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Depth of Craters formed from Explosions

Arc7Kuroi

Arc7Kuroi

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According to this field test (page 49/56), the depth/diameter ratio is 0.25. I found this searching for new ways to find depths of craters after DT informed us that the depth = 0.18 * radius (or diameter I don't remember) was only for meteor/asteroid impacts. I'm wondering if anyone can confirm the validity of this pdf or make better sense of it, I only skimmed it for key words and images, and haven't had time to delve into it fully myself.
 
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It's 0.18 * diameter, not radius I believe.

As for the pdf, I'm not sure. Looks valid enough at a glance, but more input would be best.
 
Damn, not even a few weeks and we're looking at upgrades of crater feats.

Assuming it's legit, of course.
 
As my signature says I'm a bit tight on time right now.
From a glance, the source seems ok. However, the 0.25*diameter equation, in particular, seems to be given some restraints.
The large yield Pacific cratering data and meteor impact crater data for crater radii greater than 10,000-15,000 ft deviate from this approximately invariant P./D anc become increasingly shallower with increasing size.
Click to expand...
It later on advises
Thus, for nuclear sources of strategic interest it seems most likely that craters in dry soil and rock are deep bowl-shaped craters, and quantitative estimates of crater radii and depths can be obtained from Equation 14 once the crater volume is known. For sufficiently wet soils and crater depths greater than about 20 ft, shallower dished-shaped craters are expected and, based on Pacific data, radii and depths might be estimated from Equation 15.
Click to expand...
It also seems to me like the ratio in Figure 21 begins to flatten around the 10km diameter mark in Figure 21, but at that point we are, data-wise, mostly talking about meteor impacts again anyway.

Sooo maybe look into how to consider those restraints. I believe the paper provides what is needed for that.
 
If so, should all calculations using this method (the depth = 0.18 * radius (or diameter I don't remember) be revised?
 
Thank you for helping out DontTalk.
 
Also, something else I just noticed. The paper suggests to not model the craters for which the 0.25*diameter applies as half ellipsoids, as we often do.
They instead appears to suggest a paraboloid model, following V ≈ 1.5*R^2*D.
The crater radii and depths provided by these best estimates are consistent with parabolically shaped craters, i.e., V ≈ 1.5*R^2*D
Click to expand...
(page 37)
 
DontTalkDT said:
Also, something else I just noticed. The paper suggests to not model the craters for which the 0.25*diameter applies as half ellipsoids, as we often do.
They instead appears to suggest a paraboloid model, following V ≈ 1.5*R^2*D.

(page 37)
Click to expand...
Hmmmmmm, no mention of pi in the formula?

EDIT: NVM, volume of a paraboloid is as follows: (pi/2) * R^2 * Height (Or in this case, depth)

So it's probably actually 1.57079 * R^2 * D, since pi/2= 1.57079

But just go with 0.5 * pi * R^2 * D, much simpler.
 
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Kinda ironic that 'splosions give a deeper crater than impact situations.
 
So, I'm guessing the 0.18 times diameter for impact-based craters (For craters that are 150+ meters wide, for lower I think I saw 0.13 times diameter, will have to re-look for sources) and 0.25 times diameter+paraboloid volume for explosion-based craters is what we go with then?

Also I found this handy paraboloid volume calculator.
 
So what are the conclusions here?
 
For craters of diameters ~10 km or less you can find their depth as .25*diameter and the volume of the crater as pi/2*Radius^2*Depth.

I gotta figure out how to extend the model beyond 10km craters, which I will do when I get the chance.
 
Okay, and why is this important? Should we update some instruction page, or do you need to make a calculation?
 
Antvasima said:
Okay, and why is this important? Should we update some instruction page, or do you need to make a calculation?
Click to expand...
Yes, I think this could be very important. If my understanding is correct, this could affect all calculations involving explosions that leave craters behind.
 
UchihaSlayer96 said:
Yes, I think this could be very important. If my understanding is correct, this could affect all calculations involving explosions that leave craters behind.
Click to expand...
Honestly, only those where scaling the depth is otherwise completly impossible. Since actually scaling the depth quite obviously the precedence.

