The strongest bites in the animal kingdom

[Wrapp]

@LonePredator 


(Reposting the post since the post I posted wasnt containing the image files that I tried to upload. Will be deleting the old post once this one makes appearance in the forum.)


Now, the issue with your latest method.
 
The method by itself, even if hypothetical, is not correct.
 
The cross-sectional surface area doesn’t increase with the power of 2 or 1/2 if the volume is by the "cube" or cube root (obviously), even if the composition/ density and everything is the same.
We are talking about 3-dimensional, biological structures here. The surface area isn’t directly affected due to the scaling in a linear way, as of the power of 2. Let me explain you why. 
By scaling the size of the tissue, we scale the whole volume of that tissue such that the surface area also increases naturally in accordance with the volume (not in a linear rate). Its not with the length/ surface area of the tissue alone. This is how its done. The tissue isnt absolutely 2-dimentional to calculate it with square alone. It even has got the volume which affects the size of it. You dont always get the same trend of variance in surface area while the weight is increased in a particular rate, right? The same logic.

   

The above is a graphical representation of the same logic. The cube A is 1 unit long on each side, and the cube B is having the dimensions of 2, 2, and 0.25 units. You could realise that the volume of both the cube and cuboid is the same, but has a different surface area. Surface area is about 1.6 times that of cube A, while the difference in volume is about 2 times. This is not the point I am trying to say though. Read completely and try understanding the point.
As a second scenario, let the cube B have the dimensions of 2, 2, 0.5 units. You get the cuboid with double the volume and double the surface area of cube A at natural sizes.Though, here the ratio of the difference between the volume and the surface area of the two structures is 0.2599 times when scaled isometrically scaled down to the volume of cube A.
Again, given the third scenario be the cube B have the dimensions of 2,2, 0.75 units. Then, you get the cuboid with triple the volume and 2.3 times the surface area of cube A at natural size. Though, here the ratio of the difference between the volume and the surface area of the two structures is 0.1217 times when scaled isometrically scaled down to the volume of cube A.  This is tabulated till volume of B = 5, with an increment in one of the dimension with a rate of 0.25 square units. The scenarios are tabulated below.



Now, scaling with allometry isn’t significant here since the jaguar isn’t going to grow to the size of the tiger or vice versa. We just are trying to see the power-to-weight ratio which determines the ease of a body to do some work. The isometric method is highly significant for calculating the ease of locomotion in biological organisms or machines, though not that significant in bite force since it ultimately is the ability to crush, and not to move with weight as a parameter. Although, since the jaguar and the tiger stay in its weight, we are calculating the power-weight ratio for the above reason. This is the only way we can compare in real life, and is sensible, and this is the very reason to compare. It’s the measure of “ease”. This is what is called pound-for-pound measure in short.

   

If you could notice in the above table, by linearly increasing the volume of the cuboid B by one cubic unit each time (doubling, tripling, etc..), the rate of change in the surface area in accordance with the volume is not linear. This means that the difference in surface area after scaling both the structures each time to the same size isometrically doesn’t change in a linear way. For more clarity, there is the graphical representation given below.


   


In the graph, you can see that the volume- Δ Surface area curve (after isometrically adjusting to equal volume) isn’t linear first of all, and isn’t having the same slope at every point. This says that the relation isn’t linear to directly scale it always by squaring or applying inverse square, at any given random point.

Here, we did it for perfect basic geometrical shape, and even then the change wasn’t linear. Then how could your method be applied to a complex biological structure in which nothing is linear by itself? We shouldn’t!





Okay, now let me check if your method goes in par with the derived values if we apply it in the same structures.
Let us now assume that the density of both the materials are the same and consistent like you assumed and has equal weights at each cubic unit volume, and the surface area as the force since you took the function of bite force as a direct variable of surface area (that’s the reason you took the square root). Now, applying your method.
Your method is as follows:
 
Since cuboid B is 3 times heavier than cube A, 3/1 = 3
(3)^1/2 = 1.44224957031 (Cuboid B’s mass score in relation to cube A )
 
14/6 = 2.333333333333333 (3 c.u. unit volume cuboid B and 1 c.u. unit volume cube A in the table)
(14/6)^1/3 = 1.52752523165 (Cuboid B’s surface area score with relation to cube A )
Now, as per the logic you said in your method, the result must be as follows:
(1.44224957031/ 1.52752523165) *100 = 94.41739752816475%

The above result says that the cuboid B is having only 94% of the surface area of the cube A when cuboid B with thrice the volume is isometrically scaled to the volume of A, which is absolutely wrong. The cuboid B is having larger surface area if isometrically scaled. Then how come you get a value which says of cuboid B having less surface area than cube A? (Actual value = 112%). You don’t get a cuboid with thrice the volume of cube A and 94% surface area of cube A, even if you vary the dimensions in any ways isometrically. There is no other cuboid with the same volume and surface area as of cuboid B, varying the dimensions with infinity many combinations isometrically. By this simple test, we can infer that the method by itself is wrong.
 
Thus, you cant scale a jaguar’s and the tiger’s bites to same weight with squaring or taking the square root and dividing by the inverse of ratio of difference between the weights. The method isn’t correct. The difference in ratio is itself because of the difference in composition. Then how can you hypothesise them being of the same composition? The hypothesis is itself biased towards tiger while you say them to be of similar composition. Even if you take such an hypothesis, the method is wrong. Hope you get it.
 
As a revision, to put it in simple terms, the surface area doesn’t vary with the power of 2 or 1/2, but with something less than 2 (ideally something between 1 and 2) since there is the volume (in addition, variable density as an unpredictable factor) too taking up the share positively in the size increase. Hope you get it.
 
 
 
Thanks,
Wrapp. Reply