# Montgomery College Programming Question Lecture 2, Rocket ScienceSome physical problems are easy to solve numerically using just the basic equations of phy

Montgomery College Programming Question Lecture 2, Rocket ScienceSome physical problems are easy to solve numerically using just
the basic equations of physics. Other problems may be very difficult.

Consider a specific model rocket with a specific engine.
Given all the data we can find, compute the maximum altitude
the rocket can obtain. Yes, this is rocket science.

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Most physics computation is performed with metric units.
units and equations

Estes Alpha III
Length 12.25 inches = 0.311 meters
Diameter 0.95 inches = 0.0241 meters
Body area 0.785 square inches = 0.506E-3 square meters cross section
Cd of body 0.45 dimensionless
Fins area 7.69 square inches = 0.00496 square meters total for 3 fins
Cd of fins 0.01 dimensionless
Weight/mass 1.2 ounce = 0.0340 kilogram without engine
Engine 0.85 ounce = 0.0242 kilogram initial engine mass
Engine 0.33 ounce = 0.0094 kilogram final engine mass

Thrust curve
Total impulse 8.82 newton seconds (area under curve)
Peak thrust 14.09 newton
Average thrust 4.74 newton
Burn time 1.86 second

Initial conditions:
t = 0 time
s = 0 height
v = 0 velocity
a = 0 acceleration
F = 0 total force not including gravity
m = 0.0340 + 0.0242 mass

Basic physics:

Fd = Cd*Rho*A*v^2 /2 two equations, body and fins
Fd is force of drag in newtons in opposite direction of velocity
Cd is coefficient of drag, dimensionless (depends on shape)
Rho is density of air, use 1.293 kilograms per meter cubed
A is total surface area in square meters
v is velocity in meters per second (v^2 is velocity squared)

Fg = m*g Fg is force of gravity toward center of Earth
m is mass in kilograms
g is acceleration due to gravity, 9.80665 meters per second squared

Ft = value from thrust curve array at this time, you enter this data.
index i, test i>18 and set Ft = 0.0
Do not copy! This is part of modeling and simulation.

F = Ft – (Fd body + Fd fins + Fg) resolve forces

a = F/m a is acceleration we will compute from knowing
F, total force in newtons and
m is mass in kilograms of body plus engine mass that changes

dv = a*dt dv is velocity change in meters per second in time dt
a is acceleration in meters per second squared
dt is delta time in seconds

v = v+dv v is new velocity after the dt time step
(v is positive upward, stop when v goes negative)
v+ is previous velocity prior to the dt time step
dv is velocity change in meters per second in time dt

ds = v*dt ds is distance in meters moved in time dt
v is velocity in meters per second
dt is delta time in seconds

s = s+ds s is new position after the dt time step
s+ is previous position prior to the dt time step
ds is distance in meters moved in time dt

m = m -0.0001644*Ft apply each time step

t = t + dt time advances

i = i + 1

print t, s, v, a, m

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