is available for accelerating above the traction limit is also a function of v^3 and becomes a line when plotting P against v^3. The point where the blue line of P_acc crosses the vertical axis shows the engine power, even if we would only plot P_acc. The point where it crosses the horizontal axis is the maximum speed.



Any errors in my assumptions?

From a onboard telemetry video of Rosberg's pole i got a lot of time-velocity data, by analysing the frame by frame with the Windows Movie player. From that the acceleration is "(delta velocity)/3,6/(delta time)" and
"P_acc = 1,05 * 695kg * acceleration(m/s²) * velocity(m/s)". The factor 1,05 accounts for the rotational components that also need to be accelerated. I only took those data above the traction limit (=throttle at 100%) and when at the car is at a straight, seperated by if DRS was active or not.

The resulting plots( hp vs. (km/h)^3 ) are the following two, the first without DRS the second with DRS activated. From the formula in the plots, the estimated powers are 682 and 735 hp at the wheels. Not really close to each other.

The crossing points of the trend line with the x axis are located at 339 km/h and 324,4 km/h.

Blanchimont wrote:
The resulting plots( hp vs. (km/h)^3 ) are the following two, the first without DRS the second with DRS activated. From the formula in the plots, the estimated powers are 682 and 735 hp at the wheels. Not really close to each other.

I commend you for taking the effort to do this
Could we say that the average peak power is around 708hp ?
If so, then the ICE contribution is approx. 550hp when subtracting 160hp from MGU-K ?

Amazing job.

If I get this correctly the slope of the last two graphs should be different but the points at the vertical axis should be the same correct? The slope is the indication of the drag coefficient. Right?

What slope are you getting on your theoretical car ? Does it match either one of the real graphs?

321apex wrote: Could we say that the average peak power is around 708hp ?
If so, then the ICE contribution is approx. 550hp when subtracting 160hp from MGU-K ?

I think you should account for gearbox efficiency first, because the MGU-K also drives through the gearbox.

Blanchimont wrote: rho/2*cd*A = 0,655
rho/2*cl*A = 3*0,655 = 1,965
friction_coefficient = 0,015

Just curious where these numbers come from.

Also, is it right to depend on the display graphics?
and is it right to assume a drag coefficient for the car?
Any information on the incline of the track, the wind speed, the driver's right foot, the tyre grip, etc etc?

735 is too high. It would interesting if you did this for last years cars, then find a way to adjust that for beam wing removal. then do a correlation to this years straight line video.

From the qualifying broadcasts, they were saying there was a headwind down the main straight, and a tail down the back straight. I'm unsure of the magnitude, but I'd say it's important to factor this in.

Lycoming wrote:

Blanchimont wrote: rho/2*cd*A = 0,655
rho/2*cl*A = 3*0,655 = 1,965
friction_coefficient = 0,015

Just curious where these numbers come from.

I made some annotations in the first post about these numbers. I assumed the top speed without the rolling resistance would be 340km/h, from that you can calculate rho/2*cd*A and assume an efficiency, which of course can also be higher. But the nice thing about this method in theorie is that it doesn't matter how high or low the drag is, it just changes the slope of the line in the v^3 graph.

The friction_coefficient should be named rolling_resistance_coefficient and i thought that 0,015 is a good estimation as normal road tyres are in the range of 0,01 AFAIK.

321apex wrote:

Blanchimont wrote:
The resulting plots( hp vs. (km/h)^3 ) are the following two, the first without DRS the second with DRS activated. From the formula in the plots, the estimated powers are 682 and 735 hp at the wheels. Not really close to each other.

I commend you for taking the effort to do this
Could we say that the average peak power is around 708hp ?
If so, then the ICE contribution is approx. 550hp when subtracting 160hp from MGU-K ?

If we consider my calculations to be the truth then you can say that the average power in the rev range used is 708hp at the wheels, ~723hp at the crankshaft with a gearbox efficiency of 0,98. But we should be careful about that, as some said below, the wind could cause some errors, the telemetry itself doesn't need to be show the real speed.

Then i used a factor of 1,05 to include the energy that is stored in the rotating parts (tyres, wheels, brake discs, driveshafts, crankshaft, gearbox). So the error range is quite large to be certain about these power numbers. But it could be a good start, let's see if i can improve the calculations today.

PhilS13 wrote:Amazing job.

