To optimize energy consumption, or how to make F1 greener

[Ciro Pabón]

Hang on a sec Ciro, as much as I buy the two blue lines, I think the yellow one is somewhat out of whack. Why would Fizzi only use 150 Hp when accelerating from 150 km/h?
That yellow line should be flat as Ayers rock to my mind, except for the occational gear shift, and certainly not terminate at a measly 400 Hp at 300 kph?

Or am I missing something essential here?

Where are those 300 HP?

Simple: the car is not able to transmit them to the track. The friction factor for tyres is the limitation. This is very important to understand when you drive a car that has more "power than it's needed". You have to sqeeeeeze the throttle, or the tyres will spin.

For example, in Reca's graph I can see that from 42 to 43 seconds, the car goes from 124 to 163 kph, thus, the acceleration is 1.1 G. The tyre cannot develop more than this coefficient of friction. Check the acceleration row in the worksheet I made (row number 12):



Top acceleration is 1.1 Gs.

If, going at 124 kph (that is, 34 m/s) you could use the full HP, then in one second you would use 770 HP, right?

That is, you would use 770 HP * 746 watts/HP = 574.000 watts

In one second, you would be using 574.000 watts/1 sec = 574.000 joules, right?

So, Force * distance = 574.000 joules.

Force = mass * acceleration, and mass = 605 kg.

So, Acceleration * distance = 574.000 joules / 605 kg = 950 m2/s2

Now, distance = (Vf-Vo)^2/(2*acceleration), so:

950 m2/s2 = (Vf-Vo)^2/2

Thus, Vf = square root of (950 m2/s2 * 2)+ 34 m/s

Vf = 77 m/s

And acceleration = (77 m/s - 34 m/s)/1 sec = 43 m/s2

To use 770 HP, the car should accelerate at 43/9.8 = 4.4 Gs.

There is no tyre in the world that can develop a friction factor of 4. If you had one, the car would move from 124 kph to 277 kph in one second. Impossible.

The power I graphed is the effective power, not taking in account thermal losses. That's why the title of the worksheet is "Power to the wheels", not "Power to the brake". So, those 400 HP at top speed means that the engine has an efficiency of over 50%. I doubt it, I think it should be around 35%. Probably, I'm overestimating the drag.

Those numbers do not depend on the track: they're real numbers taken for an F1 car accelerating at full throttle, in a sector where (I pressume) there are no significant horizontal curves (I would be most grateful if somebody checks where the car spends seconds 42 to 51 at an Albert Park lap).

I hope you get it now.

[Belatti]

If you have got 770hp to the brake, you have at least 700hp in the wheel (assuming 10% loss in transmitting that power through the gearbox, diff and wheel bearing losses).
I understand that you cant use them all in the first 2 seconds of your chart where "G" is arround 1. What happens then? You have got 300HP in exess from the engine, and you cant use them to accelerate at a bigger G than 0.7? Where is the availiable power going to?

My guess is that your drag figures are not good, you rolling power figures decrease with speed

You are saying that when you are traveling at 290kmh moving the car through the air consumes 400HP and can not accelerate more, even if you know you have got 300 HP more availiable?

To estimate drag go backwards: lets say the car reaches its top and constantspeed (assume 320kmh) and we know the engine its giving us its full 700HP. Rolling power will not be taken into account (calculus simplicity) and inertia power is 0 (constant speed). So drag power is 700HP

700HP (521KW) = Fd*V = 5,86KN * 88,8m/s

Cd = (5,86 KN * 2) / (1,2 Kg/m3 * 88,8¨2 m2/s2 * 0,9 m2 ) = 1,37


So, mind this:




and this:

The reference area A is typically defined as the area of the orthographic projection of the object on a plane perpendicular to the direction of motion. For non-hollow objects with simple shape, such as a sphere, this is exactly the same as a cross sectional area. For other objects (for instance, a rolling tube or the body of a cyclist), A may be significantly larger than the area of any cross section along any plane perpendicular to the direction of motion. Airfoils use the square of the the chord length as the reference area; since airfoil chords are usually defined with a length of 1, the reference area is also 1.

[xpensive]

There is very much you could say about our good moderator's calculations, Belatti, but I will try to do this another way. God, how I hate trusting computers without thinking.

First of all, there is a far easier way to calculate Power vs accelleration.
To accellerate a 700 kg object from 150 to 300 km/h (41 to 83 m/s), you need 1800 kWs (W = m * v^2/2) if you disregard air-resistance. A 700 Hp engine will produce 500 kW, which means that it should take you 1500 kWs /500 kW = 3.6s to perform that accelleration at a constant rate.
If you add the air-resistance from an average of 65 m/s over say 5s, it adds another 1100 kWs, resulting in (1800+1110)kWs/500 kW = about 6 seconds.
From 150 to 300 km/h, 41 to 83 m/s, in 6 sec equals an average 7 m/s^2, or 0.7 g, which sounds reasonable to me?

Secondly, the 700 Hp we are talking, are of course the power available at the crankshaft, nothing else.
With 7-speed low-friction gearboxes, I would expect drive-shaft output not to be very far off 700 during the entire procedure, perhaps losing a few percents. Remember, a ten percent loss would mean that you have to evacuate 50 kW (equal in heat to 100 toasters) from the transmission.

Thirdly, power is force times speed, Downloading 500 kW at 150 km/h means a propulsion force coming from the rear tyres of 12 000 N. Considering CoG far at the back due to accelleration and including aerodynamic downforce, 12000 N on the rear wheels would reguire a friction coefficient of 1.0.
Is that such a big deal?

