kitakaze
Resident DJ/NSA Supermole
12 Monkeys, what are your top three secrets to effective communication with adults?
I don't think any illusion is necessary, nor extensions, nor inhuman arms. A hand in a glove is just fine.
Bigfoot: The Patterson Gimlin Film - Part 3
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12 Monkeys, what are your top three secrets to effective communication with adults?
I don't think any illusion is necessary, nor extensions, nor inhuman arms. A hand in a glove is just fine.
Aepervius
Non credunt, semper verificare
Sure vortigen.
Normally a vector can be put into x,y,z (like this see picture : 3d axis) so if you have any point in 3D, you can define it as (x,y,z) or for example (0,0,1) is 1 unit above the plan along the Z axe, whereas (1,0,0) would be 1 unit on the X axe and nothing else, (1,1,1) would be a vector where there is one unit on each axe.
very obviously if you project a vector (x,y,z) onto the plan x,y , you jsut eliminate the third coordinate (x,y,0).
There is anotehr way to see those vector, and this is usually the way we see them in film or photo , in a spheric coordinate where my theta , phi come from : spheric coordinate so we have length+theta+phi I will represent with bracket [r,theta,phi]. There is naturally a way to go from 1 coordiante to the other
x=r \, \sin\theta \, \cos\varphi \quad
y=r \, \sin\theta \, \sin\varphi \quad
z=r \, \cos\theta \quad
Visually the forshortening effect is so , we see in the photo only a projection of the various vector onto a plan, so taking the Z axis to be perpandicular to the photo film, we see 3D vector projected onto x,y with variopus theta and phi (theta and phi changing is the movement). In other word , if I take a stick , and photography it tumbling, the stick length is always the same but theta and rho change, and going to the projection over x,y you imemdiately see the x,y change.
Please note that the demonstration will not take into account perspective and distance change which add to the final result.
Remark : in my previous post I reverted the role of theta and phi but it does not change anything.
For a hand it is a bit more complicated , but imagine all finger phalange are a single vector. So the first phalange for example go from origin (0,0,0) to P1 (1,0,0) so the vector is (1,0,0) from origin and is horizontal , the second one is inclined in plan x,y and go from (1,0,0) to (2,1,0) (so the vectro is (1,1,0) if placed at origin) and the last one is vertical in the same plan and go from (2,1,0) to (2,2,0) (so vector is (0,1,0) if at origin, remember a vector is final point minus origin point).
Like this :
first phalange is made of 1 , second is diagonal made of 2 and last is horizontal made of 3.
remark my phalange 2 is much longer than my phalange 1 and 3.
Basically now, going back to [r,theta,phi] for all vector, and rotating along theta and phi to have the angle changed, and projecting AGAIN in x,y, you immediately see that vector length is changed, as well as the apparent angle between them.
Before (I put angle in degree as it is easier than radian):
vector 1 : (1,0,0) = [1,90,0]
vector 2 : (1,1,0) = [SQRT(2),90,45]
vector 3 : (0,1,0) = [1,90,90]
Now we turn by 30° vertically and 30° horizontally (theta=theta+30 and phi=phi+30)
New vector is spheric coordinate :
vector 1 : [1,120,30] = (SQRT(3)/2*1/2,SQRT(3)/2*SQRT(3),-1/2) =(SQRT(3)/4,3/4,-1/2)
vector 2 : [SQRT(2),120,75] = (SQRT(2)*SQRT(3)/2*0.258,SQRT(2)*SQRT(3)/2*0.965,-SQRT(2)/2)
vector 3 : [1,120,120] = (-SQRT(3)/2*1/2,SQRT(3)/2*SQRT(3/2),-1/2) =(-SQRT(3)/4,3/4,-1/2)
projecting into the x,y plan (getting ride of z) +numeric approx
proj vector 1 : = (0.43, 0.75,0)
proj vector 2 : = (0.63,1.18,0)
proj vector 3 : = (-0.43,0.75,0)
I don't have a picture to attach, sorry.
Now adding all together to get our hand back :
X2 Y2
0,43 0,75
1,06 1,93
0,63 2,68
Just plot it in scatter mode in open office or open document.
Comapre to the original
X3 Y3
1 0
1 1
0 1
You haven't 90° anymore but a more bent angle.
Please note that if one started from the more bent one, and turned by -30 theta and -30 phi, you would actually get a less bent hand.
This naturally NO proof that spekator did not cheat, it is a mathematical demonstration that projected angle and length change with turning in all position.
click me for a picture
Top the "3" phalange of a finger. Bottom, we rotated the whole by theta=30 and phi=30.
PS: other rotation would have made the finger apparent angle even more "unbent" (imagine the "finger" inclination going the other way around, I guess this is a rotation of about 75 theta, and 75 phi).
ETA: and before sweaty protest, the whole point is on the doll you have no reference, but looking at the reflection it is obvious the angle theta and phi change are great, the same as with a SWINGING arm the angle changes change greatly. Plus elevation, plus plus plus...
