Math for the Braindead (with apologies to Bob)

[Cormoran]

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

[Scytale]

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.

[Cormoran]

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

[Scytale]

I guess that would be the allmighty law of cosines.

[Cormoran]

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

[GlobeMaker]

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.

[Cormoran]

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