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The strongest bites in the animal kingdom
The strongest bites in the animal kingdom
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The strongest bites in the animal kingdom
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 (04-30-2022, 02:58 AM)LonePredator Wrote:(04-30-2022, 02:50 AM)Pckts Wrote:(04-30-2022, 02:02 AM)LonePredator Wrote:(04-30-2022, 01:53 AM)Pckts Wrote:(04-29-2022, 11:16 PM)LonePredator Wrote:(04-29-2022, 11:07 PM)Pckts Wrote:(04-29-2022, 10:28 PM)LonePredator Wrote:What are you not understanding here?(04-29-2022, 10:15 PM)Pckts Wrote:(04-29-2022, 09:44 PM)LonePredator Wrote:(04-29-2022, 09:36 PM)Pckts Wrote: You're not comparing averages of both specimens, you're comparing an average of one specimen to a single individual of another.WRONG! Everything you said is completely WRONG. I already made it clear above that I am comparing the AVERAGE Sumatran to the AVERAGE Jaguar. It has nothing to do with understanding, you're cutting corners to try and save face. You are disregarding real factors that came into play in determining the bite force. It's literally providing you the multiple processes used in making that conclusion. It's using real jaw muscle weights obtained against an estimated body mass on top of that, it's bringing into account bone formation and muscle distribution. All of which would need to be allometrically scaled. You are simply just using a calculation to say "what if the jaguar was double the size" but it's absolutely clear that's not the only factor used to determine it's biteforce. On top of that, you have real measurements from comparable animals not in captivity and have been explained the actual differences in their morphology. The fact that you don't seem to acknowledge this or understand it makes the rest this debate pointless. 
Comparison of bite force of multiple different crocodilian species.
 *This image is copyright of its original author
@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. (05-13-2022, 09:08 AM)LonePredator Wrote:(05-12-2022, 10:00 PM)Wrapp Wrote: @LonePredator 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. 
@Wrapp @LonePredator
To sum up your differences of interpretation, just a mathematic aspect: X + 0,125X = X + X/8 = 9/8X = Y Y - Y/9 = 8Y/9 = X (05-14-2022, 03:39 AM)LonePredator Wrote:(05-13-2022, 10:14 PM)Wrapp Wrote:(05-13-2022, 09:08 AM)LonePredator Wrote:(05-12-2022, 10:00 PM)Wrapp Wrote: @LonePredator 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. Thanks, 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.
@LonePredator
OMG! Brother, my intention was simple and clear. I dont get why would you not understand the simple point even after trying my best to make you understand the same thing multiple times. (04-28-2022, 01:22 PM)LonePredator Wrote: Now, 4/3= 1.3333 and when it is squared, that will give you 1.333^2 = 1.7777 1. Now, look at what you did. Yeah, I am saying about your old method. I am saying about the misinterpretation you did. 4/3 is absolutely allright! I didnt say anything against that. Please understand! Now, how exactly did you get 4/3? This is how you got: tiger bite force * 3/4= jaguar bite force , In the next step, tiger bite force = 4/3* jaguar bite force. This is how you got the 4/3 What is this 4/3 in actual? It is the function of tiger. You would get the value of tiger, not that of jaguars while you use 4/3. To get the value of jaguar, you must use 3/4 since it is the function of jaguar. You interchanged the function. You applied the tigers function in the equation to get the jaguars value. How can this be done? You get the tigers value if you apply the tigers function, which is the 4/3. After calculating with 4/3, the sentence you made was this: " Which means even at same mass, the Jaguar would still have only 88% of the bite force of a Tiger" But, since you calculated with the tiger's function, the 4/3, you must have written as below: " Which means even at same mass, the Tiger would still have only 88% of the bite force of a Jaguar" Else, to use your same sentence, you must have calculated with the function of jaguar, the 3/4 and showuld have made necessary changes you get by doing the calculation with 3/4. This is the mistake I am trying to convey to you from the beginning. Hope you got it atleast by now! Now, look at what you wrote in your post recently, below! (05-15-2022, 11:50 AM)LonePredator Wrote: I calculated the RELATIVE value of the Tiger’s bite to the Jaguar’s bite ON PURPOSE Exactly! You calculated the relative value of the tiger's bite to the jaguar's bite. Now, you yourself are saying so, while you said it was the relative value of jaguar to that of tiger in your old post. The sentences in both of your posts are contradicting each other. Hoping that atleast by now you get the mistake that I am trying to point