# Determining 3D position after time

I'm trying to figure out how to calculate an object's angle to a fixed point after a certain amount of time, given the following information:

1: The origin 0,0,0 is your stationary point of view (the aforementioned fixed point)
2: The distance and bearings to the object from the origin are known
3: The bearings are given in one horizontal angle and one vertical angle
4: The speed and heading of the object are known and constant

I will use the following values in an example:

1: Object speed: 100 m/s
2: Object starting distance from origin: 2000 m
3: Object bearing from origin: 1°00′00″ horizontal, 2°00′00″ vertical
4: Object heading: 130°00′00″ horizontal, 5°00′00″ vertical
5: Position to calculate after t = 60 s

I'm guessing the solution would involve the following method:

1: Determine the starting coordinates of the object, based on the horizontal and vertical bearings, and the distance.
2: Use the object's starting coordinates, its heading, and its speed multiplied by t = 60 to determine its coordinates after the specified time.
3: Determine the angle from the origin to the coordinates at t = 60.

It looks like there is a lot of trigonometry involved in this. Perhaps this is easier to solve than it looks, but to me, it appears to have a lot of steps involved in it. Or perhaps there is a better method to solving this? Some formula I'm unaware of?

• I'm not sure what any of that means. As for what I tried to do, I thought I could use the distance formula and solve for the variables, as I knew the distance and the origin coordinates, but I quickly realised that I'd be solving for three unknowns (the x, y and z of the starting coordinates). – Hiigaran Jun 6 '17 at 3:43
• The bearing from origin to the start? That's the known information listed in point three of the example. Or did you mean what I did to try and solve for the bearing from origin to the t=60 coordinates? If the latter, I didn't make it that far. – Hiigaran Jun 6 '17 at 3:51
• Ok i see the edit. This is trivial then.. Just apply the line equation and you are done. – Brethlosze Jun 6 '17 at 3:54

## 1 Answer

You list an object bearing from the origin, presumably that is at the start of your time period. That gives you an $(x,y,z)$ based on the distance which we can call $d$. Then add the vector that is the object motion in $60$ seconds. That will give a final location that depends on $d$. You still don't have any information that allows you to measure $d$. It started somewhere on a ray from the origin and is now on a different ray, but you can't find the point along the ray. Once you find another piece of data to determine $d$ you will have a unique solution, which will involve a bunch of trig to change angles into distances. Why are you opposed to the trig? That just generates constants to multiply by.

• I thought d would be the speed of the object multiplied by the time you want to solve for. Speed of 100 m/s with t=60 would be 6000. Or did I misunderstand what you meant by missing information? – Hiigaran Jun 6 '17 at 3:45
• That tells us how far it moved, but we don't know where it started. If it was way far away at the start that much motion won't change the angles we observe it at much. If it started very near the origin, the angles at the end will be the direction of motion. – Ross Millikan Jun 6 '17 at 3:50
• Just do everything in Cartesian coordinates. Where do you measure the bearings from? If from the $x$ axis, the starting position is $(2000 \cos 1^\circ \cos 2^\circ, 2000 \sin 1^\circ \cos 2^\circ, 2000 \sin 2^\circ)$ Compute the displacement the same way. Add that to the starting position. You will need some $\arccos$ and $\arcsin$ functions to get the angles at the end. – Ross Millikan Jun 6 '17 at 4:37
• You are probably putting degrees into the spreadsheet but it expects radians. The first set looks right since $1^\circ \approx \frac 1{57} \approx \sin 1^\circ$ – Ross Millikan Jun 6 '17 at 5:08
• You compute the $x,y,z$ displacements from the motion. The total displacement is $60\cdot 100=6000$ and you apply the same trig functions to get the components. Add that to the starting location and you have the ending location. – Ross Millikan Jun 6 '17 at 14:05