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05-16-2022, 06:22 PM( This post was last modified: 05-17-2022, 04:58 AM by LonePredator )
(05-16-2022, 05:55 PM)Wrapp Wrote:
(05-16-2022, 05:51 PM)LonePredator Wrote: 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.
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.
Once again, I'm sorry I streched this argument for this long and I'll delete my previous posts now.
(What you said about the square-cube law still wasn't correct though)
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.)
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 yourself. You were still wrong about some of the very basic things of physics and yet you are too arrogant as if you are the expert. Are you here on wildfact just to show off despite the fact that you lack even high school level knowledge?
And what will you review about my method? To review my method you first need to know the concept of the square cube law (which you have no idea about).
If you really were going to correct me then why didn’t YOU do the correct calculation yourself till now? If you really knew the all the concepts then shouldn’t you have already done the correction in my calculation till now?
05-16-2022, 07:49 PM( This post was last modified: 05-16-2022, 07:54 PM by Wrapp )
(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:
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] 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).
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 '[b]' thing which appears between the lines. Its because of some kind of error.
05-16-2022, 08:01 PM( This post was last modified: 05-17-2022, 05:03 AM by 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:
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] 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).
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
05-16-2022, 08:26 PM( This post was last modified: 05-16-2022, 08:37 PM by 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:
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] 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).
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.
Let me make it a little more easier for you to understand with an example. AND READ CAREFULLY.
Let’s assume we have a cube and it’s length is 2cm on all edges so it’s volume will come out to be 8cm^3
Now let’s increase the lengths by 2 times then the new lengths would be 4cm on all edges and now the new volume will become 2^3 times the previous volume.
i.e. 8x2^3=64
And now for the SURFACE AREA. The surface area of the first cube was 6a^2 i.e. 24cm^2 and now the new surface area will increase by 2^2 times i.e. 24x2^2=96 and you can verify by foruma, the new surface area of the new cube is indeed 96cm^2
Now since the cube increased by x^3 times. i.e. 64/8=8 times (new volume/old volume) then the surface area would increase by 8^2/3 times i.e. 4 times.
And indeed, 96/24=4 so indeed the surface area did increase by 4 times.
So JUST LIKE I SAID, the volume increased by 8 times and the surface area increased by (8^2/3 = 4) times. ONCE AGAIN I PROVED THAT YOU DO NOT UNDERSTAND THE DIFFERENCE BETWEEN EQUALITY AND PROPORTIONALITY.
This much is enough to explain a high school student. If you can’t understand THIS then you really need to study high school maths. Now please think before you speak, it’s really embarrassing that you’re making such silly mistakes.
You [b]made such a silly mistake which even high school student never make.[/b]
(05-16-2022, 08:01 PM)LonePredator Wrote: 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 proportion.
You are making me laugh again and again! Say that you dont get what is being done! I showed this guy with the exact same methodology and this guy beats around the bush for no reason.
Do you have any idea on what you are doing? You are adjusting the jaguar's weight (volume in this case) to double and then seeing at what bite force value you get form it.
Just the same. Adjusting the volume of the cuboid to two times keeping the proportions the same and seeing the surface area you get after it. Do you understand what this means? Are you blinded seeing the decimals? Then say that! Dont say its wrong.
We are having the parametre of the volume. That is the thing to what we must scale first. You are gonna scale the jaguar to the twice its weight first, and only then you are gonna measure/ assume the length or surface area of it whcih you will get as a result after scaling the weight to two times isometrically. Just like that length and the surface area of that new cuboid is something we measure/assume from after sacling it to twice the volume first. Its the result, not your given parametres, man! Yo! This is getting real funny! Look at how you think!
(05-16-2022, 08:01 PM)LonePredator Wrote: 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 that whole topic because it proved that you don’t even understand high school level physics. Please explain yourself first before trying to correct me
I dont know how to make you hear my laughter! Nothing to say!
(05-16-2022, 08:26 PM)LonePredator Wrote: Let me make it a little more easier for you to understand with an example. AND READ CAREFULLY.
Let’s assume we have a cube and it’s length is 2cm on all edges so it’s volume will come out to be 8cm^3
Now let’s increase the lengths by 2 times then the new lengths would be 4cm on all edges and now the new volume will become 2^3 times the previous volume.
i.e. 8x2^3=64
And now for the SURFACE AREA. The surface area of the first cube was 6a^2 i.e. 24cm^2 and now the new surface area will increase by 2^2 times i.e. 24x2^2=96 and you can verify by foruma, the new surface area of the new cube is indeed 96cm^2
Now since the cube increased by x^3 times. i.e. 64/8=8 times (new volume/old volume) then the surface area would increase by 8^2/3 times i.e. 4 times.
