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Thinker Thunker size comparison of Patty

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wiiawiwb

Where's Gigantofootecus when you need him? 

 

I would think there are precalculated formulas that would assign percentages for how a particular width appears diminished as depth is increased. In other words, if an object is 'x" inches wide, if you move it back 10', it will appear .83x. If there are such charts, and we do know how far McLarin was from a particular tree, we should be able to apply that precalculated percentage to get his width as though he were next to the tree. Then from there we can precisely calculate a tree width.

 

Anyone know of such charts?

 

Do we know how many feet behind that one tree that McLarin was?

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Bigfoothunter
2 hours ago, wiiawiwb said:

Where's Gigantofootecus when you need him?

 

 Gigantofootecus was an asset to this site in my view. He took issue with a mod that he thought was applying a double standard when he got a warning for referencing another member as a scoftic. As we currently have witnessed - the differences between a skeptic and a scoftic is quite clear. He was one of the few members who was skilled in Photogammetry, thus he brought something of value to the discussions. I will email Gigantofootecus and try to get him to come back.

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Bill

One has to make their own chart using the lens formula: O divided by A equals D divided by F

O is the object size (height of a man, width of a tree trunk, etc)

A is the aperture size (height of the measurable size of that object in the film image, based on a full frame height of 0.3" for the aperture, so you measure the object in the film image as a percentage of the frame height, keeping in mind you must use a true full frame image, not a crop)

D is the distance from camera lens optical center to actual object.

F is focal length of the lens (if for example, if the lens is a 25mm or 1" lens, then the object distance is 1" less than the measured distance from object to the film plane or film gate. The focal length is the offset of the lens optical center from the film gate/plane)

 

You need three numbers to solve for the fourth.

 

If you are solving for distance, it is the object size multiplied by the focal length, divided by the aperture size.  (for "how far away is the object or person? Like How far away is Jim McClarin, from Green's camera, given we know his height.)

If you are solving for the object size, it is the distance multiplied by the aperture size, divided by the focal length  (for how tall is Patty)

If you are solving for focal length, it is the distance multiplied by the aperture size, divided by the object size.  (for what lens is on the camera)

If you are solving for the aperture size, it is the focal length multiplied by the object size, divided by the distance. (for how big in the object going to appear in the film)

 

The big challenge is making sure your three numbers are correct. Or you can do a chart with estimates, like, "If Patty is this far away, with this lens, how big is she? And then run, if she's 100 feet away, she's this big. If she's 105 feet away, she's this big. etc.

I'm no longer offering estimates because I'm still checking numbers and still have some issues i need to clear up.I want my three numbers to be absolutely certain before I again calculate the fourth.

 

The actual technical discussion of the lens formula is in my book, in one of the appendix sections. It explains this in detail and shows example calculations,

 

Bill

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SweatyYeti
2 hours ago, wiiawiwb said:

Where's Gigantofootecus when you need him? 

 

I would think there are precalculated formulas that would assign percentages for how a particular width appears diminished as depth is increased. In other words, if an object is 'x" inches wide, if you move it back 10', it will appear .83x. If there are such charts, and we do know how far McLarin was from a particular tree, we should be able to apply that precalculated percentage to get his width as though he were next to the tree. Then from there we can precisely calculate a tree width.

 

Anyone know of such charts?

 

 

I'm glad you asked, wiia. I have been wondering that same thing, myself, for quite a long time.  :) 

 

Quote

One has to make their own chart using the lens formula: O divided by A equals D divided by F

O is the object size (height of a man, width of a tree trunk, etc)

A is the aperture size (height of the measurable size of that object in the film image, based on a full frame height of 0.3" for the aperture, so you measure the object in the film image as a percentage of the frame height, keeping in mind you must use a true full frame image, not a crop)

D is the distance from camera lens optical center to actual object.

F is focal length of the lens (if for example, if the lens is a 25mm or 1" lens, then the object distance is 1" less than the measured distance from object to the film plane or film gate. The focal length is the offset of the lens optical center from the film gate/plane)

 

 

Could you provide one mathematical example, Bill....for how much, percentage-wise, an object diminishes in size (on the film) over the course of 50'...in the case of a 25MM lens? 

 

I can try working that out, using the photogrammetry formula....but, it might be easier/quicker....for you to do the math. :) 

 

The reason why I'm asking about a 50' distance.....is because that is roughly the distance from trees TC-1/TC-2....to tree TC-4. And since those trees' widths were measured....I'd like to see if the numbers work-out in accordance with their apparent size on the film....given a 25MM lens on the camera.  

