The strongest bites in the animal kingdom

[Wrapp]

(05-13-2022, 09:08 AM)LonePredator Wrote:

(05-12-2022, 10:00 PM)Wrapp Wrote: @LonePredator 

Brother, your logic about scaling the bite force of jaguar and tiger isnt right. The jaguar has the upper hand when both are scaled to the same size.

Actually you are right. The calculation was wrong. I guess my mind was somewhere else at that time, rather than doing a square root, I did a square and rather than doing a cube root of the mass I kept it the same. Never mind.

Now, @Pckts Here is the correct calculation this time of how much would a Jaguar’s bite force be in comparison to the Tiger if both are scaled to same size (allometric variations are not taken in count here)

200kg /100kg = 2
Now we must do a cube root of 2 and we get 1.2599 (Lets call this number the Tiger’s bodymass score in relation to Jaguar)

Now, 1000/750 = 1.333
Now we must do a square root of this and we will get 1.1547 (let’s call this one the Tiger’s bite force score)

Now these two ‘scores’ can be compared linearly so...

1.1547/1.2599 = .9165
.9165 x 100 = 91.65

This means that when scaled to the same size, the bite force of the Tiger would still be only 91.65 % that of the Jaguar (without any allometric variations)

Which means that when scaled to the same size as a Bengal Tiger, the Jaguar would have an 8.3% stronger bite (as long as the given bodymass in that study is correct with its respective bite force)

Yeah, I appreciate that you made corrections. But that wasnt the exact mistake. Even your previous method favoured jaguar. Let me explain that mistake you did in the previous calculation. And also, the method you used now isnt right, and I will be explaining it here. Hope you read the full reply and try to understand.


The major fault which turned the actual result of your previous calculation upside-down was this:
 
While you started by squaring 4/3, the 4/3 is the tiger’s bite when looking at it from the jaguar’s perspective.
i.e., j = ¾ t (as said in the research by Adam Hartstone-Rose and c.o.)
à 4/3 j = t
Let the above ‘j’ and ‘t’ represent the bite force of jaguars and tigers at their natural weight. So this is how you are proceeding with the 4/3. Now, you square it as a function of surface area and bring it to half, and then make it in percentage.
 
[([(4/3)^2]/2)*100] = 88.88888888888887%
 
Exactly as you got in your calculation. Everything is right.
But, you said that the jaguar is having only 88% bite force of tiger. But the logic is the other way round. Your calculation says that the tiger has only 88% bite of the jaguar at equal size, and does not say about the jaguar having 88% of bite force of tiger.
Check the logic behind the equation once again by yourself and you will get it.
Whatever you calculated, it was a function of ‘t’. It was the value of tiger with reference to jaguar, not that of jaguar with reference to the tiger, like you have previously assumed.
Thus, your equation proves that the jaguar still has 11% (11.11111111111113%) higher bite force than the tiger at same size, taking jaguar as a benchmark reference. Another point of view is that the jaguar has 12.5% higher bite force than the tiger at same size, taking tiger as a benchmark reference.
Your equation says that the jaguar still has a higher bite force at same size. It was because of a small misunderstanding that you made this mistake. Everyone is bound to make mistakes, right? It’s a human thing. And I appreciate your genuinity to recalculate and share it here. Thanks. Reply