Trigonometry and Interplanetary travel - Help please?

[djnekkid]

Hi guys.

Im trying to solve a slight problem. Basicly, how far would Duna be from Kerbin, at the 'optimal launch time'. This diagram would probably discribe the problem better then words, and the real question is, how long is the DK-line??

Edited September 22, 2012 by djnekkid
wrong link

[Centauri]

Hi djnekkid,

Using trigonometry, DK can be solved, albeit with a lot of maths. In this case, however, we cannot just use a single sine formula to solve for the missing side, as triangle SDK is not a right triangle. So, to solve for oblique triangles, we must see what information we have and use either the (law of sines) or the (law of cosines + the law of sines) to solve the triangle.

For simplicity, I'm renaming the points of the triangle A, B, and C starting from point S and going around clockwise. Therefore, their opposite sides of each angle are their lowercase letters (e.g., SD is now side c).

First, we must see how much information we have. In this case, we know two sides and their inclusive angle (SAS). So, we will need to use a tag-team of the law of cosines to solve for a (the missing side), then use the law of sines to solve for the rest of the triangle (which isn't necessary, as you only wanted the missing side).

The law of cosines states that:

(a^2) = (b^2) + (c^2) - (2bc(cosA))

(I have yet to figure out how to insert a formula into vBulletin, so bear with me )

So, by adding a big ol' square root bar over both sides:

(a) = sqrt[(b^2) + (c^2) - (2bc(cosA))]

Given the information, we can now work the equation out:

(a) = sqrt[(4.296E14) + (1.850E14) - (2(13599840)(20726195)(cos(44.36)))

(a) = sqrt[(4.296E14) + (1.850E14) - (2(13599840)(20726195)(0.715))

(a) = sqrt[(4.296E14) + (1.850E14) - (4.031E14)]

(a) = sqrt[(2.114745161E14)]

(a) = 14542163.39

There you have it! Your missing side is approximately 14542163 meters.

I hope this is correct. If I'm wrong, my freshman ego is going to take a major hit

[djnekkid]

The awnser seemed to be about 9,5million, using this formula: DK*=*root( SD2*+ SK2*-(2*SK*SD) cos S )

[fallingzombie]

just stay awake in math class.

[djnekkid]

Hi Centauri

I did get the awnser from some of my facebook-friends, and one of them also got the roughly 14,5mill km awnser, and two others got the 9,5mill awnser. And my excel (well, openoffice calc) spreadsheet also gave 9,5mill. The guy who got the same awnser as you saied his calculator were set on "GOM", whatever that means. But I guess it would be either 9,5 or 14,5mill km that is correct

In my spreadsheet is cos44,36 0,92 and not 0,715. So it might be some degree-setting somewhere perhaps?

[mossman]

I ran through your math, Centauri, and got the same answer (With some precision loss - I did some rounding) so you're correct unless we're both wrong.

Also, this thread has motivated me to actually study for my trig exam next week instead of blowing stuff up all day - Thank you for that.

[Centauri]

I used my graphing calculator for working out the problem, and there are two modes for numerical values: radian and degree. Degree is for trigonometric equations, so I set that beforehand. I got 0.715 on both times I checked the problem.

Based on your answer of 0.92, I'm thinking you used the radian setting in your spreadsheet. I switched the mode back to radian on my calculator and entered cos(44.36) and got 0.92.

Hope this helps!

[djnekkid]

I ran through your math, Centauri, and got the same answer (With some precision loss - I did some rounding) so you're correct unless we're both wrong.

Also, this thread has motivated me to actually study for my trig exam next week instead of blowing stuff up all day - Thank you for that.

Im also starting to think that 14,5mill km is the correct awnser. As the math from some eletrical study I did about 15 years ago I seem to remember that both sin(45) and cos(45) is equal to 'half of the square of two', aka aprox 0,705. And 0,715 sounds about right when this degree is slightly lower, the number would be slightly higher

[Centauri]

Im also starting to think that 14,5mill km is the correct awnser. As the math from some eletrical study I did about 15 years ago I seem to remember that both sin(45) and cos(45) is equal to 'half of the square of two', aka aprox 0,705. And 0,715 sounds about right when this degree is slightly lower, the number would be slightly higher

Yeah, I remember from Geometry class a couple of years ago that sin(45) and cos(45) equals (sqrt[2])/(2), or roughly 0.705, as you said. Again, I hope I helped you with this problem

@mossman: No problem Last week, I had an inclined plane mechanical advantage quiz in the design lab, and there were a ton of trig problems. Thank god that they're always right triangles!

[Wild Card]

Actually, degree or radians work well for trig. I prefer radians. 2pi equals 360 deg. pi equals 180 deg. Etc. Has to do with the circumference of a circle with a radius of 1, the size of the arc you are using for a certain angle will equal a radian measurement. If you really like trig, you'll prefer radians to degrees, they make much more sense.