And indeed, 96/24=4 so indeed the surface area did increase by 4 times.
That is enough to explain a high school student. If you can’t understand THIS then you really need to study high school maths.
You are caught red handed now. This is what I wanted from you! Just this!
What exactly are you doing here? Any idea, my friend?
Are you scaling the jaguar to double its volume and then finding the bite force of it, or are you scaling the other parametres such as the length and the surface area and finding the volume which will result from it? Make it clear to yourself first on what you are doing!
I didnt expect that I have to say all this in such a detail to someone like you!
05-16-2022, 08:43 PM( This post was last modified: 05-16-2022, 08:54 PM by LonePredator )
(05-16-2022, 08:35 PM)Wrapp Wrote:
(05-16-2022, 08:01 PM)LonePredator Wrote: 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 proportion.
You are making me laugh again and again! Say that you dont get what is being done! I showed this guy with the exact same methodology and this guy beats around the bush for no reason.
Do you have any idea on what you are doing? You are adjusting the jaguar's weight (volume in this case) to double and then seeing at what bite force value you get form it.
Just the same. Adjusting the volume of the cuboid to two times keeping the proportions the same and seeing the surface area you get after it. Do you understand what this means? Are you blinded seeing the decimals? Then say that! Dont say its wrong.
We are having the parametre of the volume. That is the thing to what we must scale first. You are gonna scale the jaguar to the twice its weight first, and only then you are gonna measure/ assume the length or surface area of it whcih you will get as a result after scaling the weight to two times isometrically. Just like that length and the surface area of that new cuboid is something we measure/assume from after sacling it to twice the volume first. Its the result, not your given parametres, man! Yo! This is getting real funny! Look at how you think!
(05-16-2022, 08:01 PM)LonePredator Wrote: 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 that whole topic because it proved that you don’t even understand high school level physics. Please explain yourself first before trying to correct me
I dont know how to make you hear my laughter! Nothing to say!
(05-16-2022, 08:26 PM)LonePredator Wrote: Let me make it a little more easier for you to understand with an example. AND READ CAREFULLY.
Let’s assume we have a cube and it’s length is 2cm on all edges so it’s volume will come out to be 8cm^3
Now let’s increase the lengths by 2 times then the new lengths would be 4cm on all edges and now the new volume will become 2^3 times the previous volume.
i.e. 8x2^3=64
And now for the SURFACE AREA. The surface area of the first cube was 6a^2 i.e. 24cm^2 and now the new surface area will increase by 2^2 times i.e. 24x2^2=96 and you can verify by foruma, the new surface area of the new cube is indeed 96cm^2
Now since the cube increased by x^3 times. i.e. 64/8=8 times (new volume/old volume) then the surface area would increase by 8^2/3 times i.e. 4 times.
And indeed, 96/24=4 so indeed the surface area did increase by 4 times.
That is enough to explain a high school student. If you can’t understand THIS then you really need to study high school maths.
You are caught red handed now. This is what I wanted from you! Just this!
What exactly are you doing here? Any idea, my friend?
Are you scaling the jaguar to double its volume and then finding the bite force of it, or are you scaling the other parametres such as the length and the surface area and finding the volume which will result from it? Make it clear to yourself first on what you are doing!
I didnt expect that I have to say all this in such a detail to someone like you!
JUST READ THE POST ABOVE. I increased the volume of the cube by x^3 times i.e. 8 times and the surface area automatically increased by x^2 (8^2/3) times i.e. 4 times.
NOW CORRECT THIS (warning: if you try, you’ll end up embarrassing yourself a lot more)
The same thing happens with the Jaguar which increases in volume by 2 times so it’s surface area increased by 2^2/3 times (assuming same proportions as before)
Tell me the truth, how old are you? Are you in high school? AND ONCE AGAIN YOU AVOIDED THE WHOLE THING. WHAT IS YOUR JUSTIFICATION FOR THE STATEMENT YOU MADE WHICH WAS THE OPPOSITE OF THE SQUARE CUBE LAW If you are not a liar then address it first.
(05-16-2022, 08:43 PM)LonePredator Wrote: JUST READ THE POST ABOVE. I increased the volume of the cube by x^3 times i.e. 8 times and the surface area automatically increased by 8^2/3 times i.e. 4 times.
