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Photogrammetry 101: Solving The Trackway Triangle

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This is part 2 in my Photogrammetry 101 series, which focuses on measuring the PGF. 90% of photogrammetry is solving triangles. The trackway triangle spans frames 288 to 480 to the camera, like this:

post-337-0-47411100-1437333203_thumb.jpg

Sides 'c' and 'b' represent the distances to frames 288 and 480 from the camera, respectively. Side 'a' will be derived from Patty's step length, angle A will be measured on the film, angles B and C must be derived using photogrammetry.

The trackway triangle represents 17 steps according to the following graphic:

post-337-0-60668400-1437333990_thumb.jpg

For this exercise we are assuming that Roger used the K100 film camera with a 25mm lens. In fairness, this exercise should be repeated for alternate lens configurations, which I will leave others to do. For me the question was could I construct a model that worked for a 25mm lens. Let's see.

A 25mm lens has a full frame width of 23.26 degrees. 17 steps represents 18 degrees (measured on the graphic above).

In my last Photogrammetry 101 topic, Distance From the Camera, I used the foot ruler to estimate distances from the camera to Patty for frame 352. This time we are going to use Patty's step length to estimate her distances from the camera and solve the trackway triangle. Then we can compare both methods to see if we are on the right track.

You will have to trust me regarding the photogrammetry but there are 2 main principles of geometry used to solve the trackway triangle, namely the Sin and Cosine Laws:

Sin Law:

sin A / a = sin B / b = sin C / c

Cosine Law:

a² = b² + c² − 2bc cos A

b² = a² + c² − 2ac cos B

c² = a² + b² − 2ab cos C

It turns out we don't have enough information to derive the distances from the camera. However, there is a direct inverse relationship between the size of an image and its distance from the camera. We can use this relationship to potentially give us angles B and C then we can complete the trackway triangle.

We must estimate the size difference ratio between images of Patty in frames 288 and 480 (F480/F288). This will also be the distance ratio from the camera (F288/F480). I found that I had to scale Patty down in F288 by 71% to match her size in F480.

post-337-0-74362700-1437355390.gif

So now the trick is to solve the equation where a=58 feet, A=18 degrees, c=?, b=.71 x c, by using the sin and cos laws to solve the trackway triangle. The triangle in the graphic below is to scale where 5 pixels = 1 foot. Use a pixel ruler to confirm the distances.

post-337-0-44712500-1437334326_thumb.png

According to this model Patty was 118 feet from the camera at frame 352, which is very close to the distance from the camera using the foot ruler. It is interesting to note that Patty would be 119 feet from the camera using the foot ruler if she was 6'1" tall. Regardless, it is clear that the trackway model for a 25mm lens will not fit Green's measurements. But the discrepancy is only 15 feet, which could easily be attributed to a misplacement of Roger's position while he filmed the 17 steps. Otherwise, how would they know where Roger stood at any given time?

I also added trees TC-1 and TC-2 to the model. They fit perfectly within the criteria that they were 12' apart (important: center to center) and TC-1 was 10' closer to the camera than TC-2, and they were exactly 3 degrees apart on film. When the trees were added to the model Patty indeed came within a few feet of TC-2. The next installment of the Photogrammetry 101 series will explore this further.

Keep in mind this was not a formal exercise since no error analysis was done. This was intended to demonstrate the methodology required to solve the trackway triangle. To be fair the same exercise should be done for a 20mm lens and varying step lengths. Also, it would be interesting to do the same exercise for Green's footage. The next step is to track each step to fine tune the model. Patty surely didn't walk in a straight line or take steps exactly 41" apart. Pretty close tho.

Next: Photogrammetry 101: The TC-1 TC-2 Tree Triangle

Edited by Gigantofootecus

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JustCurious

You know, just the other day I ran across my old trig slide rule from high school.  I wonder what I did with it?  Not to nit pick, but it's the Sine Law - sin is the mathematical abbreviation for sine.

 

A while back I was thinking that this was a geometry problem, but knew that whole 'distance from the camera' being an unknown is the monkey wrench in the whole equation.  I was too lazy to go beyond thinking through a couple of scenarios, but quickly realizing option 1 would not work because of 'x', option 2 would not work because of 'y' and option 3 is where I quit thinking about it.  (This from someone who used to be able to calculate these things in my head!!!)

 

I, for one, appreciate your thought and efforts on this problem.  Thank you for presenting it!!

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SweatyYeti

^

 

Make that two. Thanks, Giganto. :)

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Guest

You know, just the other day I ran across my old trig slide rule from high school.  I wonder what I did with it?  Not to nit pick, but it's the Sine Law - sin is the mathematical abbreviation for sine.

 

A while back I was thinking that this was a geometry problem, but knew that whole 'distance from the camera' being an unknown is the monkey wrench in the whole equation.  I was too lazy to go beyond thinking through a couple of scenarios, but quickly realizing option 1 would not work because of 'x', option 2 would not work because of 'y' and option 3 is where I quit thinking about it.  (This from someone who used to be able to calculate these things in my head!!!)

 

I, for one, appreciate your thought and efforts on this problem.  Thank you for presenting it!!

 

Of course it's the Sine Law. But as a computer programmer I am always typing sin and cos.

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SweatyYeti

One thing I'm very unclear on, Giganto....is the horizontal and vertical 'angle-of-views', for any given lens size.

 

At the bottom of your graphic above...you have written..."Full Frame angular width (25MM lens) = 23.26 degrees."

