Torque and Power required to accelerate a car from 0-60MPH

[robinfisichel]

Hi All

I have a typical automotive problem here i am being confused by;

I have a three wheeled vehicle with two driven wheels and i am trying to spec a motor/power supply for it. To do this i want to calculate the torque and therefore power required by the motor.

The process i have followed is;

Fc=Force at tyre contact patch
Fr=Sum of resistive forces
Fi=sum of inertial forces
Fg=gravititational forces =0 (car on a level road)

Fc = Fr+Fi

and then to find torque;

T=Torque (max torque negating transmission losses etc)
r=tyre radius

T=F.r

and for power;

P=power
$$\omega$$=rotational velocity

P=$$\omega$$.T

Now the problem I am having is that the inertial forces required to accelerate the vehicle are quite large and the torque I am getting is much too large.

The vehicle stats are as follows;

Mass=410kg
Wheel radius = 0.33m (13inch)
Final velocity=26.6m/s (60mph)
Initial Velocity=0
0-60mph Time = 10s

Aero
Cd=0.15 (drag coef x)
Frontal area=1.74m^2
Density of air=1.225kg/m^3

For simplicity I've moddeled this as one wheel so don't need to worry about how many wheels are driven and the distribution of masses on those wheels.

My answer is telling me that i need around 430 Nm of torque (which is ridiculously high for a car like this) and conversley the power required is only 35 Kw. (around 50bhp)?

I may be making an incorrect assumption somewhere here, but if anyone can help that would be great.

(I also modeled in some transmission lossess into mine and rolling resitance of the tyres, as well as inertial forces of spinning tyres but these values were very small compared to the Aerodynamic drag and force required to accelerate the vehicle).

Cheers

 

[Bob S]

Here are some very approximate numbers:

1) Your acceleration is ~2.66 m/s², or ~0.3 g, so little chance of burning rubber.

2) Power P (watts) = M·v·a (mass · velocity · acceleration) = ~410·26.6·2.66 = 29,000 watts = 39 HP at 60 mph

Your tire ω at 60 mph = 26.6 m/s /0.33 m = 80 radians/s = 770 RPM

so wheel torque at 60 mph = 29,000 watts / 80 radians/s = 362 Newton meters (not including air drag). This is NOT engine torque.

The air drag power at 60 mph = P_d = ½·ρ·A·C_d·v³= 3000 watts.

Bob S

 

[robinfisichel]

Here are some very approximate numbers:

1) Your acceleration is ~2.66 m/s², or ~0.3 g, so little chance of burning rubber.

2) Power P (watts) = M·v·a (mass · velocity · acceleration) = ~410·26.6·2.66 = 29,000 watts = 39 HP at 60 mph

Your tire ω at 60 mph = 26.6 m/s /0.33 m = 80 radians/s = 770 RPM

so wheel torque at 60 mph = 29,000 watts / 80 radians/s = 362 Newton meters (not including air drag). This is NOT engine torque.

The air drag power at 60 mph = P_d = ½·ρ·A·C_d·v³= 3000 watts.

Bob S

Thanks Bob s, obviously since i modeled some losses into mine the required power and torque are slightly greater but in the same order of magnitude.

I thought this wasnt engine torque, how do i then translate this torque to the engine? I should say this is an ELECTRIC motor, which i confess i know little about in terms of gearing ratios etc.

My ultimate aim here is to spec the power and voltage requirements for the motor, and also specify torque and power curves.

 

[Bob S]

You will want maximum power near 60 mph, with wheel RPM = 770 RPM. You will need a gear ratio (including differential) to match to the motor RPM at maximum output power. So if for example the motor will put out 35 kW (47 HP) at 3000 RPM, the gear ratio should be about 3.9:1. The gear ratio is also a torque multiplier; the motor torque will have to be ~360 Nm/3.9 = 92 Nm at 3000 RPM..

The electric motor will have nearly constant torque at lower RPM, meaning nearly constant acceleration; ~ 2.66 m/s², independent of velocity. Unless you want more torque at lower velocities, you won't need a gearbox. Note that the acceleration is related to the power (see above) by the equation P = M·v·a, so at constant acceleration, the power from the motor is lower at lower velocities, although the torque and acceleration is constant. If you do use a gearbox, you could go up to ~0.5 g's without burning rubber, more if you have all-wheel drive.

Bob S

 

[willyadventur]

Hi Bob

I have searched the internet for a long time for answers about torque, and you seem to know your stuff. Also I agree with the gear box selection and you seem to fit in well with the 1/4 mile speed trap method of estimating Hp. (I’ve also come across many American sites that the quote torque in Kg/m, N/m and ,odd things like that).

I would like to ask the following.

1) It takes ‘X’ amount of torque to get the vehicle moving from rest. I realize you are using a gear box, but the engine RPMs will still be low at this point, so how does the formula "P (watts) = M•v•a (mass • velocity • acceleration) = 39 HP at 60 mph", fit with that?. (No low end torque very little initial acceleration)

2) The engine will probably produce max torque at, say, 1500rpm. Ok, you use a gear box to multiply the torque, but don't we also need to consider the point where we achieve maximum torque?. ( I think this would affect the acceleration time curve or, are we only concerned with the average torque over time – Calculus 101?)

3) Comparing a diesel engine to the petrol engine used above. A diesel engine producing the same amount of horse power at 3000rpm will produce considerable more torque at lower RPM's. Will this have any effect on acceleration, the formulas and/or gearbox selection? Also many sites say the extra torque is a benefit of a diesel, can you explain in detail why? (Towing capacity. Heavy trucks that need to accelerate from rest and climb hills, Etc).

Finally; when selecting a gearbox some manufactures ask for the torque reserve of the flywheel. Do you know how to calculate it per their requirements?

Best regards
Willy

 

[mrspeedybob]

If you use a gearbox you will have torque multiplication at lower speeds so I don't think constant acceleration is going to be a valid assumption.

 

[rcgldr]

The electric motor will have nearly constant torque at lower RPM.

For a typical dc electrical motor, torque decreases as rpm increases. Note sections 3.1 and 3.2 in this article:

http://lancet.mit.edu/motors/motors3.html

The minimum power to accelerate to some speed within some amount of time involves involves constant power and a continuously variable transmission with accleration that decreases as speed increases. Given the constraint of an ungeared electrical motor, and the fact that the motor's power output will vary with speed, reaching a maximum at about half of peak rpm (see section 3.2 in the article above), will make the math a bit more complicated.

You could just plug in all the parameters into a spreadsheet, using numerical integration (like Runge Kutta 4) to calculate speed versus time (in a series of rows), and adjust the motor parameters until you accomplish 0 to 60 mph in 10 seconds with minimal motor requirements. For practical purposes, you'd probably want a bit more than the minimal spec motor.

If there is no aerodynamic drag and no other losses, infinite grip, and power is constant, then you get these idealized equations:

a = acceleration
v = velocity
t = time
f = force
m = mass
p = power

a = dv/dt = f / m = f v / (m v) = p / (m v)
v dv = (p/m) dt
1/2 v^2 = p t / m
v = sqrt(2 p t / m)
p = v^2 m / (2 t)

subsituting the numbers above (except that 60 mph = 26.8224 m/s)
minimum power with no drag or losses and infinite grip = 14748.5434 watts = 19.7781225 horsepower