# Watts accelerating mass in a frictionless environment

1. Apr 25, 2015

### jFlower

< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

I have read every post I've found on here about this subject, but for some reason the answer is elusive - I suspect due to an issue with units.

Here's the problem.
vInitial = 0 meters / sec
Mass = 1600kg (about the weight of a car)
Power going into mass = 74600watts (100hp)
Frictional forces = 0

So, I calculate acceleration = 74600 / 1600, or 46.6 mps^2 <- this must be the problem

Then to apply the acceleration, I'm running a computer simulation running at 60hz, so time is 1/60, like so

timeStep = 1/60
velocity = velocity + acceleration * timeStep

Within 0.6 seconds or so velocity hits 100kph (27.7meters per second) which makes sense since ~0.6 times 46.6 = 27.7). What am I doing wrong?

Last edited by a moderator: Apr 25, 2015
2. Apr 25, 2015

### Dr.D

This must be the problem!! You got that correct!!

Power/mass has no significance at all.

Power is force*velocity.

You need to rethink the problem, getting your equations correct.

3. Apr 25, 2015

### jFlower

Thanks, it sounds like I'm on the right track to figuring this out. So does it help to say the object is a motorcycle (that weighs the weight of a car) and the watts are going into a single rear wheel with a diameter of 700mm? What would the velocity be for this - wheel angular velocity?

The only input measurement I have to work with is watts, and I'm not sure how to get a force value from that.

4. Apr 25, 2015

### Dr.D

Try this approach:

Change in kinetic energy = work done on the system

If the machine starts with zero velocity and goes up to some later velocity V2, we have

(1/2)*m*(v2^2 - v1^2) = work done = integral of power

Now v1=0, and if the power is constant (a doubtful assumption), then

(1/2)*m*v2^2 = P*t

where P = power and t = elapsed time. Now, this would suggest that the longer you let it run, the faster it will go, but we know this is not true. What has been omitted is the friction (windage, internal friction, rolling resistance, etc), but at least as a start, over a short time interval, this is approximately true.

5. Apr 25, 2015

### jFlower

So, taking "(1/2)*m*v2^2 = P*t" and trying to solve for acceleration (in m/s^2), I'd want to solve for V2 with a t of 1.0 right? So: acceleration = sqrt((2 * powerInWatts) / mass) ?

And then I'd update my velocity normally, v = v+acceleration * timeStep

So now an object with the weight of a car and 100hp goes 0->100kph in 2.27 seconds with no friction. Still seems a bit faster than I'd expect. Did I miss anything else?

PS, In the case of my equations, "powerInWatts" will be the power left over after drag is subtracted out. I've removed all that from the math for the moment to track down my issue, as I was certain in the acceleration calculations and not my drag calculations.

6. Apr 25, 2015

### PeroK

Constant power implies decreasing acceleration.

To maintain a constant acceleration, you need increasing power.

You need to rethink your model.

7. Apr 25, 2015

### jFlower

The power
The current power input into the system is given to me in watts. I'm not trying to maintain constant acceleration, I'm simply trying to calculate instantaneous acceleration given instantaneous watts and accelerate the object based on that for a very small time step (1/60th of a second).

If there is no friction I don't see why it wouldn't just linearly accelerate at the same rate forever, so I'm not sure I see what you're trying to say.

8. Apr 25, 2015

### PeroK

Even without friction, it won't linearly accelerate forever. The acceleration will decrease as you increase speed.

If you're trying to calculate the time to do 0-100kph, you don't need to calculate instantaneous acceleration.

Instead, you get it from the Energy/velocity against time equation that Dr D gave you.

9. Apr 25, 2015

### haruspex

As posted above, power = force * velocity. At each time step, you have the current velocity so can compute the current force. From that you get the instantaneous acceleration.
But note there is a problem right at the start. With zero velocity it seems you will have infinite acceleration. How do you think that should be resolved?

10. Apr 25, 2015

### jFlower

I'm not trying to calculate 0-100kph, I'm trying to calculate instantaneous acceleration at any given moment given an input wattage (which is really the remaining wattage after all drag is taken out). The 0-100kph calculation is just an easy sanity check on the maths.