 

Is that angle determined by the lens itself...or by the size of the camera's aperture opening? The vertical 'angle-of-view' is smaller than the horizontal....and that is due to the smaller vertical dimension of the aperture, correct? 

So, I'm unclear on what the differences in 'HAV' and 'VAV' would be between the 20MM and the 25MM lenses....if one, or both of those angles is determined by the aperture opening...rather than by the lens itself.

 

 

I'm wondering, because I'd like to try working with the Photogrammetry formula...but I'd like to get a better understanding of how this all works, first. :)

Edited by SweatyYeti

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Guest

Hi Sweaty,

 

I will post a Photogrammetry 101: Angle of View and The Lens Equation next. For now, in a nutshell:

 

The FOV (field of view) of a lens depends on the size of the lens and the camera's aperture, which restricts the FOV according to the aperture's dimensions. For example, here is what Roger was looking at when he filmed Patty.

 

post-337-0-04909500-1440867623_thumb.jpg

 

The circle represents the full FOV of the lens. The rectangular aperture restricts the light that gets projected onto the film and sets the HAV and VAV. All these variables get incorporated into the full lens equation.

 

post-337-0-83404000-1440868078_thumb.gif

 

With all standard 16mm cameras the VAV seems to be consistent at 7.4mm. We should always go with the VAV because the HAV is variable. Just look at Green's footage of McClarin. Green inexplicably removed the camera's aperture plate before shooting McClarin. The aperture on the film clearly defines the magazine aperture, not the camera's, which holds the film reel in place. Just another riddle, wrapped in a mystery, inside an enigma.

 

Cheers

GF

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SweatyYeti

^

 

Thanks for the explanation, Giganto. :) That's what I was wondering about.

 

I still have a question or two, but I'll wait for your lengthier post on the 'Angle Of View'....and see if that answers them.  

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Drew

 

 

With all standard 16mm cameras the VAV seems to be consistent at 7.4mm. We should always go with the VAV because the HAV is variable. Just look at Green's footage of McClarin. Green inexplicably removed the camera's aperture plate before shooting McClarin. The aperture on the film clearly defines the magazine aperture, not the camera's, which holds the film reel in place. Just another riddle, wrapped in a mystery, inside an enigma.

 

 

How did you arrive at the bolded conclusion?

I don't think I've heard this before.

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SweatyYeti

^

 

I hadn't heard that Green shot McClarin, either... :huh:

 

Why...just because he walked outside of Patty's path??

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Drew

Yeah, he kept saying "Should I follow these deep prints back here, or the shallow prints over by the tree?"

 

Green had enough of that.

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Bigfoothunter

With all standard 16mm cameras the VAV seems to be consistent at 7.4mm. We should always go with the VAV because the HAV is variable. Just look at Green's footage of McClarin. Green inexplicably removed the camera's aperture plate before shooting McClarin. The aperture on the film clearly defines the magazine aperture, not the camera's, which holds the film reel in place. Just another riddle, wrapped in a mystery, inside an enigma.

 

Cheers

GF

 

I went to John Green and photographed the inside of the camera for Bill Munns and I seem to recall that it had something to do with the aperture. Munns would be better to explain what and why he requested this of me for my memory is not clear on this.

 

I do however recall Green not even being certain if the camera he had now was the same one he used with Jim at the film site.

Yeah, he kept saying "Should I follow these deep prints back here, or the shallow prints over by the tree?"

 

Where did that quote come from?

 

Jim told me that he could still see plaster residue on the ground from where the cast were made by Titmus. Jim walked next to the residue and then after that it was a bit of guesswork. I think once Jim got past the plaster residue and was then guessing - John probably felt that anything after that was worthless. Of course John wasn't on the same line of sight as Roger, nor at the same distance to Patty as Roger was.

Edited by Bigfoothunter

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Drew

Thanks Bill, I have been doing some searches and found this info as well from Bill Munns:

 

http://bigfootforums.com/index.php/topic/2428-the-munns-report/page-106#entry582705

 

GF

The problem is Green's camera now does have the aperture plate, and he did not have enough camera mechanics knowledge to take it out and put it back.

I am exploring the idea John actually used a Bell & Howell 200 with the camera ID tab removed,(a real simple procedure I intend to do on one of the B&H 200 cameras I have), and then run film to see the image area resulting. That B&H camera comes standard with a 20mm lens.

Anyways, lots of work to be done yet.

Bill 

 

 

Hopefully you can see why i'm confused.

Edited by Drew

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SweatyYeti

Yeah, he kept saying "Should I follow these deep prints back here, or the shallow prints over by the tree?"

 

Green had enough of that.

 

 

Fascinating, Drew....though, this sad event tarnishes my image of John Green... :popcorn:

 

 

(Just kidding, of course). :)

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Bill

Daniel Perez has posted photographic evidence showing exactly what Camera John Green used to film Jim McClarin, and it is a Revere magazine camera, an "Arte Deco" model. It cannot be confused with John's Keystone camera. The aperture on the Revere is stock, not modified. The unusually wide aperture going into the left side sprocket area is the characteristic of magazine cameras in general, as determined by the purchase and examination of many reels of home movies from the 40's, 50's and 60's.

 

Given that these magazine cameras have not been widely used for about 40 years, and there is practically no technical literature about their design and engineering, the prior technical confusion is reasonable all around.

 

So please forget the Keystone camera, forget the B&H 200 theory, and forget the idea of a DIY modified camera. They are no longer applicable to this discussion.

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