Mathematical Formula for Navigation

Hi,
Is there a mathematical formula to calculate heading to fly from a radialdistance to another?

That’s a toughie. I would combine the math for converting two points into Cartesian (X,Y) and then find the relative vector of the two points X1,Y1  X2,Y2. Then the usual trickery for finding the direction from point 1 to point 2. I don’t know any trig trick to get output to be 0360° without breaking it up into the four quadrants with if/then statements.
Assuming two locations are given as
(R1, A1), (R2, A2)Cartesian positions are
(X1, Y1), (X2, Y2)
(R1Cos(A1), R1Sin(A1)), (R2Cos(A2), R2Sin(A2))Relative Cartesian position is
(X1X2, Y1Y2)
( R1Cos(A1)  R2Cos(A2), R1Sin(A1)  R2Sin(A2) )Relative polar position (direction component only)
Quadrant 1
A12 = 0° + ( ArcTan (Y1Y2/X1X2) )Quadrant 2
A12 = 180° + ( ArcTan (Y1Y2/X1X2) )Quadrant 3
A12 = 180° + ( ArcTan (Y1Y2/X1X2) )Quadrant 4
A12 = 360° + ( ArcTan (Y1Y2/X1X2) )Despite the worry that math angles start with 0° right and increase counterclockwise and compass angles start at 0° up increase clockwise the translations from coordinate system A to B and B to A cancel so it’s alright. If you’re using this on any capable device or language, the http://en.wikipedia.org/wiki/Atan2 function is a serious timesaver.

If you can do that in Cartesian coordinates, can’t it be done in Polar coordinates? If so, would that allow you to derive distance and radial from one point to another?

Interesting question. But why?

If you can do that in Cartesian coordinates, can’t it be done in Polar coordinates? If so, would that allow you to derive distance and radial from one point to another?
We did this kind of thing in my maths class at the start of the year… in polar coords, I mean. Can’t remember much of it now…


This may make for a good paper exercise for, say, a mathematics class, but applying it to the real world . . . good luck. Winds aloft will add variables which, while flying, will have to be compensated for ‘on the fly.’

This is called fix to fix.
Did it using radials and dme.
Mostly estimates.

What you want is called indeed a fixtofix, there is no formula for it as far as I am aware, personally I know the ‘pencil technique’.
For instructions on how to fly a fixtofix, look up AFI 11217, it might be hard to find information on it since it has long been removed from the USAF syllabus since advanced avionics have taken over and made the method illegal/obsolete. 
11217 is superceded, I don’t know about it being illegal…

We did this kind of thing in my maths class at the start of the year… in polar coords, I mean. Can’t remember much of it now…
Too long ago for me to remember as well. Seems like there ought to be something. They are all fixed points in space. It’s just the origins that are offset. We certainly describe them as polar coordinates (15 miles at 175* or 175 radial at DME 15).

well on a flat plane its just simple trigonometry… sin equals opposite over adjacent and all that…

Interesting question. But why?
Main reason is to track the shortest course to save fuel and time.
And other one is just love for math 
you have love for mentally calculating trig ratios?

IRL, we are using graphical (mental) resolution… Like for the bulls.
http://server.3rdwing.net/public/Bureau_75th/Bullseye%20LockOn%20A10.pdf
(Sorry, doc is in french, but looking at the pic should be enough)

If you can do that in Cartesian coordinates, can’t it be done in Polar coordinates? If so, would that allow you to derive distance and radial from one point to another?
I don’t really know what you mean “do it in polar.” What OP wants is a function which maps a 4space (two coordinates) to 1space (one angle). The functional dependence of the result on the inputs is one mechanism only, regardless if its expressed in equivalent ways. The final form of the function is completely devoid of the interim X and Y (I didn’t in my post simply because of the space it takes up). It’s a function f(x1, x2, x3, x4) = y.
Here’s the substitution for X and Y with their original input identities just to show just how long the equation gets:
Q1 A12 = 0° + ( ( R1Cos(A1)  R2Cos(A2) ) / ( R1Cos(A1)  R2Cos(A2) ) )
Notice there is nothing Cartesian about the above equation. The fact it made a temporary excursion into XY land is obscured when all is said and done. You might think it looks suspiciously written but since that describes the unique solution, there’s no other way to express the above that isn’t a trivial algebraic equivalent expression.Previously I said it was a 4space to 1space functional mapping. That’s not strictly true as the relative bearing between the points is invariant on scaling the picture. The direction between BZ 148 40nm and BZ 063 60nm is the same as between BZ 148 1nm and BZ 063 1.5nm. The direction depends on really the ratio of R1 and R2 so it’s really a mapping from 3space (A1, A2, R1/R2) to 1space (A12). The problem can be slightly simplified further if only only cares about the difference A1A2, that is the output direction will be relative to the first angle instead of north which is easy to fix at the last minute. So, excluding the northorientation fix at the last minute, it’s really a map from R2 to R1; f(X1X2, X3/X4) = y for example.
I’ll see if I can construct some graphs using the 2space mapping described above.

I don’t really know what you mean “do it in polar.” ……
Well, first, just to clarify, what I meant by “” was that I am not sure if what I’m suggesting makes sense.
My thinking goes like this: We are presented information in Polar Coordinates  Distance, Angle. I guess I’m asking why add the ‘extra’ step of converting that to Cartesian Coordinates  X,Y? If you’re starting with Polar Coord, working in a world that typically gives you Polar Coords and ending in Polar Coords, then why not just do the distance/direction calculation in Polar Coords?
That’s my thinking … the initial “” disclaimer still applies.
Although if you had BZ of the various points in question …

I didn’t mean to be harsh, just sharing what I was thinking. I too thought pretty hard about how to express a solution without resorting to x and y components. I invoked the Law of cosines; see if this goes anywhere.
Take the position vectors of the current location (A) and destination (B). Define a vector that travels from A to B. The direction of C is the goal. To inform us about C we take the dot product of vector C with the unit vector in the 0° direction (i) and compare that with the Law that says that C*i = Ci cosine Θ. C dot i is just the xcomponent of C. C is the length of C. The length of i is one by definition. And Θ is the goal.
So
Cx = C cosine Θ
If
Cx = BxAx
C = sqrt( (BxAx)^2 + (ByAy)^2)Cosine Θ = (BxAx) / sqrt( (BxAx)^2 + (ByAy)^2)
We know how to write expressions of A and B in the x and y directions with sine and cosine. Knowing Theta gives double solutions that only requires one additional step to narrow the real answer within the sets of two even or oddnumbered quadrants. For example 23° lets you know it’s either in quadrant II or IV; +41° let’s you know it’s in quadrant I or III.
Again, we end up with the components of the original two vectors defined by their components in the 0° direction and their components orthogonal to the 0° direction. This seems very Cartesian but it’s really just a consequence of orthogonality. We needn’t call them X and Y, just the direction which is aligned totally with the reference direction and that which totally isn’t. Sine and cosine themselves a defined by their mutual orthogonality and so you’ll always run into this component breakdown I fear. It seems inescapable unless you can find angular functions which are not trigonometric.

IRL, we are using graphical (mental) resolution… Like for the bulls.
http://server.3rdwing.net/public/Bureau_75th/Bullseye%20LockOn%20A10.pdf
(Sorry, doc is in french, but looking at the pic should be enough)
This is how I was taught to do it (nice pdf DJ!).
I can’t help on the math. I suck at math!