- 4,927
- 2,850
Pfft! I just did what everyone sensible did and did a little...
stay losing
Follow along with the video below to see how to install our site as a web app on your home screen.
Note: This feature may not be available in some browsers.
Pfft! I just did what everyone sensible did and did a little...
stay losing
I mean, true, but for some reason, people aren't able to grasp the Pen Tool even though all you do is just click-and-drag. I'm saying that because the Pen Tool is the closest thing to a "magnetic lasso tool" on GIMP.So true!
but with GIMP I do not have to push people to black-flag software in order to do one (1) calculation. It's not like it's hard to do this on GIMP, if you know how to do it. In this instance there's literally no reason to recommend Adobe.
How the HELL can I calculate this?
People do that? I just drew over pictures with Adobe Illustrator and used a script to determine an object's area.
GIMP's available for free. Just download it.I only have paint to use sadly
Okay, so the Bernard the Bear glacier feat... Here's how you'd wanna do it, as Bernard pulled with enough force to break the glacier from its foundation.bump
Alright thanks I will try itOkay, so the Bernard the Bear glacier feat... Here's how you'd wanna do it, as Bernard pulled with enough force to break the glacier from its foundation.
1. Determine the radius of the glacier from Bernard's size. Not the summit; just whatever the best measurable point of the midsection would be. The closer to the foundation, the better.
2. Calculate the cross-sectional area of the glacier using the measurement you got. The glacier is mostly cylindrical, so calculate the area of a circle.
3. Multiply the area by the shear strength of ice. I prefer you use an academic source because the Wiki may not have the most reliable references. 1 pascal is 1 newton per square meter, by the way.
4. Divide the result you got by 9.80665 and you'll get the force exerted in kilograms of force.
So I tried it but I cannot find the result and I think I made a mistake.Okay, so the Bernard the Bear glacier feat... Here's how you'd wanna do it, as Bernard pulled with enough force to break the glacier from its foundation.
1. Determine the radius of the glacier from Bernard's size. Not the summit; just whatever the best measurable point of the midsection would be. The closer to the foundation, the better.
2. Calculate the cross-sectional area of the glacier using the measurement you got. The glacier is mostly cylindrical, so calculate the area of a circle.
3. Multiply the area by the shear strength of ice. I prefer you use an academic source because the Wiki may not have the most reliable references. 1 pascal is 1 newton per square meter, by the way.
4. Divide the result you got by 9.80665 and you'll get the force exerted in kilograms of force.
Flashlight getting Class M and 9-A:Now THIS is how it's done.: https://vsbattles.fandom.com/wiki/U...t:_Bernard_Pulls_a_Glacier_Off_Its_Foundation
bumpI posted a thread, but I thought to bring my query here too. Kagura's feat considering the original quote only says "an astounding amount of heat is released, causing the elimination of the the entire area", gives varying results depending upon the method of calculation:
Using specific heat equations gets 6-C to High 6-B results (5.36 Gigaton of TNT to 171.53 Teratons of TNT )
Using nuclear fireball calculator gets 6-C results (8.54 Gigatons of TNT to 8.6 Gigatons of TNT)
Using radiation equations gets High 7-C to 5-A results (637.88 Kilotons of TNT to 4.208 Yottatons of TNT)
I just wish to know, which is viable?
Hey, I only did the AP part to shut you up; the calc applies to Lifting Strength and Lifting Strength only.Flashlight getting Class M and 9-A:
Me somehow getting 10-C:
Thank you
Theres no way I can use it as AP? Hmm…Hey, I only did the AP part to shut you up; the calc applies to Lifting Strength and Lifting Strength only.
It was an extended application of force (i.e push/pull over a good amount of time vs a quick strike), so no.Theres no way I can use it as AP? Hmm…
He couldnt move it in the first half of the feat and only did the actual feat when he pushed himseif and accidently fartedIt was an extended application of force (i.e push/pull over a good amount of time vs a quick strike), so no.
...which only serves to further reinforce the fact that it was an extended push rather than a single strike viable for AP calculations. If you saw someone in the gym really struggle to push a weight up but they barely did it, for example, you wouldn't be able to use that for AP. If they launched it into the air, then sure.He couldnt move it in the first half of the feat and only did the actual feat when he pushed himseif and accidently farted
No what I mean is imagine someone pushing a weight and not moving it at all then they suddenly get more serious and lift it. Thats how Bernards case worked...which only serves to further reinforce the fact that it was an extended push rather than a single strike viable for AP calculations. If you saw someone in the gym really struggle to push a weight up but they barely did it, for example, you wouldn't be able to use that for AP. If they launched it into the air, then sure.
