Explosion Formula Formatting

[SeijiSetto]

pretty minor thing
ground based explosion formula is written as such:
W = R^3*((27136*P+8649)^(1/2)/13568-93/13568)^2

am i interpreting it right as this

 

[Stryker861]

Setto couldent you have just asked this in the calc discussion thread?

 

[KLOL506]

Setto couldent you have just asked this in the calc discussion thread?

This is the Calc Group Discussion Thread LOL Or one of them, really.

 

[Stryker861]

This is the Calc Group Discussion Thread LOL Or one of them, really.

Moreso meant the general one but either works

 

[Shmeatywerbenmanjenson]

pretty minor thing
ground based explosion formula is written as such:
W = R^3*((27136*P+8649)^(1/2)/13568-93/13568)^2

am i interpreting it right as this

We really should provide a pic on the page for all complex calculations like this as I feel it would make it easier for one to copy it down correctly as just using it as is results in some errors when copying it to other calculators and the like

 

[ImmortalDread]

I could help with this task, but use math formula feature from fandom instead of using a "picture".

 

[GarrixianXD]

We really should provide a pic on the page for all complex calculations like this as I feel it would make it easier for one to copy it down correctly as just using it as is results in some errors when copying it to other calculators and the like

I prefer not. It's honestly all common sense. However, seems like not everyone here took math and physics.

Nvm, I take this back.

 

[Antoniofer]

Hell no.

This is the correct formatting:

Negative, the formula in OP is the correct one. Notice that when p = 0, then W = 0, something that does not happens in your equation.

 

[GarrixianXD]

Notice that when p = 0, then W = 0, something that does not happens in your equation.

I find your explanation quite confusing.

 

[Antoniofer]

Just replace the p from the equation with the number 0, then proceed to resolve as usual: because r^3*((27136*0+8649)^(1/2)/13568-93/13568)^2 ==> r^3*(8649^(1/2)/13568-93/13568)^2, and 8649^(1/2) = 93, then r^3*((93 - 93)/13568)^2 ==> r^3*0, that naturally end ups being W = 0. Simple logic: if no overpressure, then no explosive; and if no explosive, then no overpressure.

 

[GarrixianXD]

Just replace the p from the equation with the number 0, then proceed to resolve as usual: because r^3*((27136*0+8649)^(1/2)/13568-93/13568)^2 ==> r^3*(8649^(1/2)/13568-93/13568)^2, and 8649^(1/2) = 93, then r^3*((93 - 93)/13568)^2 ==> r^3*0, that naturally end ups being W = 0. Simple logic: if no overpressure, then no explosive; and if no explosive, then no overpressure.

Uhmmm… How did you go from r^3*(8649^(1/2)/13568-93/13568)^2 to r^3*((93 - 93)/13568)^2?

Wouldn’t it be r^3*(93/13568-93/13568)^2 instead?

 

[SeijiSetto]

Uhmmm… How did you go from r^3*(8649^(1/2)/13568-93/13568)^2 to r^3*((93 - 93)/13568)^2?

Wouldn’t it be r^3*(93/13568-93/13568)^2 instead?

8649^(1/2) is 93
(A/B) - (C/B) = (A - C)/B
therefore (93/13568) - (93/13568) = ((93 - 93)/13568) = 0

 

[GarrixianXD]

8649^(1/2) is 93
(A/B) - (C/B) = (A - C)/B
therefore (93/13568) - (93/13568) = ((93 - 93)/13568) = 0

Hold on a minute…

So r^3*(8649^(1/2)/13568-93/13568)^2 is equivalent to r^3*(93/13568-93/13568)^2. Simply it you’ll get r^3*(13568*93/13568-93)^2. What am I missing here?

 

[SeijiSetto]

Hold on a minute…

So r^3*(8649^(1/2)/13568-93/13568)^2 is equivalent to r^3*(93/13568-93/13568)^2. Simply it you’ll get r^3*(13568*93/13568-93)^2. What am I missing here?

the way i'm reading it, (8649^(1/2) / 13568) and (93/13568) are separate fractions. i am not entirely sure though, which is what this thread is meant to clarify

 

[GarrixianXD]

the way i'm reading it, (8649^(1/2) / 13568) and (93/13568) are separate fractions. i am not entirely sure though, which is what this thread is meant to clarify

Yeah, that not only doesn’t follow the order of operations but you pretty much subtracted a denominator by a numerator which wouldn’t work.

 

[SeijiSetto]

Yeah, that not only doesn’t follow the order of operations but you pretty much subtracted a denominator by a numerator which wouldn’t work.

no operations have actually been done. i rearranged the formula

 

[GarrixianXD]

My CPU burned out.

Anyways, back to @Antoniofer's point. Since there needs a medium for an explosion to take place, wouldn't it just place a restriction on variable p to p > 0? By mathematical derivations, the equation posted by Setto is definitely incorrect, as the end being timed by 2 rather than being squared itself is enough to prove my point.

 

[Shmeatywerbenmanjenson]

I prefer not. It's honestly all common sense. However, seems like not everyone here took math and physics.

Its for ease of use when copying to other calculators to help calc members and normal members alike speed up the process of calcing

Absolutely worth it in the long run

Also funny you say that when posting an incorrect expression of the formula

I dont appreciate your not so subtle jabs

Here is the correct formula

100 is equal to R or the radius

1.37895 is equal to P or the pressure in Bars

This is an example of an explosion with a radius of 100 meters

 

[GarrixianXD]

Here is the correct formula

I'm pretty sure the 93/13568 fraction has to go on the bottom. Otherwise, it would rather be written as R^3*((27136*P+8649)^(1/2)-93/13568) rather than R^3*((27136*P+8649)^(1/2)/13568-93/13568)^2

Its for ease of use when copying to other calculators to help calc members and normal members alike speed up the process of calcing

Absolutely worth it in the long run

I dont appreciate your not so subtle jabs

Yeah, true, it is easier to read. My apologies if I'm jabbing over here just thought that bracket was self-explanatory.

