Quarter mile 10s car minimum power requirement

[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).

 

[ivtec259]

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.

I seem to be lost in the imperial system. I am going to stick to the metric system then, and those who want to can convert to other units if they want to. I just thought it would be polite to show the mass in lbm or whatever is the right unit for mass.

 

[ivtec259]

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

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

That would be a good idea, thanks for the velocity formula.

 

[rcgldr]

I seem to be lost in the imperial system.

1 lbm = (1/32.174) slug, based on average gravitational acceleration = 32.174 ft / (sec^2). Doing the entire problem in imperial units, and reducing amax from 10 m / sec^2 to 9.80665 m / sec^2 (== 32.174 ft / sec^2), I get 538.8 imperial horsepower.

Alternate approach (separating stage 1 and 2) (the earlier equations are also shown). In these equations, m is mass, a₁ is constant acceleration for stage 1, f₁ is constant force for stage 1, t₁ is the time at the end of stage 1, start of stage 2, v₁ is velocity at the end of stage 1, start of stage 2, x₁ is the distance at the end of stage 1, start of stage 2, p is the power at the end of stage 1, and constant for all of stage 2, v is velocity during stage 2, and x is (total) distance during stage 2.

$v_1 = a_1 \ t_1$
$p = f_1 \ v_1 = m \ a_1 \ v_1 = m \ a_1 \ a_1 \ t_1$
$t_1 = \frac{p}{m \ a_1^2}$
$x_1 = \frac{1}{2} a_1 \ t_1^2$
$v = \sqrt{\frac{2 p}{m}(t-t_1) + v_1^2} = \sqrt{\frac{p}{m}(2 t - \frac{p}{m \ a_1^2})}$
$x = \frac{m}{3 p}(v^3 - v_1^3) + x1 = \frac{m}{3p} \left( \frac{p}{m} (2t- \frac {p}{m \ a_1^2} ) \right)^{3/2} + \frac{p^2}{6 \ m^2 \ a_1^3}$

 

[jbriggs444]

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.

That's 106,485 poundals, not pounds. The "poundal" is the force required to accelerate one pound mass at a rate of one foot/sec². It is roughly 1/32 of a pound force.

Yes, the car would weigh 106,485 poundals.

 

[OldYat47]

I was gone for a few days but wanted to get back on this. A constant value for horsepower results in reducing acceleration, inversely proportional to velocity. Constant acceleration requires constantly increasing horsepower, proportional to velocity. But the equations presented solve for one value of horsepower with one acceleration value. Both cannot be constant.

t1 = p / (m*a^2) has an infinite number of values for p and a that "work".

 

[ivtec259]

I was gone for a few days but wanted to get back on this. A constant value for horsepower results in reducing acceleration, inversely proportional to velocity. Constant acceleration requires constantly increasing horsepower, proportional to velocity. But the equations presented solve for one value of horsepower with one acceleration value. Both cannot be constant.

t1 = p / (m*a^2) has an infinite number of values for p and a that "work".

You are absolutely correct. But P and a are never constant at the same time in that formula. The formula is only valid for t>t1 (stage 2) where P is constant and a is not.
Maybe it is more pedagogic to name a a_max and P P_max instead.

The acceleration value in the formula is not the acceleration in stage 2, it is the acceleration in stage 1. Thus it is a constant.

 

[rcgldr]

t1 = p / (m*a^2) has an infinite number of values for p and a that "work".

m (1500 kg) and a (10 m / s^2) are constants defined for stage 1. I updated post #38, using subscripts for the constants (except for m == mass) which separates the math into stage 1 and stage 2.

 

[cjl]

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.

You seem very confused about units here.

3307 pounds weight = 1500 kgf = about 14.7kN.
3307lbf in Earth's gravitational field means the object has a mass of 3307lbm
1500kgf in Earth's gravitational field means the object has a mass of 1500kg
3307lbf/32.2 (gravity in ft/s^2) = 102.7 slugs (not pounds)
1500kgf/9.81 doesn't really return a meaningful result in any way

 

[OldYat47]

OK, so acceleration is a single defined value. Now you calculate a single value for power at, apparently t1. Since speed is increasing acceleration must be decreasing (constant power) so that says that as the tires "hook up" (less spinning) the car's forward acceleration slows. That's unlikely. And if acceleration is defined as a single value then power must be lower for 0<t<t1 (proportional to velocity) than for t1, so power at t1 is not a minimum.