Had to update the calc due to technical errors such as the buildings not really covering the entire area of the Kyoto prefectureI posted a thread, but I thought to bring my query here too. Kagura's feat considering the original quote only says "an astounding amount of heat is released, causing the elimination of the the entire area", gives varying results depending upon the method of calculation:
Using specific heat equations gets 6-C+ to 6-A results (3.87 Gigaton of TNT to 1.513 Petatons of TNT )
Using nuclear fireball calculator gets 6-C results (8.54 Gigatons of TNT to 8.6 Gigatons of TNT)
Using radiation equations gets 9-C+ to High 6-A results (14277.5075 Joules of TNT to 11.97828 Petatons of TNT)
I just wish to know, which is viable?
Is it possible to calculate this?
How much bigger is body compared to the foot? By foot you mean only the feet part? no legs?Probably. You have people you can scale to the size of the foot, scale the foot to the size of the robot, then do standard explosion calc.
You can measure that in the first shot. Hell, you wouldn't even need to measure the foot, just use the people there, assume an average height and pixel scale to the size of the explosions.How much bigger is body compared to the foot? By foot you mean only the feet part? no legs?
Since theres 2 explosions which one should I use? The second one is biggerYou can measure that in the first shot. Hell, you wouldn't even need to measure the foot, just use the people there, assume an average height and pixel scale to the size of the explosions.
I'd calc both but the first did the major damage so the second wouldn't really be able to be applied. It'd be like saying "well character A beat character B to near-death and C oneshot B after that, thereforce C > B". It doesn't work.Since theres 2 explosions which one should I use? The second one is bigger
Why did I get like 5 meters from pixel scaling for the second explosin tfI'd calc both but the first did the major damage so the second wouldn't really be able to be applied. It'd be like saying "well character A beat character B to near-death and C oneshot B after that, thereforce C > B". It doesn't work.
It should be a little over 9 metres, 9.3-ish. Using the ground based formula gives me 0.0647 tons, or 9-A.Why did I get like 5 meters from pixel scaling for the second explosin tf
I got High 8-CIt should be a little over 9 metres, 9.3-ish. Using the ground based formula gives me 0.0647 tons, or 9-A.
I just used the standard ground-based explosion formula, AND I took the frame right before the final flash that filled the screen. It seemed like mostly light to me so I chose to ignore it, but yeah. I'd recommend using air-based explosion [((x/0.28)^3)/1000 where x is radius in km and the result is in megatons], honestly. Gave me 1.13028067052 tons, so 8-C+.
I will just add yours since you seem to know alot moreI just used the standard ground-based explosion formula, AND I took the frame right before the final flash that filled the screen. It seemed like mostly light to me so I chose to ignore it, but yeah. I'd recommend using air-based explosion [((x/0.28)^3)/1000 where x is radius in km and the result is in megatons], honestly. Gave me 1.13028067052 tons, so 8-C+.
Can you upload your pixel scale to fandom? So I can use it on the blogI just used the standard ground-based explosion formula, AND I took the frame right before the final flash that filled the screen. It seemed like mostly light to me so I chose to ignore it, but yeah. I'd recommend using air-based explosion [((x/0.28)^3)/1000 where x is radius in km and the result is in megatons], honestly. Gave me 1.13028067052 tons, so 8-C+.
For the 1.13 ton result, I used your pixel scaling.Can you upload your pixel scale to fandom? So I can use it on the blog
As for your first question, no I don't think so. Swords and other sharp objects are just noted to be sharp, AP doesn't change. For example, two identical people with equal AP. One has a sword. Obviously the one wielding the sword would win due to the fact that it's sharp and thus it can kind of """bypass durability""", but that's it.Stabby stab:
Is there a way to calculate the 'increase in power' of an attack from its concentration over a surface?
To give an example, Krillin's Kienzan/Destructo disc. Or a more real one, if I used a pocket-knife to stab somebody, I wouldn't really be exerting more energy, just concentrating the same amount into a smaller area.
Is there a consistent formula for this?
Similarly, what about the increase in density? For example, if I use brass knuckles, the mass barely changes; it's the strength of the material, and the fact that the impacting object is more rigid than my fist, that really increases the damage (since my fist/arm would deform somewhat on impact, softening the blow). Is there a mathematical way to account for that?
Piercing damageAs for your first question, no I don't think so. Swords and other sharp objects are just noted to be sharp, AP doesn't change. For example, two identical people with equal AP. One has a sword. Obviously the one wielding the sword would win due to the fact that it's sharp and thus it can kind of """bypass durability""", but that's it.