out. (05-15-2022, 11:50 AM)LonePredator Wrote: And the next thing, you claimed linear scaling. WHEN did I say I was doing linear scaling? I’m NOT scaling the Jaguar to the same LENGTH as the Tiger, I’m scaling the Jaguar to the same MASS as the Tiger and mass is NOT linear in relation to length, mass is cubic since it’s directly proportional to the three dimensional volume (but you are assuming a linear relation for some reason and I don’t know why) 2. Awesome! Where exactly did I do the lienar sacling? Or where did I do something relative to scaling the Jaguar to the same LENGTH as the Tiger? My friend, I scaled everything based on the volume. i.e., mass. Did I scale anything in relation to its length? None of the scaling was relative to the length, but was to the volume/mass. I write it down in the simplest way possible and I dont get how you arent able to perceive things in the way it is said. (05-15-2022, 11:50 AM)LonePredator Wrote: And your talk about a complex 3d shape. No! It doesn’t matter because when you scale up a Jaguar the Jaguar would still have the same shape. A cube scaled to a different size will still be a cube while a Jaguar scaled would still be a Jaguar and still have the same proportions as before (only hypothetically of course as every individual Jaguar has different proportions) 3. The above one confirms that you didnt understand anything. Yes, its true that the jaguar scaled up or down will have no difference in the composition. But are you sure about the composition of the tiger? So you think that the composition, structural dimentions, and everything are same between the jaguar and tiger. You scale the surface area and body mass alone to the equal, and you dont bother about making the composition the same. Is this even right? It cant be done that way since the composition (muscle mass/ density) of the tiger is different to that of the jaguars'. You cant equate the mass surface area alone and assume the composition too to be equalised. There are a dozen of parametres apart from body mass and surface area to consider in the equation. No, you didnt get what I mean even this time and you are gonna come again saying the same. Brother, please take some time and think on it. I said about the complexities which will arise when comparing two different structures of different compositions and ratios. This was the reason I took the cube and cuboid to signify the difference in the compsition in the first place. (05-15-2022, 11:50 AM)LonePredator Wrote: And it’s VERY OBVIOUS that a graph of a cubic value against a linear value will NOT produce a straight slope, that’s pretty obvious and everybody already knows that. That’s not really some new knowledge. Why did you assume that I was assuming a linear relation between surface area and volume/mass (which I never did)? 4. Agin?! Linear value? What is this linear value? Its a 2-dimentional value, the surface area. You did the same. You took the surface area and the mass. Even I took the surface area and the volume as mass since mass is a direct the consequence of volume and density. And in this case, this goes right in gell with your hypothesis of both the objects being of the same densities. Just the exact same relation. If you are saying about differentiating the values, then I did it to show the unpredictability of the change in surface area to change in volume/ mass/ density. I differed the volume/ mass of the cuboid and showed the differential value, not the difference. The diffrerence isnt the point here. You cant form a universal equation which differentiates the mass-surface area relation between any object. You may be able to do it for a specific object, not for any and every object, just like the case with the jaguar and the tiger. (05-15-2022, 12:17 PM)LonePredator Wrote: And THIS IS IMPORTANT. Your comparison between the cube and cuboid of equal volume proves that you do not understand the difference between ‘equal to’ and ‘proportional to’. Do you understand the difference between proportional and equal? 5. This in turn proves that you think that the jaguar and tigers are having the muscles with same proportions/ densities. This was the very sole reason for me to take the cube and the cuboid, my friend! I very well do understand the difference between 'equal' and 'proportional'. The whole point of my comparison was to say that you dont get anything with such an equation while the objects in comparison are of two different dimentions/ proportions/ densities. This is similar to that in s.no.3. (05-15-2022, 12:17 PM)LonePredator Wrote: Next, you are also for some reason assuming that the ratio of growth in volume and ratio in growth of surface area should be same as per my method BUT IT IS NOT AND I NEVER SAID IT IS. 6. Oh my god! Is that what you could understand? I am fearing that all my effort is going in vain since what you are perceiving is