NOW CORRECT THIS (warning: if you try, you’ll end up embarrassing yourself)
Tell me the truth, how old are you? Are you in high school? AND ONCE AGAIN YOU AVOIDED THE WHOLE THING. WHAT IS YOUR JUSTIFICATION FOR THE STATEMENT YOU MADE WHICH WAS THE OPPOSITE OF THE SQUARE CUBE LAW If you are not a liar then address it first.
Really? Again? You are caught already. You are rediculing yourself again and again trying to prove nothing, ridiculing your own reputation.
I have to teach you right from the basic logics of mathematics! You made me do that!
Now, say me which is the parametre you are gonna scale to? You are gonna scale the jaguar's weight to two times isometrically. By doing so, you will get the surface area of the jaguar which was scaled to two times its weight isometrically. This is exactly what was done!
You dont first scale the surface of the jaguar to two times its surface area isometrically first and take the volume you get after it, prodigy!
What did you do in your demonstration? I am laughing so hard by now!
Would you want to find the weight of the jaguar after double the bite force isometrically, or would you want to find the bite force of the jaguar after doubling the weight isometrically?
If first one is the case, then what you demonstrated is correct, but will unfortunately make your equation wrong according to yourself. If the second one is the case, then that is exactly what I did and hence your equation will be rendered unreliable as already proved.
Just use your intellectual!
(05-16-2022, 08:43 PM)LonePredator Wrote: I increased the volume of the cube by x^3 times i.e. 8 times
What did you mean by the above sentence? You increased the volume of the cube by x^3 times? Is this even a logical mathemetical term? Did you mean to cube the side such that you got the volume? Wasnt that waht you meant? Its getting really funny!
05-16-2022, 09:11 PM( This post was last modified: 05-16-2022, 09:22 PM by LonePredator )
(05-16-2022, 09:03 PM)Wrapp Wrote:
(05-16-2022, 08:43 PM)LonePredator Wrote: JUST READ THE POST ABOVE. I increased the volume of the cube by x^3 times i.e. 8 times and the surface area automatically increased by 8^2/3 times i.e. 4 times.
NOW CORRECT THIS (warning: if you try, you’ll end up embarrassing yourself)
Tell me the truth, how old are you? Are you in high school? AND ONCE AGAIN YOU AVOIDED THE WHOLE THING. WHAT IS YOUR JUSTIFICATION FOR THE STATEMENT YOU MADE WHICH WAS THE OPPOSITE OF THE SQUARE CUBE LAW If you are not a liar then address it first.
Really? Again? You are caught already. You are rediculing yourself again and again trying to prove nothing, ridiculing your own reputation.
I have to teach you right from the basic logics of mathematics! You made me do that!
Now, say me which is the parametre you are gonna scale to? You are gonna scale the jaguar's weight to two times isometrically. By doing so, you will get the surface area of the jaguar which was scaled to two times its weight isometrically. This is exactly what was done!
You dont first scale the surface of the jaguar to two times its surface area isometrically first and take the volume you get after it, prodigy!
What did you do in your demonstration? I am laughing so hard by now!
Would you want to find the weight of the jaguar after double the bite force isometrically, or would you want to find the bite force of the jaguar after doubling the weight isometrically?
If first one is the case, then what you demonstrated is correct, but will unfortunately make your equation wrong according to yourself. If the second one is the case, then that is exactly what I did and hence your equation will be rendered unreliable as already proved.
Just use your intellectual!
(05-16-2022, 08:43 PM)LonePredator Wrote: I increased the volume of the cube by x^3 times i.e. 8 times
What did you mean by the above sentence? You increased the volume of the cube by x^3 times? Is this even a logical mathemetical term? Did you mean to cube the side such that you got the volume? Wasnt that waht you meant? Its getting really funny!
That is a thing called algebra my friend, do you also not know what algebra is?
The thing is that you have NEVER studied maths and physics in your entire life and you are trolling here. I said that when volume increases by 'v' times then the surface area increases by v^2/3 times and YOU SAID IT'S WRONG. You started saying all sorts of nonsense like red handed and what not.
I'll tell you in one sentence what I did in my calculation and then you answer me with yes or no.
If the volume of a cube (or any particular shape) increases by 2 times then it's surface area will increase by 2^2/3 times. YES OR NO? YOU SAID NO which means you never studied maths in your entire life.
(05-16-2022, 09:11 PM)LonePredator Wrote: I'll tell you in one sentence what I did in my calculation and then you answer me with yes or no.
