Difference between revisions of "Optimal ground launch flight path"
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<math>\sum F=ma=m\frac{dv}{dt}=\frac{dmv}{dt}=\frac{dp}{dt}</math> | <math>\sum F=ma=m\frac{dv}{dt}=\frac{dmv}{dt}=\frac{dp}{dt}</math> | ||
− | That is the force is the rate momentum changes with time. Consequently, the force exerted on the rocket by the exhaust is equal to the velocity multiplied by the mass flow. Taking into gravity the fores acting in the direction of flight | + | That is the force is the rate momentum changes with time. Consequently, the force exerted on the rocket by the exhaust is equal to the velocity multiplied by the mass flow. Taking into gravity the fores acting in the direction of flight and dividing both sides by the mass we obtain: |
− | <math> | + | <math>a=V_e\frac{dm}{dt}+g sin(\theta)</math> |
== Optimal Flight Path == | == Optimal Flight Path == |
Revision as of 01:59, 13 August 2009
Contents
Introduction
The current derivation of the rocket equation on wikipedia (Aug 13 2009) seems to be overly cumbersome: http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation
The rocket equation basically comes down to the basic law of Newtonian dynamics. That is the sum of the forces equals the mass multiplied by the acceleration.
<math>\sum F=ma</math>
The product of the rocket exhaust velocity multiplied the mass flow is the rate momentum is leaving the rocket. Because momentum must be conserved the rocket gains an equal and opposite amount of momentum as the exhaust. A more convenient form of newtons law is:
<math>\sum F=ma=m\frac{dv}{dt}=\frac{dmv}{dt}=\frac{dp}{dt}</math>
That is the force is the rate momentum changes with time. Consequently, the force exerted on the rocket by the exhaust is equal to the velocity multiplied by the mass flow. Taking into gravity the fores acting in the direction of flight and dividing both sides by the mass we obtain:
<math>a=V_e\frac{dm}{dt}+g sin(\theta)</math>
Optimal Flight Path
The optimal flight path occurs when the vertical component of the thrust cancels out gravity and the remaining part of the thrust is due to horizontal acceleration. This can be written as:
<math>g-\frac{v^2}{R}cos(\theta)-T sin (\theta)=0</math>
where: g is the gravitational acceleration R is the distance from the center of the earth T is the thrust per mass.
Using trig identities:
<math>g+/-\frac{v^2}{R}\sqrt{1-sin^2(\theta)}-T sin (\theta)=0</math>
Rearranging:
<math>+/-\frac{v^2}{R}\sqrt{1-sin^2(\theta)}=T sin (\theta)-g</math>
Squaring both sides:
<math>\frac{v^4}{R^2}(1-sin^2(\theta))=T^2 sin^2 (\theta)-2gTsin(\theta)+g^2</math>
Rearranging:
<math>(T^2+\frac{v^4}{R^2}) sin^2 (\theta)-2gTsin(\theta)+g^2-\frac{v^4}{R^2}=0</math>
This equation can be solved using the Quadratic Equation.
A Visual Basic Function to Caculate the Optimal Flight Path
Bellow is a visual basic function used in RocketDesignSpreadsheet the optimal thrust angle.
Function thrustAngle(V, T, Optional h, Optional g)
Re = 6378.137 * 1000 ' Radious of The Earth
If IsMissing(h) Then
h = 0
End If
If IsMissing(g) Then
Gc = 0.00000000006674 ' N(m/kg)^2Gravational Constant
Mearth = 5.9736E+24 ' Mass of the earth
g = Gc * Mearth / (h + Re) ^ 2
End If
R = h + Re
a = (T ^ 2 + V ^ 4 / R ^ 2)
b = -2 * g * T
c = g ^ 2 - b ^ 4 / R ^ 2
des = b ^ 2 - 4 * a * c
If des < 0 Then
thrustAngle = 0
Else
r1 = (-b + Sqr(des)) / (2 * a)
r2 = (-b - Sqr(des)) / (2 * a)
theata1 = WorksheetFunction.Asin(r1)
theata2 = WorksheetFunction.Asin(r2)
resid1 = g - V ^ 2 / R * Cos(theata1) - T * Sin(theata1)
resid2 = g - V ^ 2 / R * Cos(theata2) - T * Sin(theata2)
If Abs(resid1) < Abs(resid2) Then
thrustAngle = theata1
Else
thrustAngle = theata2
End If
End If
End Function