Meteor radiant angle. Is this computation right?

[solarblast]

I've written a Python program to simulate a meteor falling into the Earth's atmosphere. It's fairly straightforward. I give it the lat/long of two observing stations, and the lat/lng where the meteor begins along with an angle from north for the direction. An initial height above the beginning lat/lng is given along with the fall angle.

To define the fall angle, think of a tangent line to the Earth at the meteor's lat/lng, and another line parallel to it at the height of the meteor. Call it the height line The fall angle is measured using the height line and the radial line from the Earth's center to the height of the meteor. The origin is the intersection of the two lines, and the path of the meteor, a line, starts at the origin and moves at an angle downward from the origin. The angle is the fall angle. Typically, a fall angle is 0 (skips off the Earth's atmosphere) or moves just a few degrees into the atmosphere.

To test the program's output, I use the the radial line to the start of the track to another point on the track several degrees away. The separation angle is determined by two radial lines separated by an arbitrary angle. The length of the first line is r+h0, where r is the Earth's radius (earth is taken as a sphere), and h0 is the height of the meteor at the start of its path. The second line has a length r+h1. h1 is generally taken as less than h0.

To compute the fall angle, take the two lines as vectors, P1 and P2. P2-P1 is the path vector. How do I find the fall angle? My first take at this is to use |P1|*|P2-P1|*cos(phi) = P1 dot (P2-P1), where phi is the angle between P1 and P2-P1. If I specify, the fall angle as 10, 15 or whatever degrees, I find phi to be 90 degrees plus the specified fall angle. I seem close, but no cigar. Perhaps I need a third vector P3 that is perpendicular to P1 to form P3-P1 and P2-P1. Comments?

 

[solarblast]

Ah, I can post pix. See attached. It should be clearer what I'm trying to do.

 

[IsometricPion]

Since line BC is parallel to a line tangent to Earth's surface at point F, the four angles at the intersection of lines BC and AD are all right angles. This means you can use a little trigonometry and geometry to get the angle (you have two legs of a Pythagorean triangle and the angles of a triangle add up to 180^o). Hint: if you know how far apart E and F are along Earth's surface, you know the fall angle.

 

[solarblast]

Since line BC is parallel to a line tangent to Earth's surface at point F, the four angles at the intersection of lines BC and AD are all right angles. This means you can use a little trigonometry and geometry to get the angle (you have two legs of a Pythagorean triangle and the angles of a triangle add up to 180^o). Hint: if you know how far apart E and F are along Earth's surface, you know the fall angle.

It may look that way, but BC is not parallel to the tangent at F.The fall angle is roughly 20 to 30 degrees in the drawing. Draw it at 5 degrees and you will see it is not parallel.

 

[IsometricPion]

It may look that way, but BC is not parallel to the tangent at F.

That makes things slightly more complicated. Having drawn my own version based on your description in the OP, and comparing it with yours, I think I have an answer. Since line BD is perpendicular to line AD, angle ABC is the complement of the fall angle. So, $$\vec{P_1}\cdot(\vec{P_2}-\vec{P_1})=|\vec{P_1}||\vec{P_2}-\vec{P_1}|\cos(\phi)=|\vec{P_1}||\vec{P_2}-\vec{P_1}|\sin(\alpha)$$ where $$\alpha$$ is the fall angle.

 

[solarblast]

I'll check it out. I think I came to something close to that above, or maybe a post elsewhere. I didn't check out though whether it got the "extra" 90 degrees out of the picture.

 

[solarblast]

Yes, that gives the correct result in the test case(s) I've considered in Python. -15 degrees in the last test I tried. Thanks. Now the trick is to find a more complex situation to test it on and be able to verify the result. I think a colleague might be able to verify it with some data he might have for a much more complex piece of software. I've mostly placed the two stations symmetrically around either side of a latitude or longitude, and the lat/lng of the start of the track somewhat similarly. I think the -15 case did have the track position in some arbitrary position.