# I Quarter mile 10s car minimum power requirement

1. Mar 31, 2016

### ivtec259

Hi everybody!

I am facing a problem in calculating the theoretical minimum power requirement P of accelerating a mass m a distance x in a time t in a two-stage acceleration process.

Stage 1: Constant force acceleration amax (maximum grip of the tires of a car limits acc.)
Stage 2: Constant power acceleration a (maximum power of the cars engine limits acc.)

Calculating either stage by itself did not prove that difficult, but calculating stage 2 when the initial velocity v1 is not equal to zero was where I hit a roadblock.

I found this thread that may be of help with stage 2: https://www.physicsforums.com/threads/formulas-for-constant-power-acceleration.639913/

I would very much appreciate any help!

2. Mar 31, 2016

### Staff: Mentor

Welcome to the PF.

What is the context of your question? Are you working on a personal project? Self-studying physics? Taking a physics class?

What is your physics background? Are you in school now? Have you studied calculus yet? How about calculus-based physics?

3. Mar 31, 2016

### ivtec259

Thank you berkeman. This is a personal project you could say. I am a mechanical engineer. I have studied calculus. Not sure about calculus-based physics, but probably yes.

Last edited: Mar 31, 2016
4. Apr 1, 2016

### Kevin McHugh

You do need a mimimum of HP to get to 10 s in the quarter. However, gearing has a lot to do with transferring torque to the pavement. Depending on the weight of the vehicle, you will need close to 600-700 hp to get to 10 s. I have a 3400 pound chevelle with a 475 hp motor. With 373 gears out back, I was mid 12s in the quarter. I would suggest 410 - 456s out back to launch. You will be screaming through the traps though!

If you are trying to calculate the acceleration through the run, you will need to consider gear changes. For every gear change, the corresponding rpm drop will affect the power, and acceleration of the vehicle, until you get back into the power band of your cam, i.e. your acceleration will not be constant through the run.

5. Apr 1, 2016

### bobsmithofd

From my experience, weight, weight transfer capability, wheel hop, transmission/clutch efficiency, tire friction, are additional factors as you probably already know. Depending on the type of aspiration, you know you can boost horsepower - with lots of little tricks folks seem to have forgotten. my 57 ford, 2700lbs with me in it, turned consistent 10.9xs. I never put the car through a HP/Torque test, but I estimated the power at rear wheels in the 350 range, and the torque in the 250 ft/lbs range at launch.

6. Apr 1, 2016

### ivtec259

Thank you for your input guys. You mention a lot of practical matters that affect real world performance and some good examples with fossil cars.

But this was supposed to be a purely theoretical exercise that can give a hint to the absolute minimum power requirement as a function of m and amax.

Maybe there is no analytical solution, but there should at least be a way to calculate this with an iterative process.

Just for fun, the minimum power required with:
Constant force acceleration: P = 4*m*x^2/t^3
Constant power acceleration: P = 9*m*x^2/8/t^3

Last edited: Apr 1, 2016
7. Apr 1, 2016

### Kevin McHugh

Bear in mind, acceleration through a quarter mile is an average. A dragster requires an average acceleration of c.a. 44 m/s2 to go from 0 to 330 mph in 4.5 seconds. You can see instantaneous acceleration here: http://www.mfes.com/accel.html

8. Apr 1, 2016

### Staff: Mentor

I'm a mechanical engineer too and would typically do something like this with a spreadsheet. It avoids the math and makes changing variables easy.

9. Apr 1, 2016

### rcgldr

For stage 2, continuing from that previous thread:

$t = \sqrt[3] {\frac{9\ m\ x^2}{8\ p}}$

$t^3 = {\frac{9\ m\ x^2}{8\ p}}$

$p = {\frac{9\ m\ x^2}{8\ t^3}}$

Last edited: Apr 1, 2016
10. Apr 10, 2016

### ivtec259

Yes but how do i continue from there? I already derived that equation, and it only works when the initial velocity is zero right?

11. Apr 10, 2016

### FactChecker

Don't forget that the aerodynamic downforce starts to play a large part in the traction at the higher speeds. This is more true for a top fuel dragster, but it may become important for a sub 10 second quarter mile. (I don't think it's important at 10 seconds, but it's something to check.)

