Quarter mile 10s car minimum power requirement

[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 a_max (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!

 

[berkeman]

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 a_max (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!

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?

 

[ivtec259]

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?

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.

 

[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.

 

[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.

 

[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 a_max.

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

 

[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

 

[russ_watters]

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.

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.

 

[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}} ##

 

[ivtec259]

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}} ##

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

 

[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.)

 

[ivtec259]

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.)

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.

 

[FactChecker]

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.

I'll buy that.

 

[rcgldr]

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

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 a_max, and t1 depends on p, a_max, and m.

 

[ivtec259]

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##

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

 

[OldYat47]

Let's start with the most basic stuff:
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?

 

[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.

 

[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.

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

 

[rcgldr]

##v = \sqrt {2t \frac{P_{max}}{m} + a_{max}^2t_1^2 - 2t_1 \frac{P_{max}}{m}}##

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 a_max, and t1 depends on p, a_max, and m.

 

[Khashishi]

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

 

[rcgldr]

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

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

 

[ivtec259]

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

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

 

[rcgldr]

But t₁ is not an independent variable since ##t_1 = \frac{P_{max}}{m \ a_{max}^2}##

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

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} ##

 

[ivtec259]

So substituting for t₁ in Khashishi's last equation should resolve this issue.

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

 

[ivtec259]

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

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} ##

Well there is only one P and it is a constant, so you can call it whatever you want but P_max 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.

 

[Khashishi]

Of course, any real car, with drag and wheel inertia, will need much more power.

 

[rcgldr]

In the case of a motorcycle, it's even worse due to the relatively high ratio of aerodynamic drag versus weight compared to a car.

1999-2007 Suzuki Hayabusa, about 10 second 1/4 mile, 143 mph (230 kph) total weight (with rider) about 700 lbs (317 kg), rear wheel power about 156 hp (116 kw). This would translate into about 730 hp for a 3300 lb car, but for a car it would be less. A Corvette ZR1 with drag radials (motorcycles have sticky tires), with about 535 rwhp, runs around 10.5 seconds around 131 mph. I ran some online calculators and a 10 second 1/4 mile with a 3300 lb car (including driver), would need about 650 rear wheel horsepower, but minimum power to do this with sticky tires (drag radials) and a good driver would probably be less.

For a bike comparison, the Kawasaki ZX-14R, 740 lbs (with rider), 196 rwhp, 9.5 seconds, 150 mph with a professional rider.

 

[OldYat47]

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

Your numbers have me mystified. Working backwards, accelerating at 10 m/s^2 for 10 seconds will cover about 1,640 feet, not a quarter mile (1,320 feet). Next, 3307 pounds is a weight corresponding to about 153.4 kg mass. Lastly, knowing mass, elapsed time, and distance, what formula gave you a value for power?

 

[ivtec259]

Your numbers have me mystified. Working backwards, accelerating at 10 m/s^2 for 10 seconds will cover about 1,640 feet, not a quarter mile (1,320 feet). Next, 3307 pounds is a weight corresponding to about 153.4 kg mass. Lastly, knowing mass, elapsed time, and distance, what formula gave you a value for power?

I am sorry if i was unclear in that post. I will try to make it more clear.
This is not constant acceleration the whole quarter mile, only until the power of the engine starts to be the limiting factor. a is the maximum acceleration limited by traction of the tires. 3307 US pounds is definitely 1500 kg. Check again.
The formula I used was the one written in the post, but it is impossible to solve for P as far as I know, so you have to try different P until you get x to a quarter mile.
Note that this formula is only valid for t>t1 and x>x1 (stage 2).
Stage 1 constant a. Stage 2 constant P.

 

[OldYat47]

So the 3,307 pounds is used to calculate available traction? And yes, 3,307 (pounds weight) = 1,500 (kilograms weight, or Newtons). But 3,307 (pounds weight)/32.2 = (pounds mass) and 1,500 (kilograms weight or Newtons) / 9.8 = 153 (kilograms mass). Those are the masses we are trying to accelerate down the quarter mile.

And in all my decades I have never seen a viable formula for using mass, distance and time to calculate power. There is a good reason for that. Please show the specific equation and how the variables were plugged in.

