Skulls, Skeletons, Canines & Claws

[tigerluver]

(10-30-2016, 05:11 AM)GrizzlyClaws Wrote: Here at the post #90 is a holotype of a set of three lion canines (two upper and one lower) with all details available.

http://wildfact.com/forum/topic-big-cat-s-canines-and-claws?page=6



Here at the post #137 is the densest holotype available for the tiger canine. And it has been weighed at 67 grams, but this canine has been worn a lot. So without it, it should weigh around 70 grams.

http://wildfact.com/forum/topic-big-cat-s-canines-and-claws?page=10




If we compare these two holotypes, the length/width ratio for lion is 0.27, and the length/width ratio for tiger is 0.31.

Maybe @tigerluver could make an elaborate calculation/formula to explain the volume/density difference between the two holotypes?

Density is based on volume, which is 3-d, thus mass must be divided by a cubed measurement (ie. mm^3).

Usually we have three measurements on canines, length, width, and depth, at best.

So here are 3 ways we can get an actual density value with a cubed unit. I'll use the superior lion canines referenced in @GrizzlyClaws post to show the math.

You could divide the mass of the canine by the cube of its length:
Canine density = canine mass/(length)^3
Canine 1: 51.6 g/(106 mm)^3 = 4.33e-5
Canine 2: 51 g/(105^3) = 4.40e-5

You can substitute any other measurement and cube that as well. However, as this calculation takes into account only one measurement and forces a cubed value out of it, it is the least accurate.
 
Another way could be to force a cubed unit out of two measurements. You can take the length and multiply it by the square of a width/depth measurement as follows:
Canine density = canine mass/(length * width^2)
Canine 1: 51.6 g/(106 * 29^2) = 5.79e-4
Canine 2: 51 g/(105 * 28^2) = 6.20e-4

More accurate the previous method as two dimensions are working to create a 3-D model.

Finally, you can use a length, width, and depth measurements to get a cubed measurements without forcing anything as the last two methods did:
Canine density = canine mass/(length * width * depth)
Canine 1: 51.6 g/(106 * 29 * 20) = 8.39e-4
Canine 2: 51 g/(105*28*20) = 8.67e-4

This would be the most accurate out of what is in this post as you actually dividing a mass by a true volumetric measurement. Of course, you could buff density accuracy in others ways, but that's too complex for what we need here.

The tiger canine referenced unfortunately only shows to measurements (length = 103 mm; width = 32 mm).

We have to use the second method for comparison with the lion canines.

Tiger canine density = 70 g/(103 * 32^2) = 6.64e-4

The two lion canines average to 6.00e-4 for the same measurement, which is ~10% less dense. The actually discrepancy can still be different, as we forced a 3D measurement in this case and did not account for canine depth. We did not account for the way each canine is shaped either. Reply