How is Chubby the alligators weight estimated?
There is a massive alligator in Florida that lives near a golf course, and he is estimated to be the largest alligator in Florida history assuming the estimations of his size are correct.
Estimating his length seems fairly easy, but how would they estimate his weight?
1 answer
Chubbs, the alligator in the question, is 15 feet long, according to the OP's link. OP's source and mine: Sports Illustrated. According to this source
The state's [Florida's] biggest alligator on record is 14 feet long and 780 pounds.
The article does not say if the previous record holder (PRH) had recently ate or pooped, which might make a non-trivial difference, if the OP desires great accuracy and precision.
I'm glad the OP asked for an estimate, not a measurement, otherwise one would have to sedate the alligator and suspend him (most likely a him) in a sling attached to a spring balance.
For a zeroth order approximation for Chubbs, one would take (15/14) x 780 and get 836 pounds.
Of course Chubbs, if he had the exact morphology of the (PRH), would be larger in all three dimensions. This would give a first order approximation of (15/14)cubed x 780 = 959 pounds.
To refine that number, one would look at the largest diameter (or circumference) of both alligators and adjust accordingly. Maybe Chubbs (named after a local golfer, and not his figure) is svelte relative to the PRH -- or obese. You should be able to get a reasonable approximation to the ratio of diameters from photos, the photos normalized to a 15/14 length ratio. I don't advise using a measuring tape on an unsedated alligator.
Then the estimate of Chubbs's weight would be (15/14) x [D(Chubbs)/D(PRH)]squared x 780.
For any other alligator, use the same method with a reference alligator as close as possible in length and shape.
This post was sourced from https://outdoors.stackexchange.com/a/20790. It is licensed under CC BY-SA 4.0.
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