If the volume of a cube (or any particular shape) increases by 2 times then it's surface area will increase by 2^2/3 times. YES OR NO? YOU SAID NO which means you never studied maths in your entire life.
Absolutely a big 'yes'. Yes! Are you satisfied now? That isnt even my argument in the first place, and you dont understand even what is the issue, and for the same reason you perceive everything wrong.
Look at yourself, my friend! You arent still getting the blunder you have made.
I respect you a lot and atleast now, try to understand the thing where you are going wrong! Take your own time to read this and I positively hope you understand what I mean to say.
You are right! If you increase the length by two times on every sides, you increase the surface area by 4 times and volume by 8 times. But, this isnt even the case here to the least. Let me say you how.
Now, you could notice that the cuboid 2, after increasing the sides of the cuboid 1 two times on each side. As you said, when we increase the sides by two times everywhere, the surface area increases 4 times and the volume increases by 8 times.
Since you said about the square-cube law here, you are saying about increasing all the sides by the equal multiple, which is basically multiplying all the sides by, for example, two and getting 4 times the surface area and 8 times the volume.
Its all right that you get 4 times the surface area, 8 times the volume if you increase the side by 2 times. Now, how do you find a cuboid with twice the volume of cuboid A which is isometric to cuboid A? Thats what we have to find. Lets look at it.
What was your task? You had to find the bite force of a jaguar after isometrically scaling it to twice its volume, right? So, you are scaling the jaguar to twice the volume isometrically, and then finding the surface area which you got after doing it.
Likewise, we are gonna scale the cuboid A to twice its volume isometrically (without changing the ratios between the vertics) and then find the surface area which we get form it. Isnt this the exact same way?
If you look at cuboid A and Cuboid B, cuboid B has twice the volume of cuboid A. But the proportions arent the same. So, how do you find the measures of a cuboid with twice the volume of A and which maintains the same proportions of A? Lets do it!
Now, lets scale the cuboid A which has volume 1, to volume 2 without changing its proportions.
Ratio of vertices of cuboid A : 1:8:8
Now, the equation of this cube be: x*8x*8x = 1 (where 'x' = measure of the smallest vertice)
Now, we are gonna keep the ratio the same and equate the volume (product of the three vertices) to twice the initial volume. By this, we could find the vertices of that cuboid which has twice the volume of cuboid A and stays with the same ratio. By doing this: we get:
x*8x*8x = 2,
64x^3 = 2,
x = (2/64)^ 1/3
= (1/32)^3
Therefore, x = 0.314980262473361
As we said, x is the value of the smallest vertice.
Now, lets find out the other two vertices. For that, we must multiply x by 8 (since ratio is 1:8:8).
Lets call this cuboid the 'cuboid awesome'. By doing so, the vertices of cuboid awesome are:
Now, check for the proportion. 2.519842099786888/ 0.314980262473361 = 8.
Thus, the ratios are kept the same, and the volume of "cuboid awesome" is twice the volume of cuboid A, as we were intending to get.
This is the exact same thing I did in the comparison. If you go back and see, the measure of the cuboid I took too was the same. I just elaborated my steps with reasoning. Thats all I did.
Now, can you say this isnt isometric variation? If you say so, I am happy to hear from you about why this isnt isometric.
05-16-2022, 11:18 PM( This post was last modified: 05-17-2022, 06:41 AM by LonePredator )
(05-16-2022, 10:48 PM)Wrapp Wrote: OMG!
(05-16-2022, 09:11 PM)LonePredator Wrote: I'll tell you in one sentence what I did in my calculation and then you answer me with yes or no.
If the volume of a cube (or any particular shape) increases by 2 times then it's surface area will increase by 2^2/3 times. YES OR NO? YOU SAID NO which means you never studied maths in your entire life.
Absolutely a big 'yes'. Yes! Are you satisfied now? That isnt even my argument in the first place, and you dont understand even what is the issue, and for the same reason you perceive everything wrong.
Look at yourself, my friend! You arent still getting the blunder you have made.
I respect you a lot and atleast now, try to understand the thing where you are going wrong! Take your own time to read this and I positively hope you understand what I mean to say.
You are right! If you increase the length by two times on every sides, you increase the surface area by 4 times and volume by 8 times. But, this isnt even the case here to the least. Let me say you how.
Now, you could notice that the cuboid 2, after increasing the sides of the cuboid 1 two times on each side. As you said, when we increase the sides by two times everywhere, the surface area increases 4 times and the volume increases by 8 times.