 

 

 

25 minutes ago, Bigfoothunter said:

 

 Gigantofootecus was an asset to this site in my view. He took issue with a mod that he thought was applying a double standard when he got a warning for referencing another member as a scoftic. As we currently have witnessed - the differences between a skeptic and a scoftic is quite clear. He was one of the few members who was skilled in Photogammetry, thus he brought something of value to the discussions. I will email Gigantofootecus and try to get him to come back.

 

 

That would be great, if you could get Giganto to come back to the forum, Bill. His knowledge, and his contributions were very valuable. :) 

 

 

Edited by SweatyYeti

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Bill

Sweaty:

To use the formula, one must always use distance from camera to one object. So for two trees, you can calculate one tree from camera, to determine width, for example, and then calculate another tree width based on its distance from camera, and then compare tree widths. But rectifying the full frame image to the 0.3" aperture true size is important, and using an image 3000 pixels high (including one full image and one full black frame separation line also) is a good way to rectify the image scale.

 

One mathematical example would be the following:

 

McClarin Height Analysis (solving for Jim's distance from Green's camera, assuming a 25.4mm (1") lens on that camera

Passing second tree, he is 445 pixels high out of an adjusted frame of 3000 pixels high.  3000 pixels = 0.3"

D = O x F divided by A

O is 6.41 feet (6'5")

F is 1"

A is 0.0445"

distance is 144' from camera. Tree is 115'

 

 

Edited by Bill
addition

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SweatyYeti
43 minutes ago, Bill said:

Sweaty:

To use the formula, one must always use distance from camera to one object. So for two trees, you can calculate one tree from camera, to determine width, for example, and then calculate another tree width based on its distance from camera, and then compare tree widths. But rectifying the full frame image to the 0.3" aperture true size is important, and using an image 3000 pixels high (including one full image and one full black frame separation line also) is a good way to rectify the image scale.

 

One mathematical example would be the following:

 

McClarin Height Analysis (solving for Jim's distance from Green's camera, assuming a 25.4mm (1") lens on that camera

Passing second tree, he is 445 pixels high out of an adjusted frame of 3000 pixels high.  3000 pixels = 0.3"

D = O x F divided by A

O is 6.41 feet (6'5")

F is 1"

A is 0.0445"

distance is 144' from camera. Tree is 115'

 

 

 

 

Thanks for the example, Bill.  I will see what I can figure out, regarding the trees' apparent sizes on the film. :) 

 

 

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Patterson-Gimlin
On 12/3/2017 at 8:43 PM, norseman said:

I’m 6’2” tall, 260 lbs and I have a 11x4 inch foot...... so being 6 feet tall and having a 15x6 inch foot is unworkable as well.

 

It’s marked 12 inches. That isn’t the problem. You have two trees marked 18 inches and 8 inches respectively. Only ONE value can be correct as the trees are obviously close to the same size. 

 

ONE of those values puts a 15x6 inch foot on a film subject that is somewhere between 5-6 feet based on Redbones calculations (8”)..... the OTHER value puts a 15x6 inch foot on a 7 ish foot tall film subject according to Thinker thunkers calculation (18”).

 

Yah, I know which value makes a helluva lot more sense. If we are going to disqualify something on that diagram based on a squiggle line? You might as well throw the whole thing in the trash can.

 

Do you think the trackway is hoaxed? I have no idea how you can just dismiss the trackway entirely and not take seriously it WAS NOT made by something my size or smaller. Patty’s feet are monstrous....

 

Ill leave you with this to ponder.....

 

http://extras.denverpost.com/scolumns/kiz0314.htm

 

Shaq:

 

7’1” tall

315 lbs

size 22 shoe 

16” long feet.

7' 1"

350 lbs.

shoe size 16 or 17 depending on shoe

5" wide

13" long

Edited by Patterson-Gimlin
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wiiawiwb
6 hours ago, Bigfoothunter said:

 

 Gigantofootecus was an asset to this site in my view. He took issue with a mod that he thought was applying a double standard when he got a warning for referencing another member as a scoftic. As we currently have witnessed - the differences between a skeptic and a scoftic is quite clear. He was one of the few members who was skilled in Photogammetry, thus he brought something of value to the discussions. I will email Gigantofootecus and try to get him to come back.

