Power required to reach X velocity in X time with drag

In summary, the conversation is about a program that calculates the power needed to accelerate a car to a given speed. The program uses various equations, but the problem arises when trying to account for drag. The participants discuss using numerical integration and optimization to find the correct power needed. They also mention considering aerodynamics and tire traction in the calculations. Overall, the conversation involves complex mathematical concepts and requires further learning and approximations to solve the problem accurately.
  • #1
A_User
5
0
Hello,
I wrote a program that adds force to a car like so:
Engine Force = Power / Velocity
Drag Force = -Velocity²
Net Force = Engine Force - Drag Force = Power / Velocity - Velocity²
I'd like to determine power based on how fast I want a car of a given mass to reach a given speed, for example, 0 to 20 m/s in 10 seconds for a 1000 kg car. Here is the equation I came up with to calculate the power needed without any drag:
Power = Force * Velocity = (Mass * Average Acceleration) * (Displacement / Time)

Using it to solve the example:

Mass = 1000 kg
Time = 10 s
Average Acceleration = 20 / 10 = 2 m/s
Displacement = 0.5 * 2 * 10² = 100 m
Power = (1000 * 2) * (100 / 10) = 20,000 W to go from 0 to 20 m/s in 10 seconds for a 1000 kg car
I'm not sure if that equation is sound but it calculates the power needed in a vacuum perfectly. The problems start when I try to account for drag:

Net Force = Power / Velocity - Speed²
Net Force + Speed² = Power / Velocity
Power = (Net Force + Velocity²) * Velocity
Power = [(Mass * Average Acceleration) + (Displacement / Time)²] * (Displacement / Time) = 21,000 W

This is incorrect. When I run my program with the calculated power, it takes the car ~11.5 s to reach 20 m/s, not 10 s. By trial and error, I determined that the correct value is ~23,500 W.

I've tried all day to find the correct equation with no success. What am I doing wrong?
 
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  • #2
A_User said:
What am I doing wrong?
Since the drag is nonlinear you cannot use averaged values. You have to do actual integration (eventually numerically, so it's good that it's a program).
 
  • #3
A.T. said:
Since the drag is nonlinear you cannot use averaged values. You have to do actual integration (eventually numerically, so it's good that it's a program).
That was my hunch, but I don't know how to go about it, I never learned integrals, much less how to solve them
 
  • #4
A_User said:
That was my hunch, but I don't know how to go about it, I never learned integrals, much less how to solve them
If you integrate over time numerically, you basically simulate the car in small time steps. The simplest approach for this is Euler-Integration. This gives you the duration to reach a given speed for a given power.

In an outer optimization loop you can then find the power needed for a given duration to reach a given speed. You can use Newton-Method here.

But note that the power that goes into acceleration cannot be constant throughout the acceleration. It's zero when v=0, for example.
 
  • #5
A.T. said:
If you integrate over time numerically, you basically simulate the car in small time steps. The simplest approach for this is Euler-Integration. This gives you the duration to reach a given speed for a given power.

In an outer optimization loop you can then find the power needed for a given duration to reach a given speed. You can use Newton-Method here.

But note that the power that goes into acceleration cannot be constant throughout the acceleration. It's zero when v=0, for example.
I'll give it a try, thanks
 
  • #6
When you consider aerodynamics, the acceleration becomes a function of velocity (equation 1b).

You also have to consider the maximum tractive force you can get. You correctly identified the one based on power. But at low speeds - as noted by @A.T. - you will have to consider the one based on tire traction, which can accept less than what the power can provide.

Then you can numerically integrate this way to find the distance traveled and the time taken to do it.

All of this theory is neatly wrapped up in this simulator.
 
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  • #7
If you consider the power of the engine to be constant you would have:
##F=ma=mv'=m\frac{dv}{dt}=P/v-v^2##
##\frac{mv}{P-v^3}dv =dt##
##\int_{0}^{V} \frac{mv}{P-v^3} dv=\int_{0}^{T} dt=T##

This integral is defined, but is a very difficult integral to calculate. The easiest way to solve this problem is to make use computing methods or series.

If you really want to see why the real result you can see here in Wolfram:
https://www.wolframalpha.com/input/?i=int(m+v/(P-v^3),+v)

That monstrous function is equal to T plus the value of that function in zero, which is ##\frac{-\sqrt{3}m\pi}{18 P^{1/3}} ##
To get P you need to substitute the values of m, v and T and find the root of the below equation for P, which I think is not defined. So, anyway, you will have to make use of approximations.
 
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  • #8
Thank you everyone for your very complete answers, looks like I got a lot of learning to do!
 
  • #9
jaumzaum said:
If you consider the power of the engine to be constant you would have:
##F=ma=mv'=m\frac{dv}{dt}=P/v-v^2##
##\frac{mv}{P-v^3}dv =dt##
##\int_{0}^{V} \frac{mv}{P-v^3} dv=\int_{0}^{T} dt=T##

This integral is defined, but is a very difficult integral to calculate. The easiest way to solve this problem is to make use computing methods or series.

If you really want to see why the real result you can see here in Wolfram:
https://www.wolframalpha.com/input/?i=int(m+v/(P-v^3),+v)

That monstrous function is equal to T plus the value of that function in zero, which is ##\frac{-\sqrt{3}m\pi}{18 P^{1/3}} ##
To get P you need to substitute the values of m, v and T and find the root of the below equation for P, which I think is not defined. So, anyway, you will have to make use of approximations.

Minor nitpick - your V^2 term should have a constant attached to account for the aerodynamics of the vehicle - it will basically be equal to 1/2*frontal area*air density*drag coefficient. Aside from that I agree though.
 
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  • #10
I just used OP's equations. In the original equation he used this constant equal to one. but I agree the general formula would have a k in there.
 
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  • #11
Yes I used a placeholder constant of 1 to make things simpler until I could figure out integration
 

1. What is the formula for calculating power required to reach a certain velocity in a given time with drag?

The formula for calculating power required to reach a certain velocity in a given time with drag is P = (m * v * (v + u))/t, where P is power, m is mass, v is final velocity, u is initial velocity, and t is time.

2. How does drag affect the power required to reach a certain velocity in a given time?

Drag is a force that acts in the opposite direction of motion and increases as velocity increases. This means that as velocity increases, the power required to overcome drag also increases. Therefore, the higher the drag, the more power is required to reach a certain velocity in a given time.

3. What are the units of power in the formula for calculating power required to reach a certain velocity in a given time with drag?

The units of power in the formula are watts (W), which is equal to joules per second (J/s).

4. How does mass affect the power required to reach a certain velocity in a given time with drag?

The mass of an object affects the power required to reach a certain velocity in a given time with drag because the greater the mass, the more energy is needed to accelerate the object to a certain velocity. This means that a heavier object will require more power to reach the same velocity in the same amount of time as a lighter object.

5. What are some factors that can affect the amount of power required to reach a certain velocity in a given time with drag?

Some factors that can affect the amount of power required to reach a certain velocity in a given time with drag include the mass and shape of the object, the density and viscosity of the fluid it is moving through, and the surface area of the object. Other factors such as air temperature and pressure can also have an impact on the amount of power required.

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