Since you said about the square-cube law here, you are saying about increasing all the sides by the equal multiple, which is basically multiplying all the sides by, for example, two and getting 4 times the surface area and 8 times the volume.
Its all right that you get 4 times the surface area, 8 times the volume if you increase the side by 2 times. Now, how do you find a cuboid with twice the volume of cuboid A which is isometric to cuboid A? Thats what we have to find. Lets look at it.
What was your task? You had to find the bite force of a jaguar after isometrically scaling it to twice its volume, right? So, you are scaling the jaguar to twice the volume isometrically, and then finding the surface area which you got after doing it.
Likewise, we are gonna scale the cuboid A to twice its volume isometrically (without changing the ratios between the vertics) and then find the surface area which we get form it. Isnt this the exact same way?
If you look at cuboid A and Cuboid B, cuboid B has twice the volume of cuboid A. But the proportions arent the same. So, how do you find the measures of a cuboid with twice the volume of A and which maintains the same proportions of A? Lets do it!
Now, lets scale the cuboid A which has volume 1, to volume 2 without changing its proportions.
Ratio of vertices of cuboid A : 1:8:8
Now, the equation of this cube be: x*8x*8x = 1 (where 'x' = measure of the smallest vertice)
Now, we are gonna keep the ratio the same and equate the volume (product of the three vertices) to twice the initial volume. By this, we could find the vertices of that cuboid which has twice the volume of cuboid A and stays with the same ratio. By doing this: we get:
x*8x*8x = 2,
64x^3 = 2,
x = (2/64)^ 1/3
= (1/32)^3
Therefore, x = 0.314980262473361
As we said, x is the value of the smallest vertice.
Now, lets find out the other two vertices. For that, we must multiply x by 8 (since ratio is 1:8:8).
Lets call this cuboid the 'cuboid awesome'. By doing so, the vertices of cuboid awesome are:
Now, check for the proportion. 2.519842099786888/ 0.314980262473361 = 8.
Thus, the ratios are kept the same, and the volume of "cuboid awesome" is twice the volume of cuboid A, as we were intending to get.
This is the exact same thing I did in the comparison. If you go back and see, the measure of the cuboid I took too was the same. I just elaborated my steps with reasoning. Thats all I did.
Now, can you say this isnt isometric variation? If you say so, I am happy to hear from you about why this isnt isometric.
YOU ARE LYING. First you said it's all wrong and now you are backtracking and saying it's right. Why are you being a deceitful liar? You changed your own statement and you are backtracking now.
Once again. YOU ARE COMPLETELY WRONG. I'm getting the idea that you have absolutely NO KNOWLEDGE of maths and physics AT ALL. You have probably never studies high school level maths and physics in your entire life.
DID YOU REALLY SAY that x is the value of vertex? Do you know what a vertex is?? A vertex has no value. It's a point in the the plane/space, a point only has coordinates not values. HAVE YOU EVER STUDIED GEOMETRY IN YOUR ENTIRE LIFE? Vertexes can ONLY have coordinates. Vertexes DO NOT have a value. How can you make a ratio of 3 different vertices?
All this 'calculation' you have done here is absolute garbage nonsense.
If the volume of a cube, cuboid, sphere or any other 3d shape is proportionally increased by 2 times then it's surface area will ALWAYS increase by 2^2/3 times. THAT IS A FACT BECAUSE THIS IS THE WHOLE PRINCIPLE OF THE SQUARE CUBE LAW. If you are challenging this principle then please stop studying maths for god's sake. You have never studied maths in your whole life.
THE SQUARE CUBE LAW SAYS THAT IF VOLUME PROPORTIONALLY INCREASES BY 2 THEN SURFACE AREA INCREASES BY 2^2/3 TIMES AND YOU ARE TRYING TO SAY THIS IS WRONG??? ARE YOU SAYING THIS STATEMENT FROM GALILEO IS WRONG? YES OR NO???? ANSWER THIS SIMPLE QUESTION.
05-16-2022, 11:25 PM( This post was last modified: 05-17-2022, 05:38 AM by LonePredator )
(05-16-2022, 10:48 PM)Wrapp Wrote: OMG!
(05-16-2022, 09:11 PM)LonePredator Wrote: I'll tell you in one sentence what I did in my calculation and then you answer me with yes or no.
If the volume of a cube (or any particular shape) increases by 2 times then it's surface area will increase by 2^2/3 times. YES OR NO? YOU SAID NO which means you never studied maths in your entire life.
