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DuncP89
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This is NOT a school work/homework, it's just a problem relate to my personal interest found on a ballastics book. And I already converted all units into SI units to make it simple.
I tried but couldn't get to correct answer.
Please not a is speed of sound, K2 is a constant with equation.
The answer is 0.28 mil lower which is also 0.01575 degree lower.
A U.S. 37-MM projectile is fired with a muzzle velocity of 2600ft's (792.48m/s). The projectile weighs 1.61 Ibm(0.73kg). Assuming K2 = 0.841[unitless] and using standard sea level met data(ρ=0.0751Ibm/ft^3 (1.2kg/m^3), a = 1120 ft/s (341.4m/s))
If this weapon is used at an increased altitude and assuming the density and temperature of the atmosphere are ρ=0.06Ibm/ft^3 (0.98kg/m^3) and T=30F degree, how much higher or lower will the weapon have to be aimed to hit a target at 800 yards.
Drag force: F = 1/2ρ*S*C*V^2
S = πd^2/4 (d is diameter)
Cd = ρS/2m * K2*a/Vx = k2/Vx
Vx = Vx0*exp(-k2*t) ...t is time of flight ot any range x
Vy = (Vy0+g/k2)exp(-k2*t)-g/k2
t=x/Vx0 * ln(Vx0/Vx) / (1-Vx/Vx0)
tan∅ = tan∅0 + gx/Vx0^2((1-Vx0/Vx)/(1-Vx/Vx0)) ...∠∅ is the angle of fall and ∠∅0 is the initial launch angle
y = y0 + x*tan∅0 - (g*t^2/(2ln(Vx0/Vx)))
I first assume the first firing is flat firing, so the initial launch angle is zero, after the increased altitude, i calculated the new k2 and then apply it into these two equations:
Vx = Vx0*exp(-k2*t) ...t is time of flight ot any range x
Vy = (Vy0+g/k2)exp(-k2*t)-g/k2
I assume the muzzle velocity is Vx at the halfway of the range since Vy and halfway of range is zero, i then manipulate these two equations to a form of Vy0/Vx0, and the do a inverse tan to get the initial launch angle.
but it doesn't work and i realized that muzzle velocity is not Vx at halfway since the moment the projectile went out, the velocity is changing. and i was stuck here
I tried but couldn't get to correct answer.
Please not a is speed of sound, K2 is a constant with equation.
The answer is 0.28 mil lower which is also 0.01575 degree lower.
Homework Statement
A U.S. 37-MM projectile is fired with a muzzle velocity of 2600ft's (792.48m/s). The projectile weighs 1.61 Ibm(0.73kg). Assuming K2 = 0.841[unitless] and using standard sea level met data(ρ=0.0751Ibm/ft^3 (1.2kg/m^3), a = 1120 ft/s (341.4m/s))
If this weapon is used at an increased altitude and assuming the density and temperature of the atmosphere are ρ=0.06Ibm/ft^3 (0.98kg/m^3) and T=30F degree, how much higher or lower will the weapon have to be aimed to hit a target at 800 yards.
Homework Equations
Drag force: F = 1/2ρ*S*C*V^2
S = πd^2/4 (d is diameter)
Cd = ρS/2m * K2*a/Vx = k2/Vx
Vx = Vx0*exp(-k2*t) ...t is time of flight ot any range x
Vy = (Vy0+g/k2)exp(-k2*t)-g/k2
t=x/Vx0 * ln(Vx0/Vx) / (1-Vx/Vx0)
tan∅ = tan∅0 + gx/Vx0^2((1-Vx0/Vx)/(1-Vx/Vx0)) ...∠∅ is the angle of fall and ∠∅0 is the initial launch angle
y = y0 + x*tan∅0 - (g*t^2/(2ln(Vx0/Vx)))
The Attempt at a Solution
I first assume the first firing is flat firing, so the initial launch angle is zero, after the increased altitude, i calculated the new k2 and then apply it into these two equations:
Vx = Vx0*exp(-k2*t) ...t is time of flight ot any range x
Vy = (Vy0+g/k2)exp(-k2*t)-g/k2
I assume the muzzle velocity is Vx at the halfway of the range since Vy and halfway of range is zero, i then manipulate these two equations to a form of Vy0/Vx0, and the do a inverse tan to get the initial launch angle.
but it doesn't work and i realized that muzzle velocity is not Vx at halfway since the moment the projectile went out, the velocity is changing. and i was stuck here