# Calculate speed at a distance from power

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1. Mar 12, 2016

### petterg

I'm failing at a setting up the formula for calculating speed of a car after a set distance, when accelerated with a given power. I hope someone is able to see my mistake.

The scenario is:
I'm having an under powered car, and the need to get a boat trailer up from a local boat ramp frequently. The problem is not the car power, but the clutch ability to get the power to the wheels. The car is able to get the trailer up, but I know the clutch will burn out pretty soon if I don't find a way to reduce the load when starting.

My plan is to create a kind of car ramp with the same angle as the slope of the boat ramp, and put them behind the trailers wheels (under water line) when backing down the ramp. That way the trailer wheels will start on leveled surface when the car does the hill start.

Now I'm trying to calculate how long theses car ramps need to be. To do so, I'll need to calculate the speed the car/trailer will have when the trailer wheels comes off the ramp.

The information I know is:
The hill angle is H=10 deg
Weight on wheels starting in the hill is M1=1600kg
Weight on wheels starting horizontally on the leveled car ramps is M2=1000kg

I assume the power transferred by the clutch when not fully engaged, the power that is accelerating the car/trailer is P=60kW.

The length of the ramps is S, and the goal is to find the speed (V) as a function of the driven distance S. At the starting point speed is 0.

Other variables used in formulas:
t: time
a: acceleration

The formulas I'm trying to use:
(1) Kinetic energy (0.5*m*v^2) at S: (energy of car + energy of trailer)
Ek = 0.5 * M1 * V^2 + 0.5 * M2 * V^2

(2) Potential energy (m*g*h) at S: (energy of car + energy of trailer)
Ep = M1 * g * sin(H) * S + M2 * g * 0 = M1 * g * sin(H) * S

E = Ek + Ep = P * t

(4) Acceleration vs distance
S = v0*t + 0.5 * a * t^2 = 0.5 * a * t^2

(5) Force vs acceleration (m*a)
F = (M1+M2) * a + M1 * sin(H) * g

(6) Power vs Force
P = F * S/t

-------------------
Trying to make use of the formulas:
(7) Rewriting (6)
F = P * t /S

(8) Rewriting (4)
a = 2 * S/t^2

(9) Insert (8) into (5)
F = (M1+M2) * 2 * S / t^2 + M1 * sin(H) * g

(10) Solving (7) and (9) for t
P * t /S = (M1+M2) * 2 * S / t^2 + M1 * sin(H) * g
P/S * t ^3 - M1 * sin(H) * g *t^2 - (M1+M2) * 2 * S = 0
Substituting x = M1 * sin(H) * g * S/(3*P) and y = x^3 + (M1+M2) * S^2 / P
t = (y+(y^2+(-x^2)^3)^(1/2))^(1/3) + (y-(y^2+(-x^2)^3)^(1/2))^(1/3) + x

(11) Inserting (1) and (2) into (3) and solving for V:
0.5 * (M1+M2) * V^2 + M1 * g * sin(H) * S = P*t
V^2 = 2*P*t / (M1+M2) - M1 * g * sin(H) * S / (M1+M2)
V = [ 2*P*t / (M1+M2) - M1 * g * sin(H) * S / (M1+M2) ]^0.5

Now, when inserting numbers into (10), say S=1m, I get that the longer it takes to drive that 1m, the faster the speed. That doesn't make sense! The same thing happens when inserting numbers in (11).

2. Mar 12, 2016

### Staff: Mentor

I doesn't look like homework, but it is still homework-like, so the homework forum is probably better.

Your acceleration is not constant, the approach you tried requires a constant acceleration.

The power the cludge can transmit depends a lot on speed, I don't think the assumption of a constant power is useful. Anyway, if you want to follow that:
Constant power needs a differential equation to be solved.
To lift the boat at a given speed v, you need a power of M1*g*v*sin(10°). The remaining power is used to accelerate both boat and car: a*(M1+M2)=P-M1*g*v*sin(10°)
Acceleration is the derivative of velocity, so we get a differential equation:
$$\frac {dv}{dt} = \frac{P}{M_1+M_2} - \frac{M_1 g v \sin(10°) }{M_1+M_2}$$
Introducing new variables b=P/(M1+M2) and c=(M1+M2)/(M1*g*sin(10°)) this equation can be written as
$$\frac {dv}{dt} = b - \frac{v}{c}$$
which can be solved by $$v(t)=bc(1-\exp(-t/c))$$
With the known velocity, you can find the time needed to travel 1 meter, and therefore the speed at that point.

Last edited: Mar 12, 2016
3. Mar 12, 2016

### haruspex

There could be a problem with your plan. The boat has to be deep enough in the water to float, right? If the trailer is on this levelling ramp, the car will have to move further into the water. Does the geometry work out?
Another option would be a short bungee rope, so the car can get a bit of a speed up before taking the full weight. Could combine this with blocks under the water to stop the trailer going deeper in consequence of the rope and ensure the rope goes slack. However, the rope would give less control over the trailer's line as you go up and down the ramp.

4. Mar 12, 2016

### petterg

Thanks MFB. I'll be 40 soon. It's about time I start doing some homework ;)

I'm aware a constant power is a major approximation. However, for the short distance in this calculation I hope it's not changing very much. I'm hoping building ramps like this will allow for the clutch to be fully engaged before the trailer wheels leave the ramps, hence be able to transfer more power when the full weight of boat starts to lift.

Your approach is way easier than my attempt. I like the idea of subtracting the power needed to lift the car from the total. I had to read up on differential equations to understand how you did it. However the solution V(t)=bc−exp−t/c doesn't behave like I'd expect when inserting numbers. I assume there's a need to add a constant to make V(0)=0. That constant has to be -bc+1, which makes the solution become V(t) = 1 - exp(-t/c).
Putting it all into an spread sheet calculating V and S for t in steps of 0.1s then show that after 2 seconds the speed is 0.877m/s and the traveled distance is 1.16m. That sounds reasonable. But if I change the weight of the car, i.e make it 600kg lighter, the acceleration is slower, only 0.818m/s and traveled 1.03m at the 2second mark.
Also, the speed after infinite time is 1m/s. Somewhere there must be an error.

@haruspex
Geometry works out - actually improves. This is salt water where trailer hubs/breaks has to be kept above waterline. There's a winch to pull the boat onto the trailer.

5. Mar 12, 2016

### Staff: Mentor

Should have been bc(1-exp(-t/c)), sorry.
The speed after infinite time is just given by the power needed on the incline, P/(M1*g*sin(10°)).

6. Mar 13, 2016

### petterg

Thanks. That changed things significantly. Changes in weight behaves as expected. Still there must be something wrong, cause it says the car will drive at 19.3m/s at the 2 second mark, with a top speed just above 22m/s. It reaches 50% of top speed at t=0.7s. In case my assumption of the power transferred is way off, I tried to set it at 1/6 of original. That makes the values of V look more reasonable, but it still reaches 50% of top speed at t=0.7s, which I don't believe is right. Is that just a sign that power really is very far from constant?

7. Mar 13, 2016

### Staff: Mentor

60 kW nearly at rest gives a huge acceleration. You get out what you plug in.

Real cars at that speed are limited by motor torque - their power will be proportional to the speed of the car, leading to a (nearly) constant acceleration.

8. Mar 13, 2016

### petterg

Thanks. I was learned more than expected from this. Guess I'll just have to make some ramps and see how it goes.