Doesn't a rocket in space with a fixed given force on it accelerate at a constant rate if it never loses mass?

11. Apr 25, 2015

### haruspex

Sure, but according to your OP the force is not constant, the power is.

12. Apr 25, 2015

### PeroK

You can calculate the velocity, then differentiate that to get acceleration

As everyone is at pains to point out Power is not the same as Force. To maintain a constant force, you need increasing power.

13. Apr 25, 2015

### jFlower

Right, so at a near stand still power = mav. So, 74600 = 100 * A * 0.00000000001, or A =74600/(100 * 0.000000000001) when the body is nearly at a stand still. This equals a huge acceleration which would rocket the object forward so quickly initially that we'd hit 100kph in a hundredth of a second??

14. Apr 25, 2015

### haruspex

Exactly, so something is going wrong... what do you think the issue might be?

15. Apr 25, 2015

### jFlower

Well, I don't know, that's sorta why I posted in the first place :) I haven't taken a physics class in over 20 years and don't exactly have the textbooks handy anymore.

16. Apr 25, 2015

### OldEngr63

OK, then go back to what DrD said:

If you differentiate that equation with respect to time, you get

m*a = F = P/v

where F = force, P = power, and v = velocity. The instantaneous acceleration is then

a = P/(v*m)

which will work just fine, provided v <> 0; it blows up for v = 0. This reflects the fact that at zero speed, you simply cannot put any power into the system. At 0 speed, all of your power, whatever there is, is going somewhere else, internal friction in the engine, heat generation in the clutch, pumping losses in the fluid system, etc.

There; you have the acceleration in terms of the power. Carry on.

17. Apr 25, 2015

### haruspex

Two things....
Numerical methods always face problems when one of the variables becomes unbounded. Consider $\int_0^1x^{-\frac 12}.dx$. The integral converges, but imagine trying to do it numerically with fixed size increments in x.
The second problem is that in reality no engine can provide constant power at all speeds; there are practical limits on the torque that can be generated.
For the frictionless case, the easy bypass is to use energy, as Dr.D suggested. After time t, you know how much energy has gone in, so you know the velocity. No need to get involved with forces.
If you also want to know distance covered, you can either solve the differential equation or use the velocity calculated from energy in a numerical time step solution.

But I see that ultimately you want to allow for drag, so you cannot deduce velocity from time and power. If the drag is zero when velocity is zero, you could use the energy method for the first time step. Once you have a nonzero velocity, numerical methods should be fine, but you might need to use a very small time increment at first, while the speed is low. Ideally you should do some analysis to figure out the optimal point in time at which to switch over.
If you wish to include frictional forces which are nonzero at rest, it's different again.

18. Apr 25, 2015

### jFlower

Ok, thanks for everybodies help. Obviously Force and Power were (are?) confused in my old mind and I think I have reasonable results now.

Every 60th of a second I do:
timeStep = 1/60
inputWatts = data from some external source (joystick throttle)
WattsAfterDrag = inputWatts - dragFromRollingResistance - dragFromCdARho - dragFromGravity;
Force = WattsAfterDrag / VelocityInMetersPerSec
Acceleration = Force / VehicleMassInKG
VelocityInmetersPerSecond = VelocityInMetersPerSecond + Acceleration * timeStep

Then 1/60th of a second later it all happens again. And again, forever.

My original post had a frictionless 1600kg w/100hp going 0..100kph in 0.6 seconds, and now it's doing it in 8.2 seconds. If I missed anything obvious I'm all ears. I'm here to learn!

19. Apr 25, 2015

### haruspex

As in, going uphill?
How do you deal with the fact that the initial velocity is 0?

20. Apr 26, 2015

### jFlower

Yes, going up (or down) a slope is factored in with that coeffecient. For velocity, I clamp it so that even if it's zero it acts like some small value (0.01m/s)

To the 150 lines of code I had in my physics simulation, this entire thread resulted in me putting a single " / VelocityInMPS" in and solving the problem. Thanks again all.