If I get this correctly the slope of the last two graphs should be different but the points at the vertical axis should be the same correct? The slope is the indication of the drag coefficient. Right?

What slope are you getting on your theoretical car ? Does it match either one of the real graphs?

You're right about the slope, the point at the vertical axis should always stay the same, the power of the engine can't change. The lower the slope and the further to the right the intersection of the line with horizontal axis, the lower the drag coefficient.

The slope of the telemetry data is
-1,9991*10^-5 [hp/(km/h)^3] and
-1,8872*10^-5 [hp/(km/h)^3]

For the modeled car (340km/h without rolling resistance and 750hp) the slope is
-0,0149 [W/(km/h)^3] which is the same as
-2,025*10^-5 [hp/(km/h)^3].

ringo wrote:Also, is it right to depend on the display graphics?
and is it right to assume a drag coefficient for the car?
Any information on the incline of the track, the wind speed, the driver's right foot, the tyre grip, etc etc?

735 is too high. It would interesting if you did this for last years cars, then find a way to adjust that for beam wing removal. then do a correlation to this years straight line video.

The display graphics are the best data that are available and most of the time i think the are pretty close to the real numbers.
For example the highest telemetry speed at the start-finish straight was 326km/h, compared to 325,3km/h from the official FIA numbers. Speed at the end of sector 2 is 261km/h vs. 263,7km/h official and at the start-finish line 291km/h vs. 290,5km/h.

I didn't assume a drag coefficient for the calculation of P_acc, this was only done for the generic car in the example. My method only uses the acceleration and the car's mass and assumes the resulting graph is a function of v^3 because P_drag is a function of v^3.

Tyre grip is not needed for my method as i only use those data when the driver is at full throttle on a straight=above the traction limit.
The sections of the track i used were from T3 to T4, from T8 to T9(a slight uphill part of the track),from T10 to T11(DRS), from T13 to T14, from T15 to the DRS zone and finally the DRS zone.

I'll have a look at the T8 to T9 part and at those parts with head or tailwind according to the info here in this thread. Stay tuned, i'll try to update this thread today later on.

Blanchimont wrote:

PhilS13 wrote:Amazing job.

If I get this correctly the slope of the last two graphs should be different but the points at the vertical axis should be the same correct? The slope is the indication of the drag coefficient. Right?

What slope are you getting on your theoretical car ? Does it match either one of the real graphs?

You're right about the slope, the point at the vertical axis should always stay the same, the power of the engine can't change. The lower the slope and the further to the right the intersection of the line with horizontal axis, the lower the drag coefficient.

The slope of the telemetry data is
-1,9991*10^-5 [hp/(km/h)^3] and
-1,8872*10^-5 [hp/(km/h)^3]

For the modeled car (340km/h without rolling resistance and 750hp) the slope is
-0,0149 [W/(km/h)^3] which is the same as
-2,025*10^-5 [hp/(km/h)^3].

Got it. The analysis is good then I think, we just need less noise in the data. The Red Bulls and maybe other teams have a speed indicator on their dash it might be better than the FOM graph. Wind was 4.2 m/s pure tail on the main straight and it's pretty flat. Maybe you can isolate that acceleration from the others and factor in the wind?

Blanchimont wrote: ..... If we consider my calculations to be the truth then you can say that the average power in the rev range used is 708hp at the wheels, ~723hp at the crankshaft with a gearbox efficiency of 0,98
Then i used a factor of 1,05 to include the energy that is stored in the rotating parts (tyres, wheels, brake discs, driveshafts, crankshaft, gearbox). So the error range is quite large to be certain about these power numbers. But it could be a good start, let's see if i can improve the calculations today.

there's 3 stages of gearing (1 a spiral) between the crankshaft and the driveshafts so 0.98 seems optimistic ??
(4 stages for the mgu-k)
the effective mass factor even in top gear would be nearer to 1.1 ??
the mgu-k inertia being a significant contributor to this ?

Cold Fussion wrote:From the qualifying broadcasts, they were saying there was a headwind down the main straight, and a tail down the back straight. I'm unsure of the magnitude, but I'd say it's important to factor this in.

PhilS13 wrote:Wind was 4.2 m/s pure tail on the main straight and it's pretty flat. Maybe you can isolate that acceleration from the others and factor in the wind?

So, which one then? Heads or tails?