not even in the first degree of what I meant to say. We can talk only to people who can understand things in the way its is, right? I am giving you a totally different thing, and what you reply to me with is completely out of the cosmos. The gorwth of surface area and volume was to convey to you that they are unpredictable. That is, we cant find the surface area of an object if the volume is given for the other, and vice versa, if the proportions of the two object arent the same. You cant make a univerasl equation which is compliant for relation between all the various structures. I have to repat the same thing again and again. (05-15-2022, 12:17 PM)LonePredator Wrote: You first need to understand that mass varies cubically with one dimensional values while surface area varies squarely with one dimensional values. That is basic level physics. 7. This is where the root of your mistake is. Where does the square-cube realtion apply? Does it apply on two objects with different proportions? Asking since you equated both of the objects with same mass and same proportions while they were both not. This is why I took a cube and cuboid to represent the difference. And I did scale it isometrically so that both the cube and cuboid had the same volume (mass), like we would like to scale the tiger down to the same mass without changing the individual proportions. The whole point of doing that was not for abiding by your equation, but was to point out the basic mistake there. Do you know something? While you assume both the tiger and jaguar have the same proportions and compositions of muscles, you defeat the whole purpose of the comparison. If that was the case, then both would proportionally have the same bite. Then why the comparison at the first place? The variance in the bite if scaled proportionally is itself because of the difference in proportions, dimentiaons, and the densities. Why would you even scale if you know that the composition, dimensions and everything are the same? (05-15-2022, 12:17 PM)LonePredator Wrote: Mass of bigger cube/mass of smaller cube ^ 1/2 8. Yes, you are right. Let me correct the mistake. Let me correct and check if it abides by your method. Let me follow your exact method line by line now. 200kg /100kg = 2 -->similarly, 3 cubic uints/ 1 cubic unit = 3 cubic unit. (mass is the consequence of volume and density [M = Vd], thus equating mass as per your hypothesis) 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) -->similarly, Now we must do a cube root of 3 and we get 1.4422 (Lets call this number the cuboid B's mass score in relation to cube A) 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) -->similarly, Now, 14/6 = 2.3333 Now we must do a square root of this and we will get 1.5275 (let’s call this one the cuboid B’s bite force [surface area] 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) -->similarly, Now these two ‘scores’ can be compared linearly so... 1.5275/1.4422 = 1.0591 1.0591 x 100 = 105.91 This means that when scaled to the same size, the surface area (bite force) of the cuboid B would be 105.91% that of the cube A (without any allometric variations) ,which again is absolutely wrong. As per the precise calculations, the result is that cuboid B will be 112.17% that of the cube A (with precise isometric variations) This says that the equation of yours only works for the two structures that are of the same composition, dimentiaons, and everything, and not for structures which vary in composition, dimentions, etc. Do you know something? While you assume both the tiger and jaguar have the same proportions and compositions of muscles, you defeat the whole purpose of the comparison. If that was the case, then both would proportionally have the same bite. Then why the comparison at the first place? The variance in the bite if scaled proportionally is itself because of the difference in proportions, dimentiaons, and the densities. Are you aware of the bfq equation? They dont take the exact square and cube roots, instead they use precise numbers to scale them with parametres as log of body mass, etc... You cant just like that get off with square root and cube root like you do it for two things with exact same proportions (which is not the case with jaguars and tigers). Why would you even scale if you know that the composition, dimensions and everything are the same? We cant simply assume such things just like that. Its for sure that the composiyion, dimentions and the very structure isnt the same between the jaguar and tiger. Thus, you can form equations only after taking in consideration all the carying factors, just exclusively for the comparison between the jaguar and tiger. Otherwise, its wrong. Hope you understand. Thanks, Wrapp.
I dont know how to make LonePredator understand things. Need someones' help to teach LonePredator the logic behind the equation he had put.