QED
Normally a vector can be put into x,y,z (like this see picture : 3d axis) so if you have any point in 3D, you can define it as (x,y,z) or for example (0,0,1) is 1 unit above the plan along the Z axe, whereas (1,0,0) would be 1 unit on the X axe and nothing else, (1,1,1) would be a vector where there is one unit on each axe.
very obviously if you project a vector (x,y,z) onto the plan x,y , you jsut eliminate the third coordinate (x,y,0).
There is anotehr way to see those vector, and this is usually the way we see them in film or photo , in a spheric coordinate where my theta , phi come from : spheric coordinate so we have length+theta+phi I will represent with bracket [r,theta,phi]. There is naturally a way to go from 1 coordiante to the other
x=r \, \sin\theta \, \cos\varphi \quad
y=r \, \sin\theta \, \sin\varphi \quad
z=r \, \cos\theta \quad
Visually the forshortening effect is so , we see in the photo only a projection of the various vector onto a plan, so taking the Z axis to be perpandicular to the photo film, we see 3D vector projected onto x,y with variopus theta and phi (theta and phi changing is the movement). In other word , if I take a stick , and photography it tumbling, the stick length is always the same but theta and rho change, and going to the projection over x,y you imemdiately see the x,y change.
Please note that the demonstration will not take into account perspective and distance change which add to the final result.
Remark : in my previous post I reverted the role of theta and phi but it does not change anything.
For a hand it is a bit more complicated , but imagine all finger phalange are a single vector. So the first phalange for example go from origin (0,0,0) to P1 (1,0,0) so the vector is (1,0,0) from origin and is horizontal , the second one is inclined in plan x,y and go from (1,0,0) to (2,1,0) (so the vectro is (1,1,0) if placed at origin) and the last one is vertical in the same plan and go from (2,1,0) to (2,2,0) (so vector is (0,1,0) if at origin, remember a vector is final point minus origin point).
Like this :
first phalange is made of 1 , second is diagonal made of 2 and last is horizontal made of 3.
Code:
------------> Y
|1
|1
| 2
V 2
X 33remark my phalange 2 is much longer than my phalange 1 and 3.
Basically now, going back to [r,theta,phi] for all vector, and rotating along theta and phi to have the angle changed, and projecting AGAIN in x,y, you immediately see that vector length is changed, as well as the apparent angle between them.
Before (I put angle in degree as it is easier than radian):
vector 1 : (1,0,0) = [1,90,0]
vector 2 : (1,1,0) = [SQRT(2),90,45]
vector 3 : (0,1,0) = [1,90,90]
Now we turn by 30° vertically and 30° horizontally (theta=theta+30 and phi=phi+30)
New vector is spheric coordinate :
vector 1 : [1,120,30] = (SQRT(3)/2*1/2,SQRT(3)/2*SQRT(3),-1/2) =(SQRT(3)/4,3/4,-1/2)
vector 2 : [SQRT(2),120,75] = (SQRT(2)*SQRT(3)/2*0.258,SQRT(2)*SQRT(3)/2*0.965,-SQRT(2)/2)
vector 3 : [1,120,120] = (-SQRT(3)/2*1/2,SQRT(3)/2*SQRT(3/2),-1/2) =(-SQRT(3)/4,3/4,-1/2)
projecting into the x,y plan (getting ride of z) +numeric approx
proj vector 1 : = (0.43, 0.75,0)
proj vector 2 : = (0.63,1.18,0)
proj vector 3 : = (-0.43,0.75,0)
I don't have a picture to attach, sorry.
Now adding all together to get our hand back :
X2 Y2
0,43 0,75
1,06 1,93
0,63 2,68
Just plot it in scatter mode in open office or open document.
Comapre to the original
X3 Y3
1 0
1 1
0 1
You haven't 90° anymore but a more bent angle.
Please note that if one started from the more bent one, and turned by -30 theta and -30 phi, you would actually get a less bent hand.
This naturally NO proof that spekator did not cheat, it is a mathematical demonstration that projected angle and length change with turning in all position.
click me for a picture
Top the "3" phalange of a finger. Bottom, we rotated the whole by theta=30 and phi=30.
PS: other rotation would have made the finger apparent angle even more "unbent" (imagine the "finger" inclination going the other way around, I guess this is a rotation of about 75 theta, and 75 phi).
ETA: and before sweaty protest, the whole point is on the doll you have no reference, but looking at the reflection it is obvious the angle theta and phi change are great, the same as with a SWINGING arm the angle changes change greatly. Plus elevation, plus plus plus...
QED
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Aepervius
Non credunt, semper verificare
Even more incredible effect : do a rotation of 90° on theta alone. You won't see a finger anymore but a single line.... And by 85° you only see a very wide angle, like 170° apparent angle.
and now I am in a hurry.....
and now I am in a hurry.....
Aepervius
Non credunt, semper verificare
Yep I think so too. low tech and easy to make. And Could bend while swinging arm, depending on the material.
But the whole point of my post is that it is actually also possible this is a perspective effect.
