Finding Impact Depth: Most Realistic Way Possible

[JoeSalerno]

I'm trying to find the impact or penetration depth of a projectile in the (reasonably) most realistic way possible. I assume this has many factors including velocity, the densities of both materials, shapes of both materials, the medium of which the projectile travels through, gravity, etc. Because I know this could be infinitely complicated by variables, I will be assuming the air will be standard temperature pressure. This should ease finding the air resistance on the projectile. Depending on the necessary equations, I may end up omitting other variables. Thanks in advance to anyone who can help.

 

[Baluncore]

Is it it an asteroid hitting the Earth, or maybe a hailstone landing on snow?
Rather than asking an all-encompassing question, could you narrow it down a bit.

What range of mass and velocity does the projectile have on impact?
What material is the target? Liquid or solid? Structured or isotropic?

 

[JoeSalerno]

Is it it an asteroid hitting the Earth, or maybe a hailstone landing on snow?
Rather than asking an all-encompassing question, could you narrow it down a bit.

What range of mass and velocity does the projectile have on impact?
What material is the target? Liquid or solid? Structured or isotropic?

To be more specific, this is for a .950 JDJ bullet impacting 4340 steel. To find the velocity of the projectile on impact, I would need to find out what effect the air has on it. We can assume the starting speed is 670 m/s and the weight is 233 grams. The material that will be impacted is 4340 steel, so it is a hardened metal alloy. I assume shape matters for aerodynamic reasons, so the projectile is 24.13mm wide and 50mm tall. For simplicity, we can assume it is a "basic" pointed bullet, the closest shape I could find to match it would be an ogive on top of a small cylinder. Hopefully this should be enough to get started, let me know if this isn't sufficient.

 

[Khashishi]

The air resistance depends how long it travels through the air.

 

[JoeSalerno]

We c

The air resistance depends how long it travels through the air.

We can assume 100 Meters for simplicity.

 

[Baluncore]

We can assume the starting speed is 670 m/s and the weight is 233 grams.

I think you can assume that bullet will still be traveling at 670 m/s after 100m.
There are ballistics tables for every bullet ever sold. The ballistics table or program will give you accurate velocity data.

 

[JoeSalerno]

I think you can assume that bullet will still be traveling at 670 m/s after 100m.
There are ballistics tables for every bullet ever sold. The ballistics table or program will give you accurate velocity data.

if we were to assume that the impact velocity was 670 m/s, what would be the next step?

 

[Baluncore]

Probably find out what the projectile energy is on impact.
Then what mass of the target material could be heated with that energy to the temperature where it ceases to be strong.
As a worst case estimate you could work out the depth of a 1" hole having that mass of steel.

 

[rootone]

Well it definitely helps to have your projectile made up a least partly from depleted Uranium,
but there are, em, 'issues' with that approach.

 

[JoeSalerno]

Probably find out what the projectile energy is on impact.
Then what mass of the target material could be heated with that energy to the temperature where it ceases to be strong.
As a worst case estimate you could work out the depth of a 1" hole having that mass of steel.

So, I need to make some major unit corrections. The projectile is 77mm long, 25.4mm wide, and weighs (get ready for it, this is why it's theoretical) 46,238 grams. We will also give it an unreasonable speed of 914 m/s. When you say that I need the energy on impact, would that be E=(1/2)*mass*velocity^2 ?

 

[Baluncore]

Yes. E=(1/2)*m*v²
I think your estimate of projectile mass at 46 kg is on the high side by a factor of 100. I would expect it to be below 1 kg.

77mm long
diam = 25.4 * 0.95 = 24.13 mm
radius = 12.065 mm
area = 457.3 mm²
max vol = 35,212 mm³ = 35.212 cc, assuming cylinder
Lead density = 11.34
max mass = 399.3 gram

 

[JoeSalerno]

Yes. E=(1/2)*m*v²
I think your estimate of projectile mass at 46 kg is on the high side by a factor of 100. I would expect it to be below 1 kg.

