There is a world somewhere between reality and fiction. Although ignored by many, it is very real and so are those living in it. This forum is about the natural world. Here, wild animals will be heard and respected. The forum offers a glimpse into an unknown world as well as a room with a view on the present and the future. Anyone able to speak on behalf of those living in the emerald forest and the deep blue sea is invited to join.
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.
No, actually that calculation wasn’t even correct in the first place. You are saying that I got it the other way around but that’s not the case.
The 4/3 thing was actually correct, the only thing that was wrong was that rather than doing a square root of the ratio of Tiger’s force to Jaguar’s force, I did a square of it. And rather than doing a cube root of the Tiger’s mass to Jaguar’s mass, I kept it the same. That was the mistake in it.
Now here, I’m doing it the right way. I AM calculating the ratio of the Tiger’s bite force score to the Jaguar’s bite force score and the Tiger’s bodymass score in relation to Jaguar’s bodymass score. That’s not a mistake, I did that on purpose.
In short, please ignore the first one about the first calculation of 12.5% as that was not correct.
The second one right here is the correct one so use that. And the 4/3 thing isn’t a mistake, that has been done on purpose and that is all fine. You can calculate the Tiger in relation to the Jaguar or the Jaguar in relation to the Tiger, it makes no difference.
As @Spalea explained, I calculated the Tiger in relation to the Jaguar while you’ll calculate the Jaguar in relation to the Tiger, both are correct, it’s just a difference of interpretation.
I know that the equation was wrong. But I tried to point out the mistake you did inside that equation. Ig you didnt get what I said. I would explain it again. If you look at the calculation I did using your old method, I calculated in both the perspectives and got two different values. That was not the point there. If you take a look at it again, I have explained your mistake in it clearly. The point was that you took the multiplying factor being a function of 't'. This means that you calculated the value of tiger, and not the jaguar's. I didnt say that the 4/3 is a mistake. I have clearly explained it in there. Think you didnt properly read it. Requesting to go through it once again. Thus, in your old equation, the reality was that the tiger was scoring less to the jaguar being only 88% the bite to that of the jaguar. It was the mistake you did in determining whose value you derived. Hope you get it. I will post the same thing here once again, and hope you read it and understand what I said.
While you started by squaring 4/3, the 4/3 is the tiger’s bite when looking at it from the jaguar’s perspective. (not saying that the 4/3 is wrong) 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. Its absolutely ok. Now, you square it as a function of surface area and bring it to half, and then make it in percentage.
Now, note the point here. The equation says that the tiger is 4/3 times the jaguar. So, the result you calculate with 4/3 is the tiger's measure. Not the jaguar's one! [([(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. This was since you mistook the 4/3 to be function of jaguar, but in reality which was a function 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’. Whatever you calculated, it was for jaguar, not the tiger. 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. (I have took result from both the perspective and none of em says of jaguar being only 88% of tiger, but says the other way round.) Your equation says that the jaguar still has a higher bite force at same size. Hope you could get it now.
And also, your latest method isnt proper for your kind information. I will be explaining the problem with your latest method in the next post in much detail.
The strongest bites in the animal kingdom
