Algebra and 3 Inches.
Seems that the better educated of us over at Cipher Mysteries are using AI as a means of solving a few of the Somerton Man mysteries as is Gordon C who is befuddling us with his native intelligence in trying to compute a man’s height by using a seventy year old image of a train ticket held in his hand. I kid you not.
I thought I’d join the party by using some very basic algebra.
Looking at the pic below and knowing Charlie is about 5’11” and Roy about 5’8” yet there they stand almost in line and both appearing to be about the same height. So we must ask, ‘How is this so?’
Answer, ‘It cannot.’
Come with me on this journey.
Let’s say for the exercise that grandmother is 5’6” and grandfather is 5’9” – difference being 3 inches, same as the two lads, and the distance grandad has to step back to appear to be the same height as grandmother is X. Now we can throw up an equivalent equation, that old friend from the high school Maths class of 1959. Remember? No? Let remind you: an equivalent equation is one that results in the same value.
What is the value of X?
Take it away Einstein.
Five foot six (66 inches) divided by X equals three inches divided by five foot nine (69 inches). Simple eh?
Then to find the value of X we isolate it on one side of the equation, which gives us the following under the laws of algebra, a subject few of us excelled in.
X equals 66 inches x 69 inches divided by 3 inches – answer being 1,518 inches.
In other words for grandad to appear to be the same height as grandmother he would have to step back 126.5 feet.
So would Carl to appear to be the same height as Roy …
I’ll get out of your way now.
The Somerton Man .. 
If you would like to prove the equation find an online calculator, divide 66 by 1512 and you will find it gives you the same answer as dividing 3 by 69, being .04036507.
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too much math for me eśe, but I see what you mean, and what does the size of the ticket have to do with SM height? GC gone mad.
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I’ve got my doubts that Carl Webb is anywhere to be found in Australia
It is assumed that because there is no trace of his death that he is Somerton man
I’d say because his brother is buried in Thanbyuzayat Myanmar, he went there to visit the brother’s grave, possibly met with misadventure or not, but never had money to return, or didn’t want to, to distance himself, had no ID and dropped off the radar.
I’d say his mental health is from his guilt and that he was compelled to visit his brother’s grave, that he likely left from Perth being a route north, perhaps up the Malay Peninsula, after he had scraped some money together. He may have never made it to his brother’s grave.
There’s no trace of him in Adelaide
I’d say Carl Webb’s life ended somewhere in Asia
As for the photo, it’s photoshopped, so it’s not possible to know what height Carl Webb is from that photo
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I think it has been given over to DA
to make the official public announcement
on 8 May at Salisbury
that Carl Webb is Somerton man
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what’s going to happen after?
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PB “SAPOL media tentatively affirms exhumed body isn’t that of Carl Webb”
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Where did you see that?
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JS on CM 29 March
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what happens after?
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John Sanders writes on CM ‘ SAPOL media tentatively affirms exhumed body isn’t that of Carl Webb .. etc’
He’s funning with you matey
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Wouldn’t it depend on the height of the camera and the distance from it (not to mention the angle relative to the people in the picture)?
Actually, thinking about it some more, I’m not even sure I follow how you came up with the equation.
66/X = 66 divided by some arbitrary variable
3/69 = height difference relative to the taller individual.
How does X become the distance between them? I think what you really want to calculate has 2 unknowns (one of which we might be able to assume)…..66/X = 69/(X+Y) where X is the distance to the camera, and Y is the distance between the two people – and even then the height should be measured only as the height above the camera.
So let’s suppose X is 10 feet (120 inches). 66/120 = 69/(120+Y) -> 7920 + 66Y = 8280 -> 66Y = 360 -> y ~=5.45
The camera being higher actually works in your favour. Let’s assume the camera is at a height of 4ft (48 inches).
In that case 18/120 = 21/(120+Y) -> 2180 + 18Y = 2520 -> 18Y = 360 -> Y = 20 inches. So we’re still closer than 2 feet.
But even so, the camera distance, height and angle is mebbe gonna make a lot of difference, so there’s not too much we can infer without accurately knowing some of those variables.
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Can’t disagree with that, and apologies, your comment was dumped in the spam folder.
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