 

[cjl]

OK, so acceleration is a single defined value. Now you calculate a single value for power at, apparently t1. Since speed is increasing acceleration must be decreasing (constant power) so that says that as the tires "hook up" (less spinning) the car's forward acceleration slows. That's unlikely. And if acceleration is defined as a single value then power must be lower for 0<t<t1 (proportional to velocity) than for t1, so power at t1 is not a minimum.

Acceleration has a maximal value that depends on the grip available from the tires. To maintain constant acceleration, power must increase as speed increases, so at some point, the limiting factor is no longer the grip available, but it is instead the engine power. At no point during this process are the tires slipping much - this assumes maximal tire grip at all times.

As for the "minimum power", you're right that power is lower when traction limited, but in order to complete the overall quarter mile in ten seconds, the car must have the power calculated at T1, so this is indeed the minimum power to complete a quarter mile in ten seconds.

 

[OldYat47]

3307lbf in Earth's gravitational field means the object has a mass of 3307lbm
1500kgf in Earth's gravitational field means the object has a mass of 1500kg
3307lbf/32.2 (gravity in ft/s^2) = 102.7 slugs (not pounds)
1500kgf/9.81 doesn't really return a meaningful result in any way

3,307 lbf (pounds force. Let's call that scale weight) means the object has a mass of 102.7 lbm (pounds mass). That would be about 2.63 slugs, not 102.7 slugs.
3,307 pounds scale weight (lbf) converted to kilograms scale weight would be 1,503 kilograms scale weight (Newtons). Note that Newtons is a force, mass times acceleration. Divide by 9.81(acceleration of Earth's gravity) and you get kilograms mass, about 153.2 kg.

 

[jbriggs444]

3,307 lbf (pounds force. Let's call that scale weight) means the object has a mass of 102.7 lbm (pounds mass). That would be about 2.63 slugs, not 102.7 slugs.

Utterly wrong.

An object whose mass is 3307 pounds mass has a weight of 3307 pounds force under an acceleration of one standard gravity.

Google will dig up a ton of references. Here is one from NIST: http://www.nist.gov/calibrations/upload/j61jab.pdf page 40:

"the standard pound force being defined as the force acting on a one-pound mass in a gravitational field for which the acceleration of free fall is 9.80665 m/s²"

 

[OldYat47]

http://www.engineeringtoolbox.com/mass-weight-d_589.html

Utterly? Now you've got me smiling. This link will take you to a good explanation of mass and weight, including SI and English units and how to calculate mass from weight and vice versa.

 

[OldYat47]

So using that minimum power for stage 1, can anyone calculate a similar power for stage 2? And can anyone say what the velocity and position along the track at the end of stage 1 are (numbers please).

 

[ivtec259]

So using that minimum power for stage 1, can anyone calculate a similar power for stage 2? And can anyone say what the velocity and position along the track at the end of stage 1 are (numbers please).

Sure, not a problem.

Let m = 1500 kg, a = 10 m/s^2, P = 535 hp = 393000 W. Then velocity (v1) = 26.2 m/s, position (x1) = 34.2 m and time (t1) = 2.62 s.

The power for stage 2 is the constant 535 hp, so I'm not sure what you mean by "calculate a similar power for stage 2".

If you want to calculate the power you have to iterate with the formula until you get x = a quarter mile.

 

[jbriggs444]

Utterly? Now you've got me smiling. This link will take you to a good explanation of mass and weight, including SI and English units and how to calculate mass from weight and vice versa.

Your reference disagrees with you:

"In the EE system 1 lb of force will give a mass of 1 lbm a standard acceleration of 32.17405 ft/s²"

 

[ivtec259]

3,307 lbf (pounds force. Let's call that scale weight) means the object has a mass of 102.7 lbm (pounds mass). That would be about 2.63 slugs, not 102.7 slugs.
3,307 pounds scale weight (lbf) converted to kilograms scale weight would be 1,503 kilograms scale weight (Newtons). Note that Newtons is a force, mass times acceleration. Divide by 9.81(acceleration of Earth's gravity) and you get kilograms mass, about 153.2 kg.