12. Apr 10, 2016

### ivtec259

From what I understand, the traction is only limited at lower speeds. At higher speeds the power of the engine is too low to create wheel-spin in a 10s car, hence the 2-stage acceleration that i talked about.
But sure in top fuel dragsters i can imagine there is a case for aerodynamic down force since there is so much power and so low weight.

13. Apr 10, 2016

### FactChecker

14. Apr 10, 2016

### rcgldr

I started too far down from the prior thread. I think you would need to start here, where v0 is the initial velocity for stage 2 (in these equations, t is really t - t0, x is really x - x0).

$v = \frac{dx}{dt} = \sqrt {\frac{2\ p\ t}{m}} + v_0$

$x = \sqrt {\frac{8\ p\ t^3}{9\ m}} + v_0 \ t$

update - Khashishi explains below why this doesn't work. The issue is that v_0 isn't an arbitrary constant, but depends on t1 and amax, and t1 depends on p, amax, and m.

Last edited: Apr 14, 2016
15. Apr 10, 2016

### ivtec259

Thank you for that! I will try to do it myself from there.

16. Apr 11, 2016

### OldYat47

You cannot directly connect power and acceleration or power and torque. Sample problem #1: You have a car weighing 3,174 pounds. It's accelerating at a constant 10 ft/sec^2. What is the power being delivered to the drive wheels? Sample problem #2: An engine is producing a torque of 100 pound feet. What's the horsepower?

Remember that power is a rate. It doesn't do anything, it's a measure of how fast something is happening. For example, a constant force on a mass produces constant acceleration (ignoring friction and drag). Constant power on a mass produces linear decreasing acceleration inversely proportional to the increase in speed (ignoring friction and drag). Twice the speed, half the acceleration.

Sample problem #3: You want your car to accelerate at a constant 26.4 ft/sec^2. How much power do you need?

17. Apr 12, 2016

### Khashishi

Power is just the change in energy over time. Energy is $1/2 mv^2$, so power is just the derivative $mv \frac{dv}{dt}$
So constant power acceleration is just
$P = mv \frac{dv}{dt}$
Letting $\frac{dv}{dt} = a$
$a = \frac{P}{mv}$
So, if you plot $a$ versus $v$, you get a straight horizontal line from 0 to $v_1$ and then $a$ goes down inversely with $v$ for values above $v_1$.
Clearly, $v_1 = \frac{P_{max}}{ma_{max}}$

Now, you want the value of $P_max$ such that $x(t_f) = x_f$
Let $t_1$ be the time when stage 2 is reached.
$v(t) = a_{max}t$, for $t<t_1$
So set $v_1=\frac{P_{max}}{ma_{max}} = a_{max}t_1$
$t_1 = \frac{P_{max}}{ma_{max}^2}$
$x_1 = \frac{1}{2} a_{max} t_1^2$
Now find an expression for $x(t)$ for $t > t_1$
There are some tricks you can use to solve the differential equation $a = \frac{P_{max}}{mv}$ which is valid for $t > t_1$
$va = \frac{P_{max}}{m}$
Note that $\frac{dv}{dt} = a$
and $\frac{d(v^2)}{dt} = 2 v \frac{dv}{dt}$ by chain rule
so $\frac{d(v^2)}{dt} = 2\frac{P_{max}}{m}$
$v^2 = \int{2 \frac{P}{m}} dt = 2t \frac{P_{max}}{m} + C$
$v = \sqrt {2t \frac{P_{max}}{m} + C}$
Now, solve for the integration constant C such that $v(t_1)=v_1$
$\sqrt {2t_1 \frac{P_{max}}{m} + C} = a_{max}t_1$
$C = a_{max}^2t_1^2 - 2t_1 \frac{P_{max}}{m}$
Now, plug it back in for v(t), valid for $t > t_1$
$v = \sqrt {2t \frac{P_{max}}{m} + a_{max}^2t_1^2 - 2t_1 \frac{P_{max}}{m}}$
Integrate to get x(t), valid for $t > t_1$. (you should probably check my math)
$x = \frac{m}{3P_{max}} \left({2t \frac{P_{max}}{m} + a_{max}^2t_1^2 - 2t_1 \frac{P_{max}}{m}}\right)^{3/2} + C_1$
Now just find a $C_1$ such that $x(t_1) = x_1$.
$\frac{m}{3P_{max}} \left({2t_1 \frac{P_{max}}{m} + a_{max}^2t_1^2 - 2t_1 \frac{P_{max}}{m}}\right)^{3/2} + C_1 = \frac{1}{2} a_{max} t_1^2$
$C_1 = \frac{1}{2} a_{max} t_1^2-\frac{m}{3P_{max}} \left({2t_1 \frac{P_{max}}{m} + a_{max}^2t_1^2 - 2t_1 \frac{P_{max}}{m}}\right)^{3/2}$
There, now you have x as a function of t, for $t>t_1$
$x = \frac{m}{3P_{max}} \left({2t \frac{P_{max}}{m} + a_{max}^2t_1^2 - 2t_1 \frac{P_{max}}{m}}\right)^{3/2} +\frac{1}{2} a_{max} t_1^2-\frac{m}{3P_{max}} \left({2t_1 \frac{P_{max}}{m} + a_{max}^2t_1^2 - 2t_1 \frac{P_{max}}{m}}\right)^{3/2}$
All you have to do now is solve for $P_{max}$. I leave that step to you.

