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Calculation Requests

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  1. I'll check the numbers and get back to you
  2. Rel usually, depending on how many circles around it they can run in a second.
  3. Island level to pulverize.
 
2) Relativistic+ at 80.21% of the Speed of Light
 
Blood is mostly water, and there's 37.2 Kilograms of the stuff in a human body.

Heat capacity for water is 4178, and boiling point is 100. 4278 minus natural body heat of 37 is 4241.

37.2 times 4251 is 157765.2 Joules, Wall level
 
DMUA said:
Blood is mostly water, and there's 37.2 Kilograms of the stuff in a human body.

Heat capacity for water is 4178, and boiling point is 100. 4278 minus natural body heat of 37 is 4241.

37.2 times 4251 is 157765.2 Joules, Wall level
Thanks
 
Hmmm could someone double-check this calc for me?

I'm calcing a feat for vaporizing a 3400 pound mech primarily made of metal with an average melting point roughly equivalent to titanium. Titanium's melting point is 1668 C= running that through a converter got me 3167699.69 joules, which I then multiplied by 3400 to get 1.07701789×1010, which I think should be High 8-C. I'm kinda worried about the fact that I just got a "melting temperature" and not a "melting temperature per pound". In addition, the calc does not factor in time at all. Does anyone see any problems with this?
 
Math is done in metric, so in our terms, that's 1542.214 Kilograms.

Vaporizing something is a matter of weight times Heat Capacity (470) times the boiling point (3287) times Latent Heat of Vaporization (8878768).

So, 1542.214×470×3287×8878768 would be...

21154117456949481.28 Joules, 5.05595541514088 Megatons, Small City level+

I would assume that trying to get temperature for melting would have made the results significantly lower
 
Okay so I did the waterjet feat and I'm hoping someone will double-check it

The character in question is capable of, essentially, mimicing the abilities of a waterjet. According to this site, the strongest waterjets move at mach three. I used google's converter to go from mach three to 1029 m/s.

I then assumed an mass of one gallon of water, and used converter to get 3.79 kilograms.

I then plugged these two values into this calculator and it returned 2.0065 × 10^6 joules, or Wall Level
 
Calculate size of the aura, use density of fire and make it a heat-change calc.
 
So you find out how large the aura is, how much mass should be in it, and how much energy would be required to heat that mass to fire temperature? Also, is there a time component?
 
No time component, aside from that, yeah. If you can tell me the radius from the character it works around I can calc it for you- don't expect 9-B or higher though unless it is absurdly high radius.
 
1012 J/(kg*K) is standard for room temperature
 
I got 9-B for a radius of 4 inches? Which I don't think is "absurdly high" but

here's my math for your convenience

Q = m (mass in gs) * c(heat capacity in joule per gram degrees celsius) * Δt(change in temperature celsius)

Flame amount is 20 cm^3 per cm of skin area this calculator the characters BSA is 1.7 meters/17000 cms, 17000*20 = 340000 c^3/0.34 m^3, according to this fire is 0.3 kgs per cubic meter, 0.34 m^3 x 0.3 kgs/m^3= 0.102 kgs or 102 grams.

Q = 102 * c * Δt

c = 1.012

Q = 102 * 1.012 * Δt

Average campfire temps(which he should be at least equivalent to) is 1,100 degrees celsius - room temperature, 23 celcius, is 1,077 celsius Q = 102 * 1.012 * 1,077 Q = 1.11172248 * 10^5 which is Wall Level
 
...

yeah except you're supposed to input mass in kg.

Let's assume a volume of an ellipsoid for your given "aura"- 4 inches off of their longest measurements. Average human is 177 cm tall so add 4 inches to half of that to get a radius of 98.66 cm.

Average man has a shoulder width of 18.25 inches, or 46.36 cm. Halve again and add the aforementioned 4 inches (10.16 cm) to get a radius of 33.34 cm.

4.12e5 cm^3 is our volume of air, but of course we have to subtract the volume of the human. That should be about 70 litres, or 70,000 cm^3, leaving us with 3.42e5 cm^3.

Density of fire is, as you said, 0.3 kg/m^3. So our mass would be about 0.1026 kg of actual fire. Also you read your sources wrong- they're 1100 F, meaning about 585 C. 585 - 23 = 562 C temp change.

Proof
You're looking at very nearly baseline 9-B with this feat.
 
Alright so question, this probably doesn't have to be a full calc, but what range of 3-B would creating Bootes Void as a side effect of a short fight be?
 
Well you really didn't give much information on it
 
1.01257959184 km is the distance of the Strom and he made it with his powers/magic and its a normal storm
 
The main problem is you give too many details

To say, too many things borrowed directly from ideas that aren't yours

If you want to make a pallet swap, pallet swap the numbers on Raven's existing calc
 
Problem is idk how the hell to calc storms, I looked at the story calc and don't know how to calc it still with a calculater on google
 
It's literally just basic math

I elaborate on every step in it, if you can't follow it, no one is really Keen on doing it for you.
 
DMUA said:
It's literally just basic math
To be fair, a lot of us are from completely different areas and lifestyles. So basic math to you might not be basic math to another person.

Anyway. I want to start working on storm calcs (And other calcs that aren't destruction ones) soon. I'm just not quite confident enough in my math skills yet to do something like that.

If someone could give me a link to an easy to understand storm calc...I'd appreciate it.
 
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