Whaaaaat. No. I think you are the one that needs to check again and then recheck, kind of like in Transformers 1 when that helicopter comes back from the dead.

3,307 pounds scale weight (lbf) is a force. If you want to convert to kilograms, you first need to convert you force into a mass. But fortunately that is easy. It will be 3,307 pounds mass (lbm) on earth. Then you just google 3307 lbm to kg.

 

[OldYat47]

That doesn't disagree with me at all. Note the use of a proportionality constant. Note that 1 lbf is defined as the force that accelerates 1 lbm 32.174 ft/sec^2. Note that gc is defined as (1 lbm)*(32.174*ft/sec^2)/(1 lbf). Note that the slug is still defined as weighing 32.174 pounds.

I'm old enough to have used all these systems in the past. You've had to be very careful, less so today. So I maintain that the car's mass is 102.78 slugs, not 3,307 pounds of anything.

 

[OldYat47]

Picture me slumping in my chair and shaking my head sadly. Suppose you put a weight on a kilogram scale and it weighs 9.81 kg. That's not entirely correct though commonly used. It actually weighs 9.81 Newtons. It has a mass of one kilogram. Refer to the linked site above.

Suppose you take a 2.2 pound weight and put it on the scale above. It will read 1 kilogram. That's it's weight, not it's mass.

Please refer any of this to your local physics instructor.

 

[ivtec259]

That doesn't disagree with me at all. Note the use of a proportionality constant. Note that 1 lbf is defined as the force that accelerates 1 lbm 32.174 ft/sec^2. Note that gc is defined as (1 lbm)*(32.174*ft/sec^2)/(1 lbf). Note that the slug is still defined as weighing 32.174 pounds.

I'm old enough to have used all these systems in the past. You've had to be very careful, less so today. So I maintain that the car's mass is 102.78 slugs, not 3,307 pounds of anything.

Picture me slumping in my chair and shaking my head sadly. Suppose you put a weight on a kilogram scale and it weighs 9.81 kg. That's not entirely correct though commonly used. It actually weighs 9.81 Newtons. It has a mass of one kilogram. Refer to the linked site above.

Suppose you take a 2.2 pound weight and put it on the scale above. It will read 1 kilogram. That's it's weight, not it's mass.

Please refer any of this to your local physics instructor.

Where are they teaching you this? Just curious. Because last I checked with my physics instructor, a kilogram is a unit of mass and nothing else. If you want to talk about forces, then talk Newton. And how can you believe that I meant a force, when I wrote m = 1500 kg. I did not write F = 1500 kgf.

You can maintain what you want, but in the USA I know they use pounds mass (lbm) for measuring the mass of most things, including cars.

 

[OldYat47]

You will be shocked if you travel outside this country to discover that things like flour, rice, coffee and dozens of other commodities are sold by the kilogram or gram weight. Even cars are quoted as weighing a number of kilograms. It's sloppy, but there you go. So take my post above to your physics instructor and see what he or she says about it. You can find scales with Newton readouts, but the vast majority are in grams and kilograms weight, not mass.

Lastly, and repeating myself, a car that weighs 3,307 pounds has a mass of about 153 kg.
Moderator note: The sentence above is patently untrue. An object whose mass is 153 kg would have a weight of about 336.6 lb.

I have to leave for today, but tomorrow I will work on the numbers presented above as solutions for t1, t2, etc.

 

[ivtec259]

You will be shocked if you travel outside this country to discover that things like flour, rice, coffee and dozens of other commodities are sold by the kilogram or gram weight. Even cars are quoted as weighing a number of kilograms. It's sloppy, but there you go. So take my post above to your physics instructor and see what he or she says about it. You can find scales with Newton readouts, but the vast majority are in grams and kilograms weight, not mass.

Oh so you mean that saying for example that "My car weighs 1500 kg" is wrong and you should say "My car's mass is 1500kg"? And by the way, I do not live in the US.

 

[berkeman]

Thread closed for Moderation...

 

[berkeman]

This thread will remain closed. You guys sound like a couple of 10 year olds arguing over marbles. This will not be tolerated at the PF. Please don't post with this kind of attitude here again. Fair warning.