Celestia Forums

Dear all,

I need a little mathematical advice, unconnected with Celestia.

I'm not even sure I'm phrasing the question right, so bear with me.

Assume a space vehicle is in open space (no planets, gravitational influences of any kind)

Assume it has expended a part of its available Delta-vee (X), to reach a straight-line speed of Y.

Now, it wishes to make a course change of Z degrees. Is there a formula that can tell me how much Delta-vee it would need to expend in order to make that course change, so that the value of Y after the manouevre is the same as before?

I figure that to execute any course change greater than or equal to 90 degrees will require expending X*2, but I may be wrong (in fact, its highly likely ).

Any help would be greatly appreciated.

Thanks,

Cormoran

I think an analythical breakdown of vectors will get you out of this easily.

To reach an angle of phi degrees between 0 and 90 with the speed Y, you need to break Y*(1-cos(phi)) on your current bearing and accelerate Y*sin(phi) at an angle of 90 degrees. At the limit, 90 degrees (sharp turn), you need to apply -Y on your current bearing and then Y at a 90 degree angle, for an absolute grand total of 2Y.

Scytale,

Many thanks for the formulae. I would have thanked you sooner, but for some reason I've been unable to get into the site.

The results are slightly counter-intuitive until you think about them (Delta-vee x 2 for a 90 degree course change). I'm now having to convince the guy I did the spreadsheet for that these are accurate. I'm not doubting Scytale's information, but I'd be grateful if I could get independent confirmation of it, just so I can shut the guy up

I think they are right, Scytale knows they are right, can anyone else tell me they are right?

Next daft question:

Simple trigonometry (I told you my math was bad); Knowing just the three lengths of the sides of an irregular triangle, A, B and C, how do I work out the angles AB, BC and CA?

Anyway, thanks again Scytale

Cormie

I guess that would be the allmighty law of cosines.

Once again, exactly what I needed!

(Sorry, I wish I'd paid more attention in geometry...)

Thanks yet again, Scytale. I owe you a pint

Cormie

Hi Cormoran,
I have confirmed that Scytale is right. Using a circle in the x,y plane, think of two vector velocities Y1 and Y2. The first one points up the y axis. The second velocity points to the right z radians, where z = Z*pi/180.

Then the change in the y coordinate is Y(1-cos(z)).

The change in the x coordinate is Y*sin(z)

As Scytale said, reverse thrust with the change in y and change in x speeds as defined in his formuli. His equations make sense to me.

The Law of Cosines is useful to remember.

C^2 = A^2 + B^2 -2ABcos(gamma)

where gamma is the angle opposite side C.

Gentlemen,

You have my undying thanks... anytime you need an asteroid belt, just yell. The rule of cosines has been implemented in my course calculation spreadsheet and works beautifully...and now I can shut up my editor about the Delta-vee too.

Drat, thats two pints I owe

Cheers guys,

Cormie