 

[ivtec259]

So the 3,307 pounds is used to calculate available traction? And yes, 3,307 (pounds weight) = 1,500 (kilograms weight, or Newtons). But 3,307 (pounds weight)/32.2 = (pounds mass) and 1,500 (kilograms weight or Newtons) / 9.8 = 153 (kilograms mass). Those are the masses we are trying to accelerate down the quarter mile.

And in all my decades I have never seen a viable formula for using mass, distance and time to calculate power. There is a good reason for that. Please show the specific equation and how the variables were plugged in.

No it is not. a is assumed to be 10 in this case, which corresponds to about 1g of acceleration. This is reasonable for normal car tires.
I thought I was clear that I am talking about mass, not force. m = mass. lbs is a unit of mass and lbf is a unit of force right? Anyway 3307 lbs is 1500 kg.

Lets say you know the max acceleration of your car, and you know the mass of the car+you, and you want to race a quarter mile in 10 seconds, but you don't know how many horsepower you will need.

##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}##

Then you will plug your a, m and t into this formula, and try different P's until you get x = a quarter mile. Then you know that it is impossible to get away with a smaller P than that, and you will probably need 15-20% more power depending on how well you shift etc. Make sense?

Se plot below where you can compare to a constant force acceleration and a constant power acceleration. Note the CA-CP plot is only valid for t>t1 ~ 2.62 s.https://dl.dropboxusercontent.com/u/97064125/plot.png

 

[OldYat47]

You are making a bad assumption that the 3,307 lbs is mass because it doesn't say lbf. How reasonable is 3,307 lbm? That would mean the car would weigh 106,485 pounds. Not critical to this discussion.

And there's something odd about the dimensional analysis of the formula. You have (power*time/mass)^3/2 X (mass/power) and wind up with distance. I don't get that. Can you check that?

 

[rcgldr]

Note that this formula is only valid for t>t1 and x>x1 (stage 2). Stage 1 constant a. Stage 2 constant P.

For graph purposes, you could use a constant acceleration formula for stage 1 and the derived formula for stage 2. For velocity:

## v(P,m,t,a) = \sqrt{\frac{p}{m}(2 t - \frac{p}{m a^2})} ##

And there's something odd about the dimensional analysis of the formula. You have (power*time/mass)^3/2 X (mass/power) and wind up with distance. I don't get that. Can you check that?

Note that power = force time speed, and force = mass x acceleration. You can check this by substituting the units for power to be kg m^2 / sec^3 .

 

[OldYat47]

That's not the formula I was talking about.
I'm referring to the one that has the (3/2) fractional exponent.

 

[rcgldr]

You are making a bad assumption that the 3,307 lbs is mass because it doesn't say lbf. How reasonable is 3,307 lbm?

3307 lbm would be 102.8 slugs.

And there's something odd about the dimensional analysis of the formula. You have (power*time/mass)^3/2 X (mass/power) and wind up with distance. I don't get that. Can you check that?

That's not the formula I was talking about. I'm referring to the one that has the (3/2) fractional exponent.

## 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} ##

Looking at the units, using kg for mass, and m == meters for distance:

## P => \frac{kg \ m^2}{ sec^3} ##

## \frac {P}{ma^2} => \frac {kg \ m^2 / sec^3}{kg \ m^2 / sec^4} => sec ##

## \left( \frac{P}{m} (2t- \frac {P}{ma^2})\right)^{3/2} => \left(kg \ sec \ m^2 / (kg \ sec^3) \right)^{3/2} => \left(\frac{m^2}{sec^2}\right)^{3/2} => \frac{m^3}{sec^3}##

## \frac{m}{3P} \left( \frac{P}{m} (2t- \frac {P}{ma^2} ) \right)^{3/2} => \frac{kg \ sec^3}{kg \ m^2} \frac{m^3}{sec^3} => m ##

## \frac{P^2}{6 \ m^2 \ a^3} => \frac{kg^2 m^4}{sec^6} \frac{sec^6}{kg^2 m^3} => m ##

Follow the same logic for the formula I posted for velocity and you end up with m/s (meters / second).