>>507797387
>>507798369
>>507799513
It weighs about 13,608kg or 30,000lbs
Impact speed is about 730 metres per second (Mach 2.15/1600mph)
m=13,608kg
v=730m/s
KE= 0.5×13,608kg×(730m/s)^2
KE=3,622,467,600Joules of kinetic energy alone
As for penetration differences
If we assume it has an effective contact radius of 12.5cm the contact area is 0.049m^2 = A_contact
Penetration Calculations
Average Force & Pressure for 60m Penetration in 5,000 psi Concrete
* Penetration Depth (d) = 60 m
* Concrete Strength = 5,000 psi
* Total KE to dissipate = 3.6249636 x 10^9 J
* **Average Force (F_avg):**
* Formula: F_avg = KE / d
* Calculation: F_avg = (3.6249636 x 10^9 J) / 60 m
* Result: F_avg ≈ 6.04 x 10^7 Newtons (60.4 MN)
* **Average Pressure (P_avg):**
* Formula: P_avg = F_avg / A_contact
* Calculation: P_avg = (6.04 x 10^7 N) / (0.049087 m^2)
* Result (Pa): P_avg ≈ 1.23 x 10^9 Pascals (1.23 GPa)
* Result (psi): P_avg ≈ 1.23 x 10^9 Pa * (1 psi / 6894.76 Pa) ≈ 178,500 psi
Average Force & Pressure for 8m Penetration in 10,000 psi Concrete
* Penetration Depth (d) = 8 m
* Concrete Strength = 10,000 psi
* Total KE to dissipate = 3.6249636 x 10^9 J
* **Average Force (F_avg):**
* Formula: F_avg = KE / d
* Calculation: F_avg = (3.6249636 x 10^9 J) / 8 m
* Result: F_avg ≈ 4.53 x 10^8 Newtons (453 MN)
* **Average Pressure (P_avg):**
* Formula: P_avg = F_avg / A_contact
* Calculation: P_avg = (4.53 x 10^8 N) / (0.049087 m^2)
* Result (Pa): P_avg ≈ 9.23 x 10^9 Pascals (9.23 GPa)
* Result (psi): P_avg ≈ 9.23 x 10^9 Pa * (1 psi / 6894.76 Pa) ≈ 1,339,000 psi
As you can see the concrete being twice as strong results in the Average Pressure being approx 7.5 times higher
7.5 times 8 = 60
So there's your answer