18. Apr 12, 2016

### OldYat47

The original poster asked for a minimum value of power that would accelerate the car as described, in two "stages". I suggest that's the wrong question. Working with force (start with the engine torque curve) is simpler. If you select some constant power for either stage then the acceleration is not constant, decreasing linearly in proportion to the speed increase, just as your equations show. And what is the power at launch, when the car is on the starting line? It's zero but the car launches anyway. Why? Because power and acceleration are not directly tied together. If you select some constant acceleration then the power will continue to increase with the speed of the car, so "minimum power" is hard to define.

Take a look at my previous post and my three sample problems. I'm not saying you are incorrect, your math is good. But that the approach is a bit off.

19. Apr 12, 2016

### rcgldr

Since everything except P is known, why not treat them as constants, and just do the calculation for phase 2, where t = 0 and x = 0 correspond to the end of stage 1 and the start of stage 2, so that v = v0 at t = 0. This should simplify the equation to:

$v = \sqrt {2t \frac{P_{min}}{m}} + v_0$

update - Khashishi explains below why this doesn't work. The issue is that v_0 isn't an arbitrary constant, but depends on t1 and amax, and t1 depends on p, amax, and m.

Last edited: Apr 14, 2016
20. Apr 13, 2016

### Khashishi

rcgldr, you can't pull the $v_0$ out of the square root like that

21. Apr 13, 2016

### rcgldr

OK. Your final equation has two variables, t1 and pmin , so is there a fixed solution?

Last edited: Apr 13, 2016
22. Apr 13, 2016

### ivtec259

But t1 is not an independent variable since $t_1 = \frac{P_{max}}{ma_{max}^2}$

23. Apr 13, 2016

### rcgldr

So substituting for t1 in Khashishi's last equation should resolve this issue. Side note - this should be Pmin?

For the others here:

$v1 = a_{max} \ t_1$
$P_{max} = m \ a_{max} \ v_1 = m \ a_{max} \ ( a_{max} \ t_1) = m \ a_{max}^2 \ t_1)$
$t_1 = \frac{P_{max}}{m \ a_{max}^2}$

Last edited: Apr 13, 2016
24. Apr 13, 2016

### ivtec259

Yes I think so, and hopefully the expression will become more pretty once you simplify it.

25. Apr 13, 2016

### ivtec259

Well there is only one P and it is a constant, so you can call it whatever you want but Pmax is more suitable in this case I think, because the power will never be more than this.

I did some algebra and the best I could do was this:

$x(P,m,t,a) = \frac{m}{3P} \left( \frac{P}{m} \left(2t- \frac {P}{ma^2} \right) \right)^{3/2} + \frac{P^2}{6m^2a^3}$

The minimum power requirement to do a quarter mile in 10s is about 535 metric horsepower with the following numbers:

m = 1500 kg ( 3307 lbs )
a = 10 m/s^2

Thank you Khashishi and rcgldr for your help with the math.
Wolfram Alpha could not solve for P, so I don't think it is possible unfortunately.

Last edited: Apr 13, 2016