Absolutely a big 'yes'. Yes! Are you satisfied now? That isnt even my argument in the first place, and you dont understand even what is the issue, and for the same reason you perceive everything wrong.
Look at yourself, my friend! You arent still getting the blunder you have made.
I respect you a lot and atleast now, try to understand the thing where you are going wrong! Take your own time to read this and I positively hope you understand what I mean to say.
You are right! If you increase the length by two times on every sides, you increase the surface area by 4 times and volume by 8 times. But, this isnt even the case here to the least. Let me say you how.
Now, you could notice that the cuboid 2, after increasing the sides of the cuboid 1 two times on each side. As you said, when we increase the sides by two times everywhere, the surface area increases 4 times and the volume increases by 8 times.
Since you said about the square-cube law here, you are saying about increasing all the sides by the equal multiple, which is basically multiplying all the sides by, for example, two and getting 4 times the surface area and 8 times the volume.
Its all right that you get 4 times the surface area, 8 times the volume if you increase the side by 2 times. Now, how do you find a cuboid with twice the volume of cuboid A which is isometric to cuboid A? Thats what we have to find. Lets look at it.
What was your task? You had to find the bite force of a jaguar after isometrically scaling it to twice its volume, right? So, you are scaling the jaguar to twice the volume isometrically, and then finding the surface area which you got after doing it.
Likewise, we are gonna scale the cuboid A to twice its volume isometrically (without changing the ratios between the vertics) and then find the surface area which we get form it. Isnt this the exact same way?
If you look at cuboid A and Cuboid B, cuboid B has twice the volume of cuboid A. But the proportions arent the same. So, how do you find the measures of a cuboid with twice the volume of A and which maintains the same proportions of A? Lets do it!
Now, lets scale the cuboid A which has volume 1, to volume 2 without changing its proportions.
Ratio of vertices of cuboid A : 1:8:8
Now, the equation of this cube be: x*8x*8x = 1 (where 'x' = measure of the smallest vertice)
Now, we are gonna keep the ratio the same and equate the volume (product of the three vertices) to twice the initial volume. By this, we could find the vertices of that cuboid which has twice the volume of cuboid A and stays with the same ratio. By doing this: we get:
x*8x*8x = 2,
64x^3 = 2,
x = (2/64)^ 1/3
= (1/32)^3
Therefore, x = 0.314980262473361
As we said, x is the value of the smallest vertice.
Now, lets find out the other two vertices. For that, we must multiply x by 8 (since ratio is 1:8:8).
Lets call this cuboid the 'cuboid awesome'. By doing so, the vertices of cuboid awesome are:
Now, check for the proportion. 2.519842099786888/ 0.314980262473361 = 8.
Thus, the ratios are kept the same, and the volume of "cuboid awesome" is twice the volume of cuboid A, as we were intending to get.
This is the exact same thing I did in the comparison. If you go back and see, the measure of the cuboid I took too was the same. I just elaborated my steps with reasoning. Thats all I did.
Now, can you say this isnt isometric variation? If you say so, I am happy to hear from you about why this isnt isometric.
First of all throw that rubbish garbage calculation of yours into the dustbin. Now I'll show you how basic geometry works.
Let's assume a cuboid with l=2, b=4, h=4 then the volume will be 2x4x4=32cm^3. Now the surface area is 2lb+2bh+2lh i.e. 8+16+8=24cm^2
Now if you increase the volume by 8 times while keeping the same shape and proportions, then all the lengths will increase by 2 times and the new volume will be 8x32=256cm^3
and the new surface area will be 4 times, 4x24=96cm^2
I DID NOT increase the lengths, the lengths AUTOMATICALLY increased by x^1/3 times WHEN I INCREASED THE VOLUME by x times and the surface area AUTOMATICALLY increased by x^2/3 times WHEN I INCREASED THE VOLUME by x times. If you still don't understand, then maths and physics are not for you.