(05-15-2022, 07:49 PM)LonePredator Wrote: I don’t know if you’re doing this on purpose or you’re just that clueless to begin with. I am not at all being arrogant or argueing for nothing. Are you feeling so? If so, I have nothing to say more on that. I am surprised that you still are so adament and not even taking an effort to open your eyes and look at it! You still stay with my "results on two different perceptions" I made to eliminate any possible excuses from you. The two perceptions arent at all the case. Just forget about the two results of the "perception" thing. Now, just look at the logic behind your old calculation. Now, please "educate" me step by step on what you did on your previous equation. i.e, the meaning of the factors you took and all. Please do it! Then I can guide you on where you are going wrong. Until then, I have realised that its impossible to make you understand things. Its simple maths, my frend! Or just leaving the ego behind and looking at your equation again will do it. (05-15-2022, 07:49 PM)LonePredator Wrote: And WHEN did I say that I assumed both Jaguars and Tigers have the same structure? Did you even understand a fraction of what I said? It’s COMMON SENSE but you seem to lack that in regard to this. I said that the bigger Jaguar would still have the same structure as the smaller Jaguar. THAT is what I said but I think you are purposefully being deceitful and misinterpreting all that I said. Oh! Its you who still didnt get what I said. Why did you even say that? For you to scale the tiger to the jaguar's weight expecting the surface area to decrease isometrically when doen with the square and the cube root, its ridiculous! This is the 5th or sixth time I am saying the same thing to you. Now, where does the square-cube rule apply? It applies between two exactly similarly-proportioned structures. Is that the case with the tiger and the jaguar, that they have the same proportions? You say that you didnt say the tiger's and jaguars have the same structure, but you use the logic that can be used only for two objects having same proportional structure. You arent gonna Omg! I dont even know how to make you understand! This is the simplest way a human can explain things. If you dont understand, I have no comments. Its like a gone case! (05-15-2022, 07:49 PM)LonePredator Wrote: It’s COMMON SENSE but you seem to lack that in regard to this It would be kind if you be a bit polite! (05-15-2022, 07:51 PM)LonePredator Wrote: I don’t intend to be harsh but you have absolutely NO CLUE what you’re talking about, the BFQ calculation is done by taking in count the regression of allometric variations with mass but we’re not going to allometrically change a Jaguar, we’re doing it isometrically. You are clearly unable to comprehend the difference between isometric equations and allometric equations because if you did, you would’ve never said this. Oh, so you do isometry between two different structures with square root and cube root, isnt it? You will get answer to this if you ask to yourself whether you may be able to apply the Galilio's law when scaling the tiger and jaguar. You are not separately scaling the jaguar isometrically to the weight of tiger and then comparing their bie forces. You arent doing that. And you cant do that since you dont have the dimentions in your hand other than just a ratio between the bite forces. We cant do that! (05-15-2022, 08:11 PM)LonePredator Wrote: You also do not know what square cube law is and you are trying to prove the square cube law wrong. If two objects are of the exact same shape and proportions (the bigger and smaller Jaguar) then the square cube law applies to it but you are saying it’s wrong? Do you understand what you are talking about? (05-15-2022, 08:11 PM)LonePredator Wrote: Before going any further, you tell me the difference between proportional and equal. Do you understand the difference between proportionality and equality? YOU DO NOT because I saw your example of cube and cuboid of same volume and that example of yours PROVES that you don’t understand the basic difference between proportionality and equality. You are getting it wrong. You think I dont know the difference between proportionality and equality? This is the problem with you that you dont get what was meant or try to understand what is done. Although, I got what you meant, but you still didnt. I get your idea. Its not because I dont understand, but because you dont understand what I mean. Your aim was to isometrically scale the jaguar to the weight of the tiger, and then calculate the difference in bite. Inst that exactly what you meant to do? Now, This is the important one. Please listen to this properly. I will elaborate your own method. Now, you are comparing a tiger and a jaguar, which means that you are having two different composions and structures. And, you include the values of both the jaguar and tiger to the equation, isnt it? So, naturally, you are dealing with two different structures with different compositions. Thus was my cube-cuboid example. You arent isometrically scaling the mass of the jaguar alone and then looking at the difference in bite force in your equation, though that was your intent. For that, you need the proper dimentions and density values of both the structures. Only then you can do so as you intended. Or else, just elaborate the logic of your equation here step by setep saying out the logic and the reason behind doing each of your steps. While you write down, you wil hopefullyl realise the flaw. (05-15-2022, 07:49 PM)LonePredator Wrote: And in such a case, we can use simple Galileo’s law to calculate the surface area if we already know the mass (and you are saying Galileo’s square cube law is wrong) Is that so? I didnt even say that the Galileo's law was wrong. Do you still use Galileo's law even after this in this case? Come on! (05-15-2022, 10:06 PM)LonePredator Wrote: You just either lack the required common sense to understand this comparison or you aren't using it. Just use your common sense, if I did assume the shape AND the same mass, then wouldn't the bite force have come out the same for both? WHICH IS NOT the