Hope you get it now, Ciro.

[alelanza]

Very interesting discussion here

I'd like to add:

Remember, a ten percent loss would mean that you have to evacuate 50 kW (equal in heat to 100 toasters) from the transmission.

AFAIK most of the transmission losses under acceleration come from overcoming the inertia of all rotational masses and only a smaller percentage goes to heat. Now, i guess with the superlight components/very high rpm's of an F1 car this might change somewhat, but i'd be surprised if most of it went to heat.
Speaking of which, does anyone have data on what kind of flywheel is used on F1 engines? if one is used at all? i did a search but it mostly turns up kers related results....

[xpensive]

Inertia of rotating components is zip in the context, besides, it doesn't matter at constant speed anyway.

Excellent analysis of the Cv Belatti, though I think Ciro's cross-section area of 0.9 is a bit on the low side.

[alelanza]

Inertia of rotating components is zip in the context

Care to elaborate? We're talking about clutch components, gearbox, differential, shafts, rear tyres/rims/brakes. All very light mass but still being accelerated to very high speeds, how can it be zip under acceleration?
Constant speed yes i agree, but then we go back to the fact that constant speed and F1 aren't compatible

[imightbewrong]

I know this is a bit off topic but I think this post is very interesting when discussing fuel consumption (MPG):
http://www.bunniestudios.com/blog/?p=257

[Belatti]

Secondly, the 700 Hp we are talking, are of course the power available at the crankshaft, nothing else.
With 7-speed low-friction gearboxes, I would expect drive-shaft output not to be very far off 700 during the entire procedure, perhaps losing a few percents. Remember, a ten percent loss would mean that you have to evacuate 50 kW (equal in heat to 100 toasters) from the transmission.

Yeah, low friction gearboxes are a beauty, dont they?

When I said we can loose 10% transmitting what goes out of the crakshaft, I´m taking into account:
- friction and inertia in gearbox
- friction and inertia in differencial
- friction and inertia in wheel bearings
- friction and inertia to overcome rims and tyres (this is large: remember is rubber we are toaling about, tyres are not aligned 0º toe)
- friction (air) and inertia to overcome in brake disks

They may all be small percentages separated, but they all sum together.

Remember you have to add rotational inertia to the longitudinal we all have been calculating for all this components.

[xpensive]

Is this really F1Technical, opinions and speculations without technical back-up? Reminds me of some of my engineers.

- Friction-torque of a 40mm ID roller-bearing under 10 kN load, is in the 0.2 Nm range, depending on type.
- Inertia of a one kg, 280 mm disc, is 0.01 kgm^2. (m*r^2/2)
- Viscous losses inside the gearbox are difficult to estimate, but 1% loss is a 5 kW heater so be careful.

Now you guys go figure something out for a change, prove me wrong, damnit?.

And Belatti, technically speaking, overcoming inertia is not a loss.

[Belatti]

Is this really F1Technical, opinions and speculations without technical back-up? Reminds me of some of my engineers.

Been working with several 1300Kg tourism racing cars recently. My mind was thinking about 330mm and 5.8Kg steel brakes among other figures.

And Belatti, technically speaking, overcoming inertia is not a loss.

True, but is power you have to discount when you calculate the linear aceleration of a 700Kg car powered by a 700HP engine.

So, tell us how much power do you "waste" while trying to accelerate a car in a straight line.

[xpensive]

Darn it Belatti, you almost cost me my lunch-break, but here we go.

Let's limit things to two relations and wheels only, when the square relationship of radius, as well as mass, reduces the relevance of carbon brakes, shafts, gears and what have you, to "zip" in the context:

- Inertia of a homogenous cylinder: J = mass * radius^2/2
- Rotating energy: W = J * angular velocity squared/2

Think 4 homogenous wheels of 600 mm diameter, a total of 24 kg, accellerating from 150 to 300 km/h in 6 seconds?
From a wheel-speed of 1326 Rpm (139 rad/s) to 2652 Rpm (278 rad/s) in six, evenly spread of course.

- Inertia is 1.08 kgm^2
- Difference in rotating energy is W(278) - W(139) = 31.3 kWs

Over 6 seconds, this equals 5 kW and change, or less than one percent of power applied.

[Belatti]

OK OK Mr, you tell me this is only 1%, that is only 0,5% that one over there is 0,1%. So, when will you sum them all and answer me what was I asking for?

[xpensive]

No Belatti, I tell you the rest iz "zip in the context", care to prove me wrong?

[Belatti]

Naaahh, I wont prove you wrong, we are way out of topic and I dont know how many toasters do you need to heat up the tires. However, if you tell me how many HP you use to do anything but moving the mass foward, I´ll believe you.

[xpensive]

I don't think we are way off topic, if MrM had been an engineer and able to penetrate the details of the F1 car's use of power and energy, surely he would have given the application of KERS another thought.

As for your toasters, with industrial gearboxes, you typically have a 97-98% efficiency.

There's by the way another way of proving Ciro's tyre-friction model wrong, where he needed a friction-coefficient of more than four to convey 700 Hp at 150 km/h.
If you consider the power of the rear-wheels being torque times angular velocity and torque as force times wheel radius, with 500 kW and wheel diameter of 600 mm at a speed of 41.7 m/s, you get a propulsion force of 12 000 N just like the energy-model used before.

Hmm, good start of the day this.