At one point you say that you agree with square cube law and on the other hand you have been trying to prove how the square cube law is wrong since the last 2 days
THE SQUARE CUBE LAW BY DEFINITION STATED THE EXACT OPPOSITE OF WHAT YOU CLAIMED. YOU STILL REFUSED TO ACCEPT THAT YOU WERE TOTALLY WRONG ABOUT IT. THIS IS THE 5th TIME I AM ASKING YOU BUT YOU ARE CAREFULLY IGNORING AND AVOIDING MY QUESTION IN FEAR OF GETTING EMBARASSED
NOW GIVE ME THE ANSWER TO THIS, YOU STATED THE EXACT OPPOSITE OF WHAT SQUARE CUBE LAW SAYS AND YOU STILL REFUSE TO ACCEPT THAT YOU WERE WRONG SO HOW WILL YOU JUSTIFY THAT?? AND IF YOU IGNORE MY QUESTION ONCE AGAIN, THAT WILL PROVE WHAT A DECEITFUL LIAR YOU ARE
You don't even understand the simple difference between proportionality and equality yet you keep arguing. The level of your knowledge of physics and maths is worse than a high school student yet you keep arguing. It’s like an illiterate person trying to teach physics to other people
05-17-2022, 06:27 AM( This post was last modified: 05-17-2022, 06:48 AM by LonePredator )
(05-16-2022, 10:48 PM)Wrapp Wrote: OMG!
(05-16-2022, 09:11 PM)LonePredator Wrote: I'll tell you in one sentence what I did in my calculation and then you answer me with yes or no.
If the volume of a cube (or any particular shape) increases by 2 times then it's surface area will increase by 2^2/3 times. YES OR NO? YOU SAID NO which means you never studied maths in your entire life.
Absolutely a big 'yes'. Yes! Are you satisfied now? That isnt even my argument in the first place, and you dont understand even what is the issue, and for the same reason you perceive everything wrong.
Look at yourself, my friend! You arent still getting the blunder you have made.
I respect you a lot and atleast now, try to understand the thing where you are going wrong! Take your own time to read this and I positively hope you understand what I mean to say.
You are right! If you increase the length by two times on every sides, you increase the surface area by 4 times and volume by 8 times. But, this isnt even the case here to the least. Let me say you how.
Now, you could notice that the cuboid 2, after increasing the sides of the cuboid 1 two times on each side. As you said, when we increase the sides by two times everywhere, the surface area increases 4 times and the volume increases by 8 times.
Since you said about the square-cube law here, you are saying about increasing all the sides by the equal multiple, which is basically multiplying all the sides by, for example, two and getting 4 times the surface area and 8 times the volume.
Its all right that you get 4 times the surface area, 8 times the volume if you increase the side by 2 times. Now, how do you find a cuboid with twice the volume of cuboid A which is isometric to cuboid A? Thats what we have to find. Lets look at it.
What was your task? You had to find the bite force of a jaguar after isometrically scaling it to twice its volume, right? So, you are scaling the jaguar to twice the volume isometrically, and then finding the surface area which you got after doing it.
Likewise, we are gonna scale the cuboid A to twice its volume isometrically (without changing the ratios between the vertics) and then find the surface area which we get form it. Isnt this the exact same way?
If you look at cuboid A and Cuboid B, cuboid B has twice the volume of cuboid A. But the proportions arent the same. So, how do you find the measures of a cuboid with twice the volume of A and which maintains the same proportions of A? Lets do it!
Now, lets scale the cuboid A which has volume 1, to volume 2 without changing its proportions.
Ratio of vertices of cuboid A : 1:8:8
Now, the equation of this cube be: x*8x*8x = 1 (where 'x' = measure of the smallest vertice)
Now, we are gonna keep the ratio the same and equate the volume (product of the three vertices) to twice the initial volume. By this, we could find the vertices of that cuboid which has twice the volume of cuboid A and stays with the same ratio. By doing this: we get:
x*8x*8x = 2,
64x^3 = 2,
x = (2/64)^ 1/3
= (1/32)^3
Therefore, x = 0.314980262473361
As we said, x is the value of the smallest vertice.
Now, lets find out the other two vertices. For that, we must multiply x by 8 (since ratio is 1:8:8).
Lets call this cuboid the 'cuboid awesome'. By doing so, the vertices of cuboid awesome are:
Now, check for the proportion. 2.519842099786888/ 0.314980262473361 = 8.
Thus, the ratios are kept the same, and the volume of "cuboid awesome" is twice the volume of cuboid A, as we were intending to get.
This is the exact same thing I did in the comparison. If you go back and see, the measure of the cuboid I took too was the same. I just elaborated my steps with reasoning. Thats all I did.
Now, can you say this isnt isometric variation? If you say so, I am happy to hear from you about why this isnt isometric.
- From cube 1 to cube 2, volume increased by 8 times so the surface area increased by 8^2/3 tunes.
- From cube 2 to cube 3, volume increased by 3.375 times so the surface area increased by 3.375^2/3 times
- From cube 3 to cube 4, volume increased by 2.73037 times so the surface area increased by 2.73037^2/3 times.