77mm long
diam = 25.4 * 0.95 = 24.13 mm
radius = 12.065 mm
area = 457.3 mm²
max vol = 35,212 mm³ = 35.212 cc, assuming cylinder
Lead density = 11.34
max mass = 399.3 gram

I tried getting as accurate of a volume as basic geometry could get me by putting the volumes of a frustrum, cylinder, and paraboloid. This gave me 24,994mm^3. The kicker however is the material. I looked at a list of some of the densest materials that were maybe semi-feasable (yet ridiculous) and decided on a tungsten alloy that's 18.5g/cm^3 (I realize that it would take a massive amount of propellant to get this thing going 3000 ft/s). I just multiplied 18.5 by 2,499.4 and got that ridiculous weight. Would the result of the energy equation be in joules?

 

[JoeSalerno]

Yes. E=(1/2)*m*v²
I think your estimate of projectile mass at 46 kg is on the high side by a factor of 100. I would expect it to be below 1 kg.

77mm long
diam = 25.4 * 0.95 = 24.13 mm
radius = 12.065 mm
area = 457.3 mm²
max vol = 35,212 mm³ = 35.212 cc, assuming cylinder
Lead density = 11.34
max mass = 399.3 gram

I just realized that the way you went from cubic mm to cubic centimeters was by multiplying by 1000. This is probably because of my ignorance, but is that because we are working in cubic instead of regular length, because I multiplied by 10.

 

[Comeback City]

I just realized that the way you went from cubic mm to cubic centimeters was by multiplying by 1000. This is probably because of my ignorance, but is that because we are working in cubic instead of regular length, because I multiplied by 10.

Yes

 

[Baluncore]

Would the result of the energy equation be in joules?

Yes, the System International is good that way.

I just realized that the way you went from cubic mm to cubic centimeters was by multiplying by 1000.

No, I divided by 1000. You must go very carefully when using mm or cm rather than metres.
There are 10 mm in 1 cm. So there are 1000 mm³ in 1 cm³

 

[JoeSalerno]

Yes. E=(1/2)*m*v²
I think your estimate of projectile mass at 46 kg is on the high side by a factor of 100. I would expect it to be below 1 kg.

77mm long
diam = 25.4 * 0.95 = 24.13 mm
radius = 12.065 mm
area = 457.3 mm²
max vol = 35,212 mm³ = 35.212 cc, assuming cylinder
Lead density = 11.34
max mass = 399.3 gram

Also, for the formula, is the mass in g or kg, and is the speed m/s?

 

[Baluncore]

Also, for the formula, is the mass in g or kg, and is the speed m/s?

The System International, SI, uses the MKSA = metres, kilogram, second, ampere units. One litre of water weighs 1 kg.
Convert everything to meters per second, kilograms and joules.
https://en.wikipedia.org/wiki/International_System_of_Units

 

[JoeSalerno]

Yes, the System International is good that way.
So, I ended up with 0.462kg which sounds much more realistic. Using the energy formula, E=0.5*0.462*(914^2) I got 193,308 Joules. Earlier you said to find out what mass could be heated to a weakened point by that much energy. How would I do this, and how would I find that temperature (malleability, melting, boiling point, or something else?)
No, I divided by 1000. You must go very carefully when using mm or cm rather than metres.
There are 10 mm in 1 cm. So there are 1000 mm³ in 1 cm³

 

[Baluncore]

Whoops, you entered your text inside the quote.

So, I ended up with 0.462kg which sounds much more realistic.
Using the energy formula, E=0.5*0.462*(914^2) I got 193,308 Joules.
Earlier you said to find out what mass could be heated to a weakened point by that much energy.
How would I do this, and how would I find that temperature (malleability, melting, boiling point, or something else?)

The data you will need on your plate alloy will be thermal capacity and the temperature at which it will lose it's strength.

Rapid compression of metal causes an instant temperature rise with shear and flow. You can suddenly deform metals that could never be deformed slowly when cold with the same force.
There are two competing effects. Firstly, you could punch a conical hole clean through a plate by delivering energy to a circular area on the surface and shearing the metal. Secondly you could heat a volume of a solid block of material so that it flows out to leave a hole in the alloy.

 

[JoeSalerno]

Whoops, you entered your text inside the quote.

The data you will need on your plate alloy will be thermal capacity and the temperature at which it will lose it's strength.

Rapid compression of metal causes an instant temperature rise with shear and flow. You can suddenly deform metals that could never be deformed slowly when cold with the same force.
There are two competing effects. Firstly, you could punch a conical hole clean through a plate by delivering energy to a circular area on the surface and shearing the metal. Secondly you could heat a volume of a solid block of material so that it flows out to leave a hole in the alloy.

