# Rockeet going from A to B with least fuel

1. Apr 3, 2009

### Georgepowell

I'm not sure I should be asking this question in Maths, but I think it is a maths question and I don't care about the physics behind it.

You are in some sort of simple space craft:

It has a mass of 1kg

To move, it has one rocket that produces a constant force of 10N in which ever direction you want. (this force does not stop)

If you start from rest at position A(0m,0m) and want to get to C(10m,-7m) via B(9m,13m). Then how do you rotate your rockets (10N force) throughout your journey to complete your journey in the fastest time possible?

There is no gravity and no air resistance.

Is this question do-able? Or is it very difficult? How would I even start?

2. Apr 3, 2009

### uart

I thinks it's a very interesting question George (I dont now how to do it - though I haven't thought about it much yet).

Note that if the force is constant then I think we can assume that the rate of fuel consumption should be constant and therefore the problem is of min fuel consumption is really just the same a that of minimum travel time. Here I'm assuming of course that we're not taking into account any mass decrease due to fuel loss.

3. Apr 3, 2009

### Integral

Staff Emeritus
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4. Apr 3, 2009

### quZz

Should we take into account that mass of the rocket is decreasing due to fuel loss? If so then what is the mass flow? I guess it is important since you cannot get it from the force only.

5. Apr 3, 2009

### uart

Yes that's exactly what I was thinking. It's significantly harder than the standard brachristochrone problem though.

BTW. Just for fun I sketched those coordinates in a tech drawing program. It turns out that they very nearly lie on a 10m radius circle. With 10N, 1kg and a 10m radius turn the maximum linear velocity is 10m/s. To accelerate to 10 m/s would take a distance of 5m if directly in a straight line, though further of course if you are trying to change direction at the same time.

Taking all the above into consideration I've made an "artists impression" of what the trajectory might look like. This was pretty unmathematical and just for fun, but I'll bet it's not too far from the actual optimal trajectory. I cant help it, it's the engineer in me. ;)

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6. Apr 5, 2009

### Georgepowell

I don't want to take into account for loss of fuel.

This is very interesting. I took a good crack at answering this question on my own. I didn't have a calculator or a computer so I couldn't do any research :grumpy:.

Here is what I did:
First I solved a more simple problem, If the rocket had to get from A to B and back to A again, It would accelerate towards B until it got half way, then it would accelerate back again until it reached A. This is obvious.

Then I considered what possible paths the rocket could follow. It could follow a parabola, or a circular path (and a whole range of other ones that I couldn't be bothered considering). I went with the circular path initially. So then I simplified the problem again so It is already travelling at a speed V at a tangent to the circle. I wanted to find the force needed to keep it going in a circle. The variables I had were:

v = Velocity of rocket
F = Force applied to rocket
m = Mass of rocket

I wanted to find F in terms of v, r and m. Here is how I did it: (I am not sure this is correct as I did it without a calculator or computer)

I split the circle into a regular polygon. The angle from the radius between each corner of the polygon is theta(i.e. an angle of 360/6 would be a hexagon). Then I found the force needed to get the rocket from the initial corner to the second corner (in terms of v, r, m and the angle). I used F=ma and S=ut+0.5at². The force needed ended up as:

$$\frac{2mv(1-cos\vartheta)}{r(1-cos^2\vartheta)}$$

Putting theta as zero didn't work, so I sloppily worked out that the limit would be 1 (as theta tends to zero).

So then I had the force in terms of m, v and r. Like I wanted. But how about if it started still and I wanted the rocket to follow the path of a circle? The force would have to start at a tangent to the circle. If the angle between the tangent to the circle and the direction of the force was theta, then (using the formula I just figured out):

$$Fsin\vartheta=\frac{2mv^2}{r}$$

I can also work out the acceleration (around the circle, not towards the centre) in terms of F.

$$Fcos\vartheta=ma$$

I have theta in terms of a, and a in terms of theta, But I want theta in terms of time. So I use the integration and differentiation techniques I learnt in C4 (maths A-level module).

$$a=\frac{dv}{dt}$$

$$\frac{Fcos\vartheta}{m}=\frac{dv}{dt}$$

$$t=\frac{mv}{Fcos\vartheta}$$

I want to get rid of the v, because it is not a constant, like m and F. Before I wrote:

$$Fsin\vartheta=\frac{2mv^2}{r}$$

so:

$$v=\sqrt{\frac{Frsin\vartheta}{2m}$$

Putting that in in the other equation and rearranging I got:

$$\frac{cos(\vartheta)}{tan(\vartheta)}=\frac{mr}{2Ft^2}$$

I can't seem to solve that for theta. But If I could then I would go on to add a value of theta that was proportional to the distance travelled round the circle. So the angle is not relative to the tangent anymore.

If I could find theta in terms of time for the circle, then I could find a solution for my origional question (perhaps not the fastest time though).

7. Apr 5, 2009

### Georgepowell

BTW the ideal solution for this problem would be:

theta = F(t)

where F(t) is a continuous function of time and theta is the angle that the rocket is pointing.

8. Apr 5, 2009

### Hurkyl

Staff Emeritus
Allow me to break this problem into two parts:

Problem 1: You want to find the fastest way from point P to point Q, under the constraints:
* You start at P at rest
* You finish at Q with a specified velocity vector
* Your acceleration has fixed magnitude

Problem 2: given points A, B, and C in the original problem, choose the velocity vector through B that yields, using the solution to problem 1, the fastest route A->B->C.

(Note that the solution to problem 1 can be reversed to solve the problem of starting with a specified velocity and ending at rest)