Anyways, once this is properly worked out I should probably put something about it on the calculations page.
 
DontTalkDT said:
Honestly, only those where scaling the depth is otherwise completly impossible. Since actually scaling the depth quite obviously the precedence.

Anyways, once this is properly worked out I should probably put something about it on the calculations page.
Click to expand...
Yeah, pretty much, use this only when you are unable to see the full depth of the crater or are unable to gauge it properly.
 
Thank you for helping out as usual DontTalk.
 
OK so this has been accepted for crater feats where the depth is not properly visible or most of it is hidden or the depth is not shown.

So basically we now wait for Arc to find the multiplier for craters wider than 10 km. Once he does that, we'll have to work on a justification in the Explosion Radius page, I think? Under a new Crater calculation section.
 
Well the paper suggests using equation 15, which states
1.2v^0.22 ≤ D ≤ 2.4v^0.22
Click to expand...
If we assume v to be paraboloid shaped, then v = (pi/2) * R^2 * D, and we get
1.2((pi/2) * R^2 * D)^0.22 ≤ D ≤ 2.4((pi/2) * R^2 * D)^0.22
Click to expand...
Since it's a factor, we can pull the D out of those exponents, which gets us
1.2((pi/2) * R^2)^0.22 * D^0.22 ≤ D ≤ 2.4((pi/2) * R^2)^0.22 * D^0.22
Click to expand...
Now we divide everything by D^0.22 and get
1.2((pi/2) * R^2)^0.22 ≤ D^0.78 ≤ 2.4((pi/2) * R^2)^0.22
Click to expand...
Then we take all sides ^(1/0.78) and get an estimate for D
(1.2((pi/2) * R^2)^0.22)^(1/0.78) ≤ D ≤ (2.4((pi/2) * R^2)^0.22)^(1/0.78)
Click to expand...

Then one can simplify that into some nice formula, but I'm too lazy for that step right now.
 
DontTalkDT said:
Well the paper suggests using equation 15, which states

If we assume v to be paraboloid shaped, then v = (pi/2) * R^2 * D, and we get

Since it's a factor, we can pull the D out of those exponents, which gets us

Now we divide everything by D^0.22 and get

Then we take all sides ^(1/0.78) and get an estimate for D


Then one can simplify that into some nice formula, but I'm too lazy for that step right now.
Click to expand...
For craters larger than 10km yes?
 
Thanks again for helping out DontTalk.
 
DontTalkDT said:
Well the paper suggests using equation 15, which states

If we assume v to be paraboloid shaped, then v = (pi/2) * R^2 * D, and we get

Since it's a factor, we can pull the D out of those exponents, which gets us

Now we divide everything by D^0.22 and get

Then we take all sides ^(1/0.78) and get an estimate for D


Then one can simplify that into some nice formula, but I'm too lazy for that step right now.
Click to expand...
Simplifying we get 1.43493*R^.564103 =< D =< 3.48952*R^.564103

Idk if we are going to go with the low end or if we will go with average, but that would imply an average D ~ 2.46223*R^.564103

So, if we use the average to find the volume for craters of diameter > 10km would be V ~ 3.86766*R^2.564103 or if we use the low end V >= 2.25398*R^2.564103
 
DontTalkDT said:
Anyways, once this is properly worked out I should probably put something about it on the calculations page.
Click to expand...
Feel free to do so when you think that it is appropriate.
 
Arc7Kuroi said:
Simplifying we get 1.43493*R^.564103 =< D =< 3.48952*R^.564103

Idk if we are going to go with the low end or if we will go with average, but that would imply an average D ~ 2.46223*R^.564103

So, if we use the average to find the volume for craters of diameter > 10km would be V ~ 3.86766*R^2.564103 or if we use the low end V >= 2.25398*R^2.564103
Click to expand...
It's probably best to go with the low end, if there is no reason another end is more likely.
 
Not sure if the explosion yield page or the calculations page is more appropriate, but I will add it somewhere.
 
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