<p>The spherical triangle with vertices at the north pole and the two
video camera locations has two sides and their included angle known.
The sides are equal to 90 degrees minus the latitudes of the camera
locations. The known angle is the difference in liongitude between the
cameras. The unknown side, between the camera sites can be found from
the law of cosines for spherical triangles, then the two unknown
angles from the law of sines.</p>
<p>Any point in their video images that's seen by both cameras at the same
time can be located in geographical coordinates (lat, long, elevation)
in a way that follows. (There must be an easier way, but this one
seems easy to follow.) Conceptually divide the spherical triangle
solved before the cameras were turned on into three triangles with the
alt/az of a point in the image known and its geographical position
(GP), for this case, inside the pole/cameras triangle. There will be
great circle segments linking all four points: 6 total, 3 of them
between the pole/cameras and another 3 from the GP of the image point
to each of them.</p>
<p>At each of the cameras there are three angles to consider between 3
great circles. The angle between the camera's local meridian and the
great circle connecting it to the other camera was found before
turning on the cameras and it's the first. The azimuth from the video
image is the second and the difference between them is the third, and
the important one at this point. We can subtract to find both of them
and already know their included side. Using the other variant of the
law of cosines find the angle between the great circles connecting the
image point to the cameras. Solve for the two sides between the point
in the image and the cameras using the law of sines.</p>
<p>Both of the two triangles with pole/image point GP/camera for
vertices now have two sides and their included angles known: the side
found in the previous steps, 90 degrees minus the camera's latitude,
and the azimuth of the image point.</p>
<p>That's enough to find the lat/long of the image point's GP. One more
step finds the elevation of the object above that GP. This one uses a
plane triangle. Its vertices are at the camera, at the Earth center
and at the object whose image we're working on. The known sides of
that triangle are the Earth radius to a camera, and Earth-radius plus
elevation to object. The known angles are distance over the ground
expressed as an angle from camera to image point GP, and altitude.
That should be enough to finish location finding, with basic trig.</p>
<p>If no one finds any errors and there's interest, I can embellish.
Sorry about no images.</p>

CC BY-SA 4.0
2023-12-26T23:49:45.773Z
2023-12-26T23:49:45.773Z
stretch
<p>There are a host of practical problems to solve.  Working backwards, to find the (average) velocity you'll need to find the position at two points and the time the meteor takes to travel from one to the other.</p>
<p>Now you can hope to find the position using surveying techniques.  That means you will need to determine the bearing (azimuth) and angle above the horizon (alt) of the meteor at a particular time. These angles will determine the direction of the meteor from a point. You will know that the meteor lies on a particular line.</p>
<p>Finding the direction of the meteor from a different point at the same time gives a second line, and the intersection of those two lines is the location of the meteor.  Do that twice and you'll get your two locations and the time between them.</p>
<p>Here all sorts of practical problems arise.  Do you know the location of the cameras and their direction? How will you determine the scale of the images, or horizon line (people setting up security cameras rarely worry about getting them horizontal. How you solve these practical problems will vary according to what information you have about the cameras, or what you can deduce from the videos.</p>
<p>Will you attempt to account for the curvature of the Earth between the cameras? It simplifies the maths if you do! What about the difference in altitude?  If you have determined the direction of the meteor, you can write that as a (unit) vector <span class="math-container">$\mathbf{\hat d}_A$</span> and a location of camera <span class="math-container">$\mathbf{r}_A$</span>, and then the meteor is on the line <span class="math-container">$\mathbf{r}_A+t\mathbf{\hat d}_A$</span>, and from the second camera <span class="math-container">$\mathbf{r}_B+t\mathbf{\hat d}_B$</span>.  And, because you have had to estimate some quantities those two lines won't meet perfectly.  So you'll need to find a point that is close to both lines.</p>
<p>And finally, remember that the meteor will be slowing down as it passes through the Earth's atmosphere, so you can refine your model to include this.</p>
<p>Working back then to determine the orbit of the meteoroid would mean then finding the motion in the frame of the Earth, and &quot;running gravity backwards&quot; to determine the path of the meteoroid back to outside of the region where Earth's gravity is strong.</p>
<p>Nothing about this is easy or automatic, but the first step is surely to determine the cameras' locations, synchronize their clocks (again security cameras don't always have clocks with sub-second accuracy) and attempt to estimate the Azimuth and Alt of the meteor.</p>

CC BY-SA 4.0
2023-12-24T16:07:12.613Z
2023-12-24T16:07:12.613Z
James K
<p>How to determine the velocity of a meteor from two video data?</p>
<p>I want to try to determine the orbit of the meteor, but before that I have to determine the meteor's entry velocity and apparent radiant and apparent velocity.</p>

CC BY-SA 4.0
2023-12-25T04:48:59.870Z
A different method than that proposed in the answer below.  For each location, use background stars to find the RA/Dec of two points on the meteor's path.  Convert that in to Alt/Az coordinates, then into geocentric rectangular coordinates.  Compute the rectangular coordinates for the cameras.  The coordinates of the camera and two RA/Dec points form a triangle, which can be used to define a plane.  Use the "intersection of two planes" to determine the meteor's path.  The time between frames can be used to compute the velocity.  Use ephemeris to compute and subtract off the Earth's velocity.
Greg Miller
CC BY-SA 4.0
2024-01-26T00:29:14.550Z
In central Europe there's the Fireball network with fixed observation sites and well-defined observation in the night hours: https://en.wikipedia.org/wiki/European_Fireball_Network - it might be able to provide more detailed observation than one could possibly obtain otherwise.
planetmaker
<orbit><meteor><speed><velocity>
Determining the speed of meteor
alphaaquilae
# Determining the speed of meteor

(Asked by: username_0 on 2023-12-24)

How to determine the velocity of a meteor from two video data?

