The strongest bites in the animal kingdom

[LonePredator]

(05-16-2022, 07:49 PM)Wrapp Wrote:

(05-16-2022, 06:22 PM)LonePredator Wrote: What you said about square cube law was COMPLETELY WRONG!

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.

And now this is the exact definition of the square cube law below:

“When an object undergoes a proportional increase in size, its new surface area is proportional to the square of the multiplier and its new volume is proportional to the cube of the multiplier.”

So the exact definition of the square cube law is EXACTLY OPPOSITE of what you said.

If you are still going to try and justify that, then you’re just ridiculing your own words.

And what will you review about my method? If you really were going to correct me then why didn’t YOU do the correct calculation yourself till now??

You are still clueless about the blunder you did, or are trying your best to argue for nothing hoping to make yourself appear correct. A false upon repeating a hundred times doesnt become a correct.
There wasnt any need to paste the meaning of the square cune law here. You arent for some reason aware even by now that whatever you did in your previous method wasnt even abiding by your own logic. You say of scaling the jaguar and come with an equation which parallels both the values of both the different structures together. There is no need for me to argue on this to make you alone understand since the point I made is clear and is understandable by all. I dont get anything by making you understand things.

Anyways, atleast in your latest re-re-corrected equation, you have done only on a single parametre of just considering the jaguar alone. This is what was supposed to be considered doing earlier. But again, 
Even your recent method is wrong. Though, this time, it is going to be much easier to prove you wrong since there arent much parametres to deal with. This is getting interesting though!

Lets do it now. Yay!

Now, take the cuboid alone. We are just considering the jaguar alone in your equation, and thus the cuboid alone.

The cuboid is of the dimentions 2,2,0.25 units. (assume this be that of jaguar's with the following dimentions)

The volume of this cube happens to be 1cu. unit
      and the surface area of the same cube = 10sq.units

      ∴ vol = 1cu. unit
         sur = 10sq. unit


We are going to scale this isometrically to two times its volume.

By isometrically scaling the initial volume (1 cu.unit) of the cuboid to double its volume (2 cu.unit), the dimentions of the new cuboid happens to be:

2.519842099786888, 2.519842099786888, 0.314980262473361
           (2)                                   (2)                                 (0.25)

Lets now calculate the surface area of this new cuboid which was isometrically scaled using the traditional method.

[b](2.519842099786888 X 2.519842099786888) *2 + ([b][b]2.519842099786888[/b] X 0.314980262473361)* 4[/b][/b]

[b]= 15.87401051964599 sq.units    <-----  surface area after isometrically adjusting the cuboid to double its weight.[/b]

[b]Now, lets keep this value aside for now.[/b]

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Now lets use your own method here to try and get the same value as above. 
(Arent you so excited to see the results? Yes, you will see it in a few lines ahead. Enjoy!)


2/1 (new mass of cuboid/old mass of cuboid) =  2

Now, as per your assumption, the surface area will increase by 2^[b]0.6299605249488654 times which means the surface area will become the following:[/b]
 
10 x 2^0.6299605249488654 = 10 x 1.54752264962 = 15.4752264962 sq units  <--- [b][b]surface area as per your method after isometrically (allegedly) adjusting the cuboid to double its weight.[/b][/b]

(Note: 10/15.87401051964599 = 0.6299605249488654, where 10 is the surface area of the old cuboid and 15.87... is the surface area of the new cuboid)

Now, as you can see, the values arent the same with the one calculated using the proper method. This straight to the point proves your method wrong. (calculated up to a minimum of 10 decimal units for precision upto 10^10).

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Now lets see the deviation of your assumption to the reality.

[b][b]([[b][b]15.87401051964599 - [/b]15.4752264962] [/b]/ 15.87401051964599) X 100 = 2.512181927512565%[/b][/b]

[b]Result: You method easily gets a variance of 2.5% when comparing even a small-scale numeral like 10 and 14. It is by common sence understood to be getting a large variance when calculating on larger numerals like 750, etc. Hence proved the unreliability in your method.[/b]

Edit: Ignore the '' thing which appears between the lines. Its because of some kind of error.

Thanks,
Wrapp.

COMPLETELY WRONG! Finally, thanks for proving my initial point that you don’t even know simple maths and physics. Now it’s much easier to show how clueless you are. First of all, are you able to see the text I wrote or is that your low knowledge of mathematics? Read what I wrote again.

First of all, HOW did you get that 0.6299 value? The value of 2^2/3 is 1.587401 while your value came out to be 1.54 something, THIS PROVES YOU CANT EVEN DO A SIMPLE CALCULATION. and then when you multiply it with the actual value of 1.5874 then you’ll get the correct value. Only if you knew how to calculate.

If length gets increased by 2 times then the surface area will increase by 2^2 times and the volume will increase by 2^3 times while you are claiming some weird decimal exponent.

This itself means that when volume/mass is multipled by two. i.e. x^3=2 then the surface area is x^2=2^2/3. Now this is a good sign that you need to study high school maths and physics.

I’ll try to make it easier for you to understand because you apparently have a very hard time trying to understand simple maths.

Now read carefully. The VOLUME here has increased CUBICALLY. (x^3) and the value of x^3 is 2. Then the length must have increased linearly (x) and then the value of increased length is 2^1/3 so then the SURFACE AREA has increased squarely i.e. (x^2) so the value of that is 2^2/3. Maybe you still don’t understand but you’re just proving your lack of knowledge. Keep going.

And why did you ignore the whole thing I said? The square cube law says the EXACT opposite of what you said. How will you justify that now? I can see that you carefully avoided and ignored my question in fear of speaking something silly and embarrassing yourself. Please explain this first before trying to correct me Reply