As long as ANY shape increases in volume proportionally, the same pattern as above will ALWAYS apply. THIS IS WHAT THE SQUARE CUBE LAW IS.
You can verify this BUT PLEASE USE A CALCULATOR. We have already seen that are incapable of making calculations because you took 3 hours and still made such silly calculation mistakes (despite taking the whole 3 hours to do your ‘calculation’)
The same way, when a Jaguar’s volume/mass increases by 2 times, then the surface area/bite force will increase by 2^2/3 times.
Now you need to explain yourself. You said the exact opposite of what the square cube law says and you refuse to accept that you were wrong. SO NOW PROVE YOURSELF. PROVE HOW YOU WERE CORRECT WHEN YOU SAID THE EXACT OPPOSITE OF WHAT SQUARE CUBE LAW SAYS
If you did understand this, then it’s possible that you might not even reply anymore because you’re too afraid of admitting that you made such embarrassing mistake which even a high school student would never do. Please go and study high school maths and physics.
- From cube 1 to cube 2, volume increased by 8 times so the surface area increased by 8^2/3 tunes.
- From cube 2 to cube 3, volume increased by 3.375 times so the surface area increased by 3.375^2/3 times
- From cube 3 to cube 4, volume increased by 2.73037 times so the surface area increased by 2.73037^2/3 times.
As long as ANY shape increases in volume proportionally, the same pattern as above will ALWAYS apply. THIS IS WHAT THE SQUARE CUBE LAW IS.
You can verify this BUT PLEASE USE A CALCULATOR. We have already seen that are incapable of making calculations because you took 3 hours and still made such silly calculation mistakes (despite taking the whole 3 hours to do your ‘calculation’)
The same way, when a Jaguar’s volume/mass increases by 2 times, then the surface area/bite force will increase by 2^2/3 times.
Now you need to explain yourself. You said the exact opposite of what the square cube law says and you refuse to accept that you were wrong. SO NOW PROVE YOURSELF. PROVE HOW YOU WERE CORRECT WHEN YOU SAID THE EXACT OPPOSITE OF WHAT SQUARE CUBE LAW SAYS
If you did understand this, then it’s possible that you might not even reply anymore because you’re too afraid of admitting that you made such embarrassing mistake which even a high school student would never do. Please go and study high school maths and physics.
Yeah. You are right. I should have done surface area*2^2/3. I am getting the answer for that cuboid by doing it with surface area*2^2/3. Sorry for taking this long.
- From cube 1 to cube 2, volume increased by 8 times so the surface area increased by 8^2/3 tunes.
- From cube 2 to cube 3, volume increased by 3.375 times so the surface area increased by 3.375^2/3 times
- From cube 3 to cube 4, volume increased by 2.73037 times so the surface area increased by 2.73037^2/3 times.
As long as ANY shape increases in volume proportionally, the same pattern as above will ALWAYS apply. THIS IS WHAT THE SQUARE CUBE LAW IS.
You can verify this BUT PLEASE USE A CALCULATOR. We have already seen that are incapable of making calculations because you took 3 hours and still made such silly calculation mistakes (despite taking the whole 3 hours to do your ‘calculation’)
The same way, when a Jaguar’s volume/mass increases by 2 times, then the surface area/bite force will increase by 2^2/3 times.
Now you need to explain yourself. You said the exact opposite of what the square cube law says and you refuse to accept that you were wrong. SO NOW PROVE YOURSELF. PROVE HOW YOU WERE CORRECT WHEN YOU SAID THE EXACT OPPOSITE OF WHAT SQUARE CUBE LAW SAYS
If you did understand this, then it’s possible that you might not even reply anymore because you’re too afraid of admitting that you made such embarrassing mistake which even a high school student would never do. Please go and study high school maths and physics.
Yeah. You are right. I should have done surface area*2^2/3. I am getting the answer for that cuboid by doing it with surface area*2^2/3. Sorry for taking this long.
Thanks,
Wrapp.
Okay but I asked you another question which you avoided yet again for the 6th time in a row. You said the exact opposite of what the definition of square cube law says and you were still claiming that you were right so tell me so tell me how and on what basis you made that claim?
Okay at what basis Orca speculated having a 19k psi based on certain incident? I'm sure they have high bite force but probably not that high?
Animals always use their strength points to their advantages. But Orca rarely use their mouth like Sharks when hunting, even with their big numbers taking down a juvenile whales took them an hour or two by slamming their big bodies
The strongest bites in the animal kingdom