Does not matter, all it means is that it DOES NOT bolster a "patty=real" scenario
.
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Aepervius
Non credunt, semper verificare
I redid the picture as the scale were not 1 to 1. And yes there is an angle (very flat) between the two first red phalange.
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SweatyYeti
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HOLY COW....
...what kitakaze THINKS.....is usually WRONG...Post #1929:
http://www.internationalskeptics.co...p=4741146&highlight=awesome+redux#post4741146
And...from Post #1925...
Concept - the fingers of the suit might be moving because they have fingers in them.
I know it's heavy.
Fun Fact....
...kitakaze THINKS spektator's and wolfy's 'doll-hand illusion' gifs are correct...and awesome!!!.......and I've demonstrated both of them to be 100% PURE, meaningless, contrived and manipulated BS.
kitakaze wrote....without his 'SPEKS' on...
Wrong...
...In the comparison image that you use, to show that 'Bob-in-a-suit matches Patty'....Bob has hand extensions in the suit...(Hey....just ask him
).Note the 60-pixel difference in length, on the arms...
"A hand in a glove is just fine"....WASN'T just fine....it was insufficient.
And everything you've just said......was just plain WRONG.
Look at dat.....kitakaze, just HAMMERED.....by a Monkey!!...
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Drewbot
Philosopher
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SweatyYeti said:
http://i172.photobucket.com/albums/w28/SweatyYeti/Fun/Pattybobbob3.jpg
How do you figures change if Bob's arm in the right hand picture were angled 15 degrees toward the camera lens?
You have not yet overcome the question of whether you are looking at the true-length of the arms in this view.
SweatyYeti
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Look at Bob's left hand, Drew. It obviously has an extension on it.
But, no problem, I can do more comparisons...to prove it.
Drewbot
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Look at Bob's left hand, Drew. It obviously has an extension on it.
But, no problem, I can do more comparisons...to prove it.
But you are looking at Patty's chest, and Bob's back. How can you compare the two?
SweatyYeti
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No, I'm looking at the length of 'Bob-in-a-suit's' left forearm.....it's ridiculously long.
But, as I said earlier.....I can do more comparisons...to prove it.
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Aepervius
Non credunt, semper verificare
Sigh. I see you have no comment on the hand experience sweaty. Did you try it for yourself ?
PS: long glove. Tada !
jeez. Is this 2003 ? did I just go back into time?
PS: long glove. Tada !
jeez. Is this 2003 ? did I just go back into time?
SweatyYeti
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Patty's fingers bend at a joint close to the end of the finger...
A simple 'long glove' wouldn't do that. TADA!!
This bending is well beyond the reach of Bob's fingers.....and Bob has yet to mention anything about any type of device in the suit-arm, to make the fingers bend at that point.
LTC8K6
Penultimate Amazing
Keep in mind that Sweaty's "finger bending" clip is not what it seems.
It's two completely seperate frames that could have been taken many minutes apart for all anyone knows...even I could manage to make rigid fingers bend if I'm allowed to stop the camera and cherry pick frames.
It's two completely seperate frames that could have been taken many minutes apart for all anyone knows...even I could manage to make rigid fingers bend if I'm allowed to stop the camera and cherry pick frames.
SweatyYeti
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Aepervius
Non credunt, semper verificare
Patty's fingers bend at a joint close to the end of the finger...
[qimg]http://i172.photobucket.com/albums/w28/SweatyYeti/handmove1ag.gif[/qimg]
A simple 'long glove' wouldn't do that. TADA!!
This bending is well beyond the reach of Bob's fingers.....and Bob has yet to mention anything about any type of device in the suit-arm, to make the fingers bend at that point.
Did you read what I wrote about math ?
Did you understand it ?
TADA !
Jeez.
What were Patterson and Gimlin doing in the few days at Bluff Creek before the shooting of the film?
I think perhaps the "squat" site identified by Titmus, where the foliage or grass was bent down, was a place that had been visited by Patterson in the days before the "shoot", a potential camera location from which Patterson had considered doing the filming. If you remember, the Roe/Kunstler/Patterson drawing shows a man concealed in foliage. This is the only significant detail in which it differs from the PGF. Perhaps, given the 25 mm lens he had, Patterson figured that he had to be closer to the subject than the "squat" site would allow. I expect Patterson looked at several sites and camera angles before making his final decision. And he did it before Bob H. arrived, so as to minimize the amount of time Bob H. would be visible in the vicinity. Only Gimlin will be able to tell us.
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SweatyYeti
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parnassus wrote:
Or, you could just continue playing 'make-believe'...and fill in the rest of the blanks yourself.
Or, you could just continue playing 'make-believe'...and fill in the rest of the blanks yourself.
SweatyYeti
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Drewbot
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Sweaty, until you can verify if you are looking at the True length of Bob or Patty arm, you are just throwing darts at a moving target.
This comparison makes it as simple as possible. Can you verify that Bob's arm is in the same plane as Patty's?
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LTC8K6
Penultimate Amazing
How long were they there? There's quite a bit of variation.
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