case. Brother, you hopefully will soon realise who fits for this sentence. Thats what I said. If you assume the same shape and proportions, there would be no need for any comparison. You understand the issue with your equation and are still not properly demonstrating youe method since you are worried about getting falsified. Okay, in that case, let me commit to this. (05-15-2022, 10:06 PM)LonePredator Wrote: I calculated the ratio of the Tiger's mass to the Jaguar's mass (the values are still kept separate) and then I did a cube root of it (since both the masses are cubically proportional to 1 dimensional values but the ratio still presents a SEPARATE value for both) This can be understood as soon as we see the equation. This isnt something to be understood only after you make it a sentence. I am asking about the logic behind including the values of the tigers and jaguars, and then saying you are isometrically scaling only the value of the jaguar using the Galilio's rule. You cant give a logic cuz there isnt one. (05-15-2022, 10:15 PM)LonePredator Wrote:(05-15-2022, 09:50 PM)Wrapp Wrote: I dont know how to make LonePredator understand things. Need someones' help to teach LonePredator the logic behind the equation he had put. Let the sensible ones understand. I am sorry to say that you are hopeless until you keep your ego away and try to understand things. There is a saying which says that a doctor is the patient for the patient in the asylum. (05-15-2022, 10:16 PM)Wrapp Wrote:(05-15-2022, 10:06 PM)LonePredator Wrote: You just either lack the required common sense to understand this comparison or you aren't using it. Just use your common sense, if I did assume the shape AND the same mass, then wouldn't the bite force have come out the same for both? WHICH IS NOT the case. You said the following and I QUOTE: "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." That is what you said and I quoted you word by word. NOW YOU TELL me what you said there is right or wrong? (hint: it's COMPLETELY WRONG) because the square cube law says the exact opposite of what you said (05-15-2022, 10:22 PM)LonePredator Wrote: You said the following and I QUOTE: Brother, are you not able to understand, or are pretending to understand nothing? In the first place, the debate is about saying that you cant use the square cube law here. Then from where do you try to counter me by using the same? First, you need to prove that the square cube law is applicable in this comparison. Only then you can take such statements. I have been saying for such a long time on why the law of square-cube cant be used here. You still dont give me an answer on why exactly you included the values of tiger and jaguar into an equation, and claiming it to be scaling the jaguar alone. Both are contradicting. This was the justification you gave me, : "I assumed that the BIGGER JAGUAR would still have the same proportions as THE SMALLER JAGUAR" You said that the jaguar doesnt change in proportion if you scale it isometrically up or down. Your statement is absolutely correct. This is correct when you scale the jaguar up or down to a particular weight. But, you say of scaling the jaguar alone, while you have an equation which differentiates both the tiger's and the jaguar's values together at the same time. But differentiating the taiger's and jaguar's values togeher in an equation, there is basically the idea of scaling the jaguar to the size of the tiger defeated since you have two variables and you do a one-dimentional regression which quantifies both the values in parallel. If you assess the functioning of your equation, you could see that this is done in there. Thus, having two pairs variables (varies on diffferentiating) in the equation, i.e., the weight and the bite force of the tiger and the weight and bite force of the jaguar, everything quantified in the equation, which in other words means that you are balancing two different structures with non-proportional values in your equation. This is the reason to take the cube and the cuboid. Now, you cant use two objects of different proportions in the square-cube law. Now, what you devised was not wrong, but was not correct for this scenario since you are having only the ratio between the weight and the bite force. You simply cant do it with the logic of square-cube law. You dont know how thick is the muscles of jaguar compared to the tiger. So, the relation between an scaling a proportionally-less bulky material to a bulky material which was initiallly smaller, and vice versa will have no relativity equation devised until and unless we know exactly how bulky the material is in comparison. Without having any information regarding this, an equation which deals with the ratio alone will do it, but will not justify the reality. Your equation is not wrong, but it cant be used in the scenario given here. We are simply in lack of data. Thanks, Wrapp. (05-16-2022, 04:56 PM)Wrapp Wrote:(05-15-2022, 10:22 PM)LonePredator Wrote: You said the following and I QUOTE: Okay so my calculation was indeed wrong. I'm sorry for stretching it for this long and arguing for this long, I'll delete the previous posts now. Finally, here is the ACTUALLY CORRECT calculation: 200kg/100kg (new mass of Jaguar/old mass of Jaguar) = 2 Now, the bite force will increase by 2^2/3 times which means the bite force will become the following: 750N x 2^2/3 = 750x1.5874 = 1190N So IF the 100kg Jaguar had a bite force of 750N then the new 200kg Jaguar will have a bite force of 1190N (But almost everything you said about the square-cube law and everything else still was TOTALLY wrong) (05-16-2022, 05:51 PM)LonePredator Wrote: Okay so my calculation was indeed wrong yet again. I'm sorry for stretching it for this long and arguing for this long, I'll delete the previous posts now. There you come! I appreciate you for the correction once again. Atleast now you are able to understand things. Anywhays, this is a quick post and will be getting back to you once again after reviewing your method. (Whatever I said about the square cube law was correct though. People who read it will understand.) |
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