Hypothetical Calculation Request

[FTW395]

In various shows that revolve around fightning we often see two combatants creating shockwaves when clashing with their fists. What I wonder is how you would calculate the force or the energy of a fist, that by clashing with another fist creates a shockwave that could shatter diamond at a range of 100 meters from the epicenter of that shockwave. If the size of the diamond is needed assume it's 1 cubic decimetre or 1 dm┬│ for short.

I would love if someone could calculate this and perhaps show me the calculation afterwards, thanks in advance.

 

[Antoniofer]

Mmm, I guess if the shockwave destroyed grass, concrete or even diamond, we should:

1) Find the distance/radious between the epicenter and the destroyed object.

2) Knowing the UTS of Sheer Strength of the object (in Pa), we can use the equation from my blog here, and then multiply the result per 0.4 due not having termal energy.

 

[Austrian-Man-Meat]

"In various shows that revolve around fighting, we often see two combatants creating shockwaves when clashing with their fists."

Okay, this I would be willing to have a look at, but the specific instance you bring in is something for another time. I'll highlight the thread; see how other members would rationalize that feat and how one would go around calculating it before jumping into it myself.

 

[FTW395]

@AMM In all honesty I only made a specific request, because I thought it'd be easier for people to calculate it if they had material to work with. You can change the metals or the distance if you like, the thing that matters most is the calculation behind it.

 

[Antoniofer]

Correction from above, instead of UTS of Sheer Strength, the values from here should be used: if the shockwave shatter glass at x distance, 1 - 2 psi is used; if is concrete, 3 - 5 psi, if was reinforced, then 10 psi.

 

[FTW395]

@Antoniofer so neither distance nor size of the matter are important?

 

[Antoniofer]

Not according to the document, at least with common glass and concrete, that is the amount of overpressure needed to shatter them.

 

[FTW395]

I see, I still would like for someone else to do the complete calculation, because I'm still pretty clueless on how to do it haha.

 

[LordXcano]

@AMM I was actually thinking about this while I was gone and the way I saw it was that you could assume all the air was dispersed in a sphere.

Like if the shockwave goes out in a 5 meter radius while they're flying you got 523.6 m^3 of air being dispersed, if it's on the ground then it'd obviously just be 261.8 m^3.

In the specific situation we have here though you'd just do the destruction of the diamond over the surface area of the shockwave.

Radius in this case is 100 meters, so that's an SA of 126,000 m^2, 63,000 m^2 since they're (presumably) on the ground. A cubic decimeter would be 0.01 m^2 on each side (I believe). Assuming it takes, eh, 1000 J to break the diamond (making this up for example) then you'd do:

63,000/0.01 = 6,300,000x

6,300,000*1000 = 6,300,000,000 J or 1.5 Tons

 

[FTW395]

@LordXcano is 1000 J a reasonable estimate to break diamond? Regardless thanks a bunch for this calc, really appreciate you took the time to make it!

 

[LordXcano]

No I just made it up. I have no idea how much it takes to break diamond.

Wikipedia lists its yield strength as 1600 MPa which would get 1,600,000 J for 1 cubic decimeter if I did that right.

That'd give 2409 Tons of TNT using the above method.

 

[FTW395]

Wow, That's a huge difference. By the way according to your calc the smaller the object is that gets cracked, (here diamond) the more energy is required to crack it. Is that correct?

By the way how did you convert MPa into J? I'm sure Pa and J aren't the same, so I wonder how you got that.

 

[Antoniofer]

1 MPa = 1 J/cc, so 1600 MPa = 1600 J/cc, however, I doubt that one overpressure able to destroy a diamond without pulverizing the rest of the ground/way.

 

[FTW395]

@Antoniofer 1 Pa = 1 J/m┬│, meaning 1600 MPa * m┬│ = XJ, this is the method I'd use for calculation the amount of energy. Thing is with my method I don't get his results for any of the given distances we've got, so I've no idea how he got to 1,600,000 J.

I have no idea what cc is, so yeah.

Edit: Nevermind made a stupid miscalculation, his results are correct.

 

[LordXcano]

cc = cubic centimeter

 

[Lina_Shields]

If the shockwave is spread out as a sphere, why not use the inverse square law method to calculate the actual energy generated by the clashing punches.

• Find the surface area of the material that is making contact with the shockwave.
• Find the radius of the shockwave from the epicentre
• Find the surface area of the shockwave (sphere) using your radius
• Find the energy required to destroy said material due to shockwave
• Find the ratio of surface area shockwave and the surface area that is making contact with the material
• Multiply ratio by the energy required to bust material that is shown.

We need specific measurements (length, width, height) of material however.

 

[RadicalMrR]

Im more interested to know the AP needed to create the shockwaves since I believe it should be way higher than the shockwaves itself.

 

[Antoniofer]

To find AP of the impact individually better to divided the result by 2; I think that using the reference from the document from above, if the shockwave destroyed the widows and doors from 20 m of distance, we can use 2 psi for reference (min) to find the yield, that is what Mythbuster used to do; not sure at the final we should multiply by 0.4 due 0 termal energy.

 

[RadicalMrR]

I wonder what the most (in)famous shockwave in fiction would yield for the AP of the actual fist clash

Range: Universal or 93 billion light years

AP: 1/3 Universal

 

[FTW395]

@LordXCano what I don't understanding is that according to your calc, you divide the surface area of the Shockwave by the surface area of the object that gets cracked. This means that if you have An object with a large lenght and width, but a smal height. You'd have very little energy.

For instance let's say we have an object of diamond with 20 x 30 x 1 (lenght x width x height). We assume the radius stays the same. Seeing how it's cubed according to you, we'd get:

63,000/(20*30) = 105

Then we calculate the energy required to destroy this specific diamond object which is:

1600*10^6*(20*30*1) = 9.6*10^11

We do this times 105 and get 1.008*10^14 J or 24,091 Tons of TNT.

Now let's do the same but for an object of diamond with 5 x 4 x 30, as in the previous example the radius stays the same. We'd have:

63,000/(5*4) = 3150

Calculate the energy needed to destroy this diamond object:

1600*10^6*(5*4*30) = 9.6*10^11

Do this times 3150 and we get 3.024*10^15 J or 77,275 Tons of TNT.

So even though the size of the object inherently stays the same, the energy has a huge difference just because of how the sizes of each side are distributed.

 

[LordXcano]

The logic behind it is basically

"If this portion of the wave carries enough energy to do this, then the rest of the wave should also have enough energy to do that. We can combine the two to get the total energy"

This is why Multi-Solar System is over a trillion times higher than regular solar system, the massive size difference in the "wave"

 

[FTW395]

@LordXcano could you also give some clarification on my previous post?

 

[LordXcano]

I'd have to think about it.