Also funny you say that when posting an incorrect expression of the formula

I'd like you to point out where I'm mathematically wrong rather than stating restrictions with the variables.

 

[KLOL506]

@DontTalkDT @Executor_N0

 

[Shmeatywerbenmanjenson]

I'm pretty sure the 93/13568 fraction has to go on the bottom. Otherwise, it would rather be written as R^3*((27136*P+8649)^(1/2)-93/13568) rather than R^3*((27136*P+8649)^(1/2)/13568-93/13568)^2

It doesn't

It needs to be separate so it can be subtracted from the first set of variables (27136*P+8649)^(1/2)

R^3*((27136*P+8649)^(1/2)-93/13568)^2
Using this formula removes one of the division signs essential to make the calculation work

I'd like you to point out where I'm mathematically wrong rather than stating restrictions with the variables.

Your placing the rest of the calculation under another set of divisions instead of letting them remain separate from each other in order to subtract the first half properly

 

[ImmortalDread]

I prefer not. It's honestly all common sense. However, seems like not everyone here took math and physics.

Why would anyone use common sense to write like this in real life? We resort to this style due to the lack of access to formatting keys on our keyboards. Otherwise, it's neither common sense nor a traditional way of writing anywhere in the world.

 

[GarrixianXD]

Why would anyone use common sense to write like this in real life? We resort to this style due to the lack of access to formatting keys on our keyboards. Otherwise, it's neither common sense nor a traditional way of writing anywhere in the world.

The parentheses around the formulas.

 

[SeijiSetto]

My CPU burned out.

Anyways, back to @Antoniofer's point. Since there needs a medium for an explosion to take place, wouldn't it just place a restriction on variable p to p > 0? By mathematical derivations, the equation posted by Setto is definitely incorrect, as the end being timed by 2 rather than being squared itself is enough to prove my point.

no it's squared
the bracket is just so tall it's hard to tell the difference
also if you thought that was an issue why not reply to ME and address me instead of saying it in an aside to someone else

 

[GarrixianXD]

no it's squared
the bracket is just so tall it's hard to tell the difference
also if you thought that was an issue why not reply to ME and address me instead of saying it in an aside to someone else

You can’t separate the complex fraction and subtract a denominator by a numerator. I’m sure I stated that before. Gotta bring the 13568 on the very bottom up top and you can only subtract the denominator section from there on.

 

[SeijiSetto]

ok so i did a little more looking.
this page lists the radius to plug into the equation to get the baseline of every tier.
so to check if my equation is right, i'll plug values in and see if i get a matching output.

Spoiler: 9-A

5kg of TNT, correct within a margin of error.

Spoiler: 8-B

basically 11 tons, correct within a margin of error.

Spoiler: 7-C

5.8 kilotons. again, correct.

Spoiler: 6-C

4.3 gigatons.

y'get the point, i don't need to do anymore. the way i've written it in the OP is, apparently, the correct way.

 

[GarrixianXD]

I haven’t actually used that formula before. If that was the case it should be written as R^3*(((27136*P+8649)^(1/2)/13568)-93/13568)^2 instead.

The original right there looks like the second level numerator is subtracted by 93 and a three level division.

 

[GarrixianXD]

Yeah, your expression actually is right. Though, anyone could’ve mistaken the formula as a complex fraction by the way it was written.

 

[SeijiSetto]

I haven’t actually used that formula before. If that was the case it should be written as R^3*(((27136*P+8649)^(1/2)/13568)-93/13568)^2 instead.

The original right there looks like the second level numerator is subtracted by 93 and a three level division.

yes, i agree it's ambiguous. that's why i made this thread lol, to clarify which but i've managed to do so myself
this formula should be put on the page in proper notation using fandom's <math> tag, this works.

Spoiler: <math> latex code

<math>W = R^{3} \times \left ( \frac{\left ( 27136P+8649 \right )^{\frac{1}{2}} - 93}{13568} \right )^{2}</math>

edit: forgot the squared at the end

 

[Antoniofer]

I don't think it was that confusing, most people know that multiplications and divisions are solved in first order (not counting what is within parentheses), so if something is written like A - B/C, people will solve B/C first; plus, have heard someone complaining or commiting that mistake.

Despite everything, fine if you want to use the picture, is aesthetically more pleasant, and simplify the (27136*P+8649)^(1/2)/13568-93/13568 into ((27136*P+8649)^(1/2)-93)/13568.

 

[DontTalkDT]

My opinion is that we do humanity a favour if we force people to learn the order of operations.
Anyway, kinda don't like pictures since you can't copy + paste them. Having a picture and text is also pretty redundant. I don't think a relevant amount of people have a problem with this formulation and those that do should really learn. But if I'm outvoted on that, just add an extra bracket to the term.

 

[GarrixianXD]

My opinion is that we do humanity a favour if we force people to learn the order of operations.

My bad. Srsly, does that not look like a triple fraction to you? Inb4 I asked Amelia if my formula was the correct expression and she said it was. So I’m not the only one who tought it was a triple fraction.

 

[DontTalkDT]

Srsly, does that not look like a triple fraction to you?

No. No it doesn't. But I spend more time looking at formulas than most people.

 

[GarrixianXD]

No. No it doesn't. But I spend more time looking at formulas than most people.

Since you’re so good at math please solve the limit on my signature

 

[DontTalkDT]

Since you’re so good at math please solve the limit on my signature

The answer is 'false'. x doesn't imply 2.

 

[GarrixianXD]

The answer is 'false'. x doesn't imply 2.

Dang, you’re good.