Finding Impact Depth: Most Realistic Way Possible

[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?

 

[Baluncore]

My only questions now are did I do my math right? Because I only got 30mm, so it doesn't sound right.

I did not bother to follow your calculations once you departed SI units for cal/g.

And also, when you mentioned 50 degrees earlier, was that for the frustum?

Energy is delivered perpendicular to the surface, over the area of the projectile, the projectile will spread out sideways as it delivers the energy which will then radiate within the steel. My guestimate is that the energy will be concentrated within a cone angle of about 50°.

Not everything is obvious. If the projectile had a hardened rod as a core it would penetrate deeper with a sharper cone. When tanks were first introduced during WW1, it was found that by reversing the bullet in the cartridge before firing, it could penetrate the light armour of the tank, while a pointed bullet traveling forwards would not.

 

[JoeSalerno]

I did not bother to follow your calculations once you departed SI units for cal/g.Energy is delivered perpendicular to the surface, over the area of the projectile, the projectile will spread out sideways as it delivers the energy which will then radiate within the steel. My guestimate is that the energy will be concentrated within a cone angle of about 50°.

Not everything is obvious. If the projectile had a hardened rod as a core it would penetrate deeper with a sharper cone. When tanks were first introduced during WW1, it was found that by reversing the bullet in the cartridge before firing, it could penetrate the light armour of the tank, while a pointed bullet traveling forwards would not.

I paid a bit more attention to the units this time, making sure not to cross anything, and I got 5.3cm. This seems like a pretty realistic answer that I'd call good. Thanks for sticking through this process with me, as I've made quite a few obvious mistakes and may have not understood things the first (or fifth) time around. I have one more question though, Is the process for impacting materials that aren't metal such as ceramic different than this because they crack before deforming?

 

[JoeSalerno]

I paid a bit more attention to the units this time, making sure not to cross anything, and I got 5.3cm. This seems like a pretty realistic answer that I'd call good. Thanks for sticking through this process with me, as I've made quite a few obvious mistakes and may have not understood things the first (or fifth) time around. I have one more question though, Is the process for impacting materials that aren't metal such as ceramic different than this because they crack before deforming?

Whoops, yet another screw up, when finding the volume of the frustum I accidentally used the density of the projectile, not the material being impacted. Still, I believe I have the equations and mechanics down.

 

[Baluncore]

For every target there is a projectile. For every projectile there is an armour. Sometimes the escalation in cost and complexity of inventory is abandoned for a simple general solution such as a heavy lead slug. Heavy lead slugs are used primarily to immobilise vehicles by disrupting brittle material such as cast-iron engine blocks and transmission line components or electrical systems. After passing through most surface layers without significant deflection or fragmentation, a heavy slow slug will fracture something solid. A heavy slug would not be used against thick armour plate.

Ceramics are brittle but will resist thermal attack. Thick heavy armour plate produced from one material is now a poor investment. It is better to mix or laminate different materials. For example, WW2 D-day landing craft used granite chips set in bitumen as protection. It would absorb bullets without producing many damaging free fragments.