Sorry about accidentally putting that last post in the quote. So, by thermal capacity I think you're saying specific heat. I found that's 486 joules/kg kelvin. Using the units you mentioned earlier there weren't any thermal ones so I assumed Kelvin would work. I'm sorry if I'm being a bit dense, but what would you call the value "of which a material loses its strength"? Is this just a general number i.e. Around 2000 Fahrenheit? For this specific steel the only thermal specs given are the melting point (2600 Fahrenheit), mean coefficient of thermal expansion (6.6), and the thermal conductivity (21).

 

[JoeSalerno]

Whoops, you entered your text inside the quote.

The data you will need on your plate alloy will be thermal capacity and the temperature at which it will lose it's strength.

Rapid compression of metal causes an instant temperature rise with shear and flow. You can suddenly deform metals that could never be deformed slowly when cold with the same force.
There are two competing effects. Firstly, you could punch a conical hole clean through a plate by delivering energy to a circular area on the surface and shearing the metal. Secondly you could heat a volume of a solid block of material so that it flows out to leave a hole in the alloy.

What exactly do you mean "the temperature at which it will loose it's strength"?
Also, sorry for abandoning this forum for over a month, got real busy and forgot about it.

 

[Baluncore]

What exactly do you mean "the temperature at which it will loose it's strength"?

As the temperature of a metal is increased it reaches a point where the material becomes softer and more malleable.
My guess for steel would be about 500°C. You can probably find data on young's modulus and the yield point at different temperatures.

 

[JoeSalerno]

I found a website that lists a bunch of the properties of 4340 steel, and some of the charts are a bit confusing. For the modulus of elasticity (young's modulus), it says 10^3 N/mm^2 is 165 at 500 Celsius. The yield strength in N/mm^2 is about 1050 at 500 Celsius (I'm estimating off a graph).

 

[Baluncore]

Well found. Having numbers is all very useful, except you have not listed for comparison what the parameters are at 20°C. Do you have a link to the site so I can see what the graph looks like ?

 

[JoeSalerno]

Well found. Having numbers is all very useful, except you have not listed for comparison what the parameters are at 20°C. Do you have a link to the site so I can see what the graph looks like ?

http://www.ssm.co.nz/sitefiles/4340.pdfhttp://www.ssm.co.nz/sitefiles/4340.pdf This is where I got my information. The graph for young's modulus doesn't go that low, but there is a bit on the yield strength at 20 Celsius. It is however That chart is kind of confusing as it mentions diameter, but no unit of measure.

 

[Baluncore]

That data looks like the properties once cooled, after a cycle of heat treatment.
"Temper 540-680°C hold for 1 hour min, at temperature, air cool. (see tempering chart)."

"Anneal 650°C – 700°C". Suggests it changes properties below 600°C.
I would guess that strength properties are lost at about 500°C.

Modulus of elasticity at different temperatures is there. It falls linearly.
100°C – 205 x 10³ N/mm²
200°C – 195
300°C – 185
400°C – 175
500°C – 165

 

[JoeSalerno]

That data looks like the properties once cooled, after a cycle of heat treatment.
"Temper 540-680°C hold for 1 hour min, at temperature, air cool. (see tempering chart)."

"Anneal 650°C – 700°C". Suggests it changes properties below 600°C.
I would guess that strength properties are lost at about 500°C.

Modulus of elasticity at different temperatures is there. It falls linearly.
100°C – 205 x 10³ N/mm²
200°C – 195
300°C – 185
400°C – 175
500°C – 165

All of the other websites with specs on this alloy only have annealed specs. Because of this, I'll assume that it's the standard to anneal before using in just about any application, so we can assume the piece being impacted has been annealed.

 

[JoeSalerno]

All of the other websites with specs on this alloy only have annealed specs. Because of this, I'll assume that it's the standard to anneal before using in just about any application, so we can assume the piece being impacted has been annealed.

I just used the points, found slope, the y intercept, and plugged in 20 for temperature. This gave me 213 x 10^3 N/mm^2

 

[Baluncore]

The implication of the modulus of elasticity to temperature relationship is to get an indication of the temperature at which the steel will collapse.