 

He was a invaluable resource here. There was a recent discussion about shoulder width and how to relate it to Patty's height. That would have been right up his alley. His epiphanic theory about Patty's ASH ratio that he posited back in 2005 was the first that I am aware of using body measurements to determine height. And now, in this thread, using a tree diameter.

 

All involve comparing known height or width of objects in the film to develop a usable model for Patty's height calculation and Gigantofootecus was a pioneer in that field, in my humblest of opinion.

 

I say that with all due respect to Bill Munns whose treatise on the PGF is encyclopedic even as he breaks new ground every month.

Edited by wiiawiwb

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Backdoc
BFF Donor
On ‎12‎/‎4‎/‎2017 at 0:42 AM, PBeaton said:

Just a couple pics ta add ta the equation. 

McClarinheightcompared.jpg

shadows.gif

 

 

This irregular terrain is another important point.  

 

A man walking in a hurried fashion while wearing a suit would seemingly have some expected difficulty doing this on some varied terrain. Do they know the drop off is even coming step to step?

 

If that man was Bob H, then it is further complicated by the nonsense one of those eyes in that suit would have a glass eye glued in on the right side during such a walk.

 

Most of the time we see the PGF being played at lower speed because we are looking for fine details.  When the film speed is sped up to what might seem like reasonable walking speed, it would be more diff to pull off a walk in a suit under varied terrain.  

 

Just think of it, it is unlikely the man in a suit really knows which spots are high or low.  A suit walker wouldn't really know which spot might be suddenly soft or suddenly hard.  I am not saying this was some kind of major obstacle course, just saying this is not some walk across a level floor in the Stanford lab.

 

The walk is easier to pull off even wearing is suit if the person doing the walk was familiar with  the walking path and could anticipate what to expect.   Once someone got to a road and/or walked the road, this is less necessary.  Up  to the that time though it would seem to me to be vital to the hoax scenario.  How does one do that without leaving traces?

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Bigfoothunter
1 hour ago, Backdoc said:

 

 

"This irregular terrain is another important point.  

 

 

 

McClarin has told me that the sandbar was higher elevated in places that in others. I believe he said there was as much as a foot difference at times. This is why I have found the height variances hard to pin down.

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PBeaton

Notice the shadows.

JohnGreenJimMcClarin.jpg

Notice how quick we loose sight of McClarin's feet here.

Image result for mcclarin at  bluff creek

13 hours ago, Patterson-Gimlin said:

7' 1"

350 lbs.

shoe size 16 or 17 depending on shoe

5" wide

13" long

haha ! :o  I mean that in a good way ! :D

 

Patterson-Gimlin, mind if I ask a favour, if you took a step mimickin' the compliant gait of the PGF subject, what's the distance ? Thanks !

 

:drinks:

Pat...

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Twist
4 hours ago, PBeaton said:

.....

 

Patterson-Gimlin, mind if I ask a favour, if you took a step mimickin' the compliant gait of the PGF subject, what's the distance ? Thanks !

 

:drinks:

Pat...

 

I'd be curious to know as well.  Based on a quick google search to figure average male stride length based on height PG "should" have around a 35-36" stride length.   

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PBeaton

The PGF subject moves quite fluidly an naturally in my opinion, clearly with a longer step/stride than 6'5" McClarin. I think it was Bill Munns who put the top two images together, I put the third/bottom image together roughly. If McClarin extended his stride to match hers, he'd get shorter, not to mention if he leaned his torso as well.

jimsasquatchwalkcomparison.jpg

Edited by PBeaton
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Patterson-Gimlin
5 hours ago, PBeaton said:

Notice the shadows.

JohnGreenJimMcClarin.jpg

Notice how quick we loose sight of McClarin's feet here.

Image result for mcclarin at  bluff creek

haha ! :o  I mean that in a good way ! :D

 

Patterson-Gimlin, mind if I ask a favour, if you took a step mimickin' the compliant gait of the PGF subject, what's the distance ? Thanks !

 

:drinks:

Pat...

I am not as good as the Patterson creature with that gait.

About 34 to 35 inches.. My girlfriend said I am big enough but not nearly as fluid as  the film subject.

What was the stride of the Patterson subject?  It certainly appears to be more than the skinny dude in the comparison film.

i am not convinced it is a creature. I am less convinced it is a six foot subject.

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norseman
BFF Donor

48 inch average.

 

 

Edited by norseman

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