I want to try to determine the orbit of the meteor, but before that I have to determine the meteor's entry velocity and apparent radiant and apparent velocity.

**Tags:** orbit, meteor, speed, velocity

> **Comments:**
> **username_1**: A different method than that proposed in the answer below.  For each location, use background stars to find the RA/Dec of two points on the meteor's path.  Convert that in to Alt/Az coordinates, then into geocentric rectangular coordinates.  Compute the rectangular coordinates for the cameras.  The coordinates of the camera and two RA/Dec points form a triangle, which can be used to define a plane.  Use the "intersection of two planes" to determine the meteor's path.  The time between frames can be used to compute the velocity.  Use ephemeris to compute and subtract off the Earth's velocity.  
> **username_2**: In central Europe there's the Fireball network with fixed observation sites and well-defined observation in the night hours: https://en.wikipedia.org/wiki/European_Fireball_Network - it might be able to provide more detailed observation than one could possibly obtain otherwise.  

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## 2 answers

### Answer by username_3 on 2023-12-26. Score: 1

The spherical triangle with vertices at the north pole and the two
video camera locations has two sides and their included angle known.
The sides are equal to 90 degrees minus the latitudes of the camera
locations. The known angle is the difference in liongitude between the
cameras. The unknown side, between the camera sites can be found from
the law of cosines for spherical triangles, then the two unknown
angles from the law of sines.

Any point in their video images that's seen by both cameras at the same
time can be located in geographical coordinates (lat, long, elevation)
in a way that follows. (There must be an easier way, but this one
seems easy to follow.) Conceptually divide the spherical triangle
solved before the cameras were turned on into three triangles with the
alt/az of a point in the image known and its geographical position
(GP), for this case, inside the pole/cameras triangle. There will be
great circle segments linking all four points: 6 total, 3 of them
between the pole/cameras and another 3 from the GP of the image point
to each of them.

At each of the cameras there are three angles to consider between 3
great circles. The angle between the camera's local meridian and the
great circle connecting it to the other camera was found before
turning on the cameras and it's the first. The azimuth from the video
image is the second and the difference between them is the third, and
the important one at this point. We can subtract to find both of them
and already know their included side. Using the other variant of the
law of cosines find the angle between the great circles connecting the
image point to the cameras. Solve for the two sides between the point
in the image and the cameras using the law of sines.

Both of the two triangles with pole/image point GP/camera for
vertices now have two sides and their included angles known: the side
found in the previous steps, 90 degrees minus the camera's latitude,
and the azimuth of the image point.

That's enough to find the lat/long of the image point's GP. One more
step finds the elevation of the object above that GP. This one uses a
plane triangle. Its vertices are at the camera, at the Earth center
and at the object whose image we're working on. The known sides of
that triangle are the Earth radius to a camera, and Earth-radius plus
elevation to object. The known angles are distance over the ground
expressed as an angle from camera to image point GP, and altitude.
That should be enough to finish location finding, with basic trig.

If no one finds any errors and there's interest, I can embellish.
Sorry about no images.


---

### Answer by username_4 on 2023-12-24. Score: 0

There are a host of practical problems to solve. Working backwards, to find the (average) velocity you'll need to find the position at two points and the time the meteor takes to travel from one to the other.

Now you can hope to find the position using surveying techniques. That means you will need to determine the bearing (azimuth) and angle above the horizon (alt) of the meteor at a particular time. These angles will determine the direction of the meteor from a point. You will know that the meteor lies on a particular line.

Finding the direction of the meteor from a different point at the same time gives a second line, and the intersection of those two lines is the location of the meteor. Do that twice and you'll get your two locations and the time between them.

Here all sorts of practical problems arise. Do you know the location of the cameras and their direction? How will you determine the scale of the images, or horizon line (people setting up security cameras rarely worry about getting them horizontal. How you solve these practical problems will vary according to what information you have about the cameras, or what you can deduce from the videos.

Will you attempt to account for the curvature of the Earth between the cameras? It simplifies the maths if you do! What about the difference in altitude? If you have determined the direction of the meteor, you can write that as a (unit) vector $\mathbf{\hat d}\_A$ and a location of camera $\mathbf{r}\_A$, and then the meteor is on the line $\mathbf{r}\_A+t\mathbf{\hat d}\_A$, and from the second camera $\mathbf{r}\_B+t\mathbf{\hat d}\_B$. And, because you have had to estimate some quantities those two lines won't meet perfectly. So you'll need to find a point that is close to both lines.

And finally, remember that the meteor will be slowing down as it passes through the Earth's atmosphere, so you can refine your model to include this.

Working back then to determine the orbit of the meteoroid would mean then finding the motion in the frame of the Earth, and "running gravity backwards" to determine the path of the meteoroid back to outside of the region where Earth's gravity is strong.

Nothing about this is easy or automatic, but the first step is surely to determine the cameras' locations, synchronize their clocks (again security cameras don't always have clocks with sub-second accuracy) and attempt to estimate the Azimuth and Alt of the meteor.


---