The projectile KE delivered on impact will heat the steel. That energy will be delivered to a cone of material behind the point of impact. Assume that cone is a frustum with the same small end diameter as the projectile, with an included angle of 50°. Given that the specific heat capacity is given as 460 J/(kg·K), and if you work on the assumption that steel will be removed if the temperature exceeds 500°C, then you can estimate the depth of penetration from the volume of the cone frustum that can be removed by heating steel to 500°C.

 

[JoeSalerno]

The implication of the modulus of elasticity to temperature relationship is to get an indication of the temperature at which the steel will collapse.

The projectile KE delivered on impact will heat the steel. That energy will be delivered to a cone of material behind the point of impact. Assume that cone is a frustum with the same small end diameter as the projectile, with an included angle of 50°. Given that the specific heat capacity is given as 460 J/(kg·K), and if you work on the assumption that steel will be removed if the temperature exceeds 500°C, then you can estimate the depth of penetration from the volume of the cone frustum that can be removed by heating steel to 500°C.

What formula or set of equations would I use to find the volume of said frustum?

 

[Baluncore]

What formula or set of equations would I use to find the volume of said frustum?

That is geometry. https://en.wikipedia.org/wiki/Cone#Volume_2
Work out the volume of the cone, then subtract the volume of the conical missing point.

Find the volume of steel you can heat, add the volume of the point of the cone smaller than the projectile diameter. Solve for depth of cone with that volume, subtract the height of the missing point.

 

[JoeSalerno]

That is geometry. https://en.wikipedia.org/wiki/Cone#Volume_2
Work out the volume of the cone, then subtract the volume of the conical missing point.

Find the volume of steel you can heat, add the volume of the point of the cone smaller than the projectile diameter. Solve for depth of cone with that volume, subtract the height of the missing point.

So, my (hopefully) last question: How do I find the volume of steel I can heat to 500 Celsius? And also, what do you mean by this "add the volume of the point of the cone smaller than the projectile diameter"

 

[Baluncore]

Compute the KE from speed and mass of projectile. The specific heat capacity is given as 460 J/(kg·K) in the Bohler data sheet.
Compute mass of steel that can be heated from 20°C to 500°C and so be removed. Convert mass of steel to volume removed.
Find depth of conical frustum that contains that volume of material. https://en.wikipedia.org/wiki/Frustum#Volume

 

[JoeSalerno]

That is geometry. https://en.wikipedia.org/wiki/Cone#Volume_2
Work out the volume of the cone, then subtract the volume of the conical missing point.

Find the volume of steel you can heat, add the volume of the point of the cone smaller than the projectile diameter. Solve for depth of cone with that volume, subtract the height of the missing point.

So, I got a bit antsy and tried to tackle this with google's help, as well as my knowledge in math.
Step 1: Finding the mass of the frustum
energy(J)=mass(g) * delta T * Cp 4340 steel(cal/g C)
193,308=m * 480 * 0.11
m=3661g
Step 2: Finding the volume of the Frustum:
density=mass/volume
18.5(g/cm^3)=3661g/v(cm^3)
v(cm^3)=197.9
Step 3: Finding the height (impact depth)
Vfrustrum= (pi/3)(h)(r^2+R^2+R*r) I had to do some substitution for R as it changes with hieght. You mentioned 50 degrees earlier, so I assumed that the triangle made in the frustum would be 50 90 40. tan(50)=(R-r)/h. I re-arranged the equation so h was the variable, as well as putting in my value for r. R= (h)tan(50)+2.74
197.9=(pi/3)(h)(2.74^2 + [(h)tan(50)+2.74]^2 + {[(h)tan(50)+2.74] * 2.74})
I simplified it to 197.9=1.49h^3 + 10.26h^2 + 23.56h
I then graphed the equation and found the x intercept which was about 3. This would seemingly mean 3cm, but that sounds too shallow. Did I do this right or did I miss something?

 

[JoeSalerno]

Compute the KE from speed and mass of projectile. The specific heat capacity is given as 460 J/(kg·K) in the Bohler data sheet.
Compute mass of steel that can be heated from 20°C to 500°C and so be removed. Convert mass of steel to volume removed.
Find depth of conical frustum that contains that volume of material. https://en.wikipedia.org/wiki/Frustum#Volume

My only questions now are did I do my math right? Because I only got 30mm, so it doesn't sound right. And also, when you mentioned 50 degrees earlier, was that for the frustum? If not, how do I know what angle the frustum is without knowing height?