7.
The main goal of the lab is to develop and program a numerical and an analytical
algorithm to calculate the rocket’s velocity and height as functions of time during the
flight for all three cases and then compare. In addition, to show step size independency
for case 3 by comparing a step size of 0.05 seconds with a step size of 0.01 seconds.
Simplifying Assumptions:
The following assumptions were made:
● Horizontal displacement is negligible. The rocket experiences vertical
displacement only because the thrust acts upon the rocket solely in the vertical
direction.
● Fd is proportional to at “low” velocity. At “low” speeds, fluid resistance is linear v
with velocity. The proportion comes from the cohesion of the layers of the fluid.
● FD is proportional to at “high” velocity. At “high” speeds, fluid resistance isv2
better resembled with a quadratic relationship with velocity. This relationship is
related to the momentum transfer between the moving object and the fluid
through which it travels.
● CD, b, g and are constants. The coefficients for aerodynamic and linear dragρ
are based on the shape of the rocket, the rocket does not change shape during
the duration of the model. The possible variance in the values for g and due toρ
location or weather conditions are negligible, especially if the model flights are
occurring in the same location.
● In Case I, thrust and mass are modeled as constants. This allows for the
calculations to be simplified, meaning an answer can be calculated within a tight
time constraint, if there is one presented by a customer.
● In Case II, thrust is variable and mass is modeled as a constant. This allows for
the calculations to be more accurate than Case I, since we are lying less. The
case is still partially simplified, meaning an answer can be calculated within a
time constraint, if there is one presented.
● In Case III, thrust and mass are variable, as they would be in a true experiment.
This allows for the calculations to be more accurate than Case II, since we are
lying even less.
PreAnalysis:
Cases II and III use variable thrust, which is modeled after the experimental thrust data.
In order to allow the integration step size to be independent from the thrust data that
was given, the data was interpolated. An eighth order polynomial curve fit was found for
the thrust data using Matlab’s basic fitting tool. The equation for the line is as follows:
8.
− .37 )x 6.6202 )x − .3765 )x 65061)x 48271)x − 2321)xy = ( 3 × 105 8 + ( × 105 7 + ( 4 × 105 6 + ( 5 + ( 4 + ( 2 3
3039.4)x − 3.091)x 1.0471)+ ( 2 + ( 7 + (
Figure 11 below shows the experimental thrust data overlaid with the eight order
polynomial curve fit.
Figure 11 Plot of the Experimental Data for Variable Thrust During the Rocket Flight and the Eighth Order
Polynomial Best Fit Curve
Imputing time in for x to the eighth order best fit equation yields the equation for variable
thrust. This equation can be seen below.
(t) − .37 )t 6.6202 )t − .3765 )t 65061)t 48271)t − 2321)tT = ( 3 × 105 8
+ ( × 105 7
+ ( 4 × 105 6
+ ( 5
+ ( 4
+ ( 2
3039.4)t − 3.091)t 1.0471)+ ( + ( 7 + (
Case III considers variable mass along with variable thrust. The mass burnoff is
proportional to the net impulse delivered by the fuel at the current point. In other words,
the percentage of mass burned off at time t is equal to the percent of area under the
thrusttime curve from takeoff to the current time t, relative to the total impulse delivered
by all the fuel.
10. Next we can use Newton’s second law to relate the thrust to the change in momentum.
By taking the limit of this relationship as Δt approaches 0, a function for Thrust relating
the change in mass with the change in time and the relative velocity of the fuel from the
rocket can be reached. The equation and its derivation are shown below.
0 = ∑
F = dt
dp
= lim
Δt→0 Δt
Δp
= lim
Δt→0 Δt
p2−p1
lim
Δt→0 Δt
m (v+Δv) + Δm(v−υ ) − (m +Δm)v r
e
r
υ lim
Δt→0
= Δt
m Δvr* − Δt
Δm e
υ = mr dt
dv
− dt
dm e
= 0
υ dt
m dv* = dt
dm e
Recall that thus(t)dt
m dv* = F
(t) υ F = dt
dm e
The thrust of the rocket is the force applied to the rocket
so the equation can be rewritten in its final form as
(t) υ T = dt
dm e
Finding the indefinite integral of this new T(t) equation we get the following.
(t) t υ t m∫
T * d = ∫
dt
dm e
* d = υ e
∫
d
Finally, taking the ratio of the indefinite integral it itself with the numerator evaluated
from the initial conditions to at a specific point, t#
, and the denominator evaluated from
the initial conditions to the end of the thrust data the following proportion is reached.
(t) dt ∫
t#
0
T *
(t) dt∫
tfinal
0
T *
=
v me ∫
m(t#)
mfuel
d
v me ∫
0
mfuel
d
= v me fuel total
v m (t#)e fuel burned
= mfuel total
m (t#)fuel burned
This proportion shows that the mass burnoff is proportional to the net impulse of the
thrust.
Variable mass is represented by the equation:
(t) (1 (t))m = massfuel,total − Φ
where was given as 5.7g and is the percent mass burned. The percentmassfuel,total (t)Φ
mass burned is found using the proportion above.
11.
(t) Mass BurnedΦ = % =
(t)dt=v m=v m∫
tfinal
0
T e ∫
0
mfuel
d e fuel total
(t)dt=v m=v m (t )∫
t*
0
T e ∫
m(t )*
mfuel
d e fuel burned *
= mfuel total
m (t )fuel burned *
where is the mass of the fuel burned at any given time , and is(t )mfuel burned * t * mfuel total
the mass of the fuel burned over the entire duration of thrust. To find the(t )mfuel burned *
integral of the equation for variable thrust was taken. The resulting equation is seen
below:
(t) − .3744 )t 0.827525 )t − .62521 )t 10843.5)tmfuel burned = ( 0 × 105 9
+ ( × 105 8
+ ( 0 × 105 7
+ ( 6
+
9654.2)t − 580.25)t 1013.13)t − 6.5455)t 1.0471)t( 5
+ ( 5 4
+ ( 3
+ ( 3 2
+ (
To find , the total time duration of thrust, 0.55 seconds, was input into themfuel total
equation for t. The result is seen below:mfuel burned
(.55) − .3744 )(.55) 0.827525 )(.55) − .62521 )(.55)mfuel burned = ( 0 × 105 9
+ ( × 105 8
+ ( 0 × 105 7
+
10843.5)(.55) 9654.2)(.55) − 580.25)(.55) 1013.13)(.55)( 6
+ ( 5
+ ( 5 4
+ ( 3
+
− 6.5455)(.55) 1.0471)(.55)( 3 2
+ (
mfuel total .1411 g= 2
The final equation for becomes:(t)Φ
(t)Φ = 2.1411 g
(−0.3744×10 )t +(0.827525×10 )t +(−0.62521×10 )t +(10843.5)t +(9654.2)t +(−5580.25)t +(1013.13)t +(−36.5455)t +(1.0471)t5 9 5 8 5 7 6 5 4 3 2
The final equation for variable mass, m(t), then becomes:
(t) 5.7)(1 ))m = ( − ( 2.1411 g
(−0.3744×10 )t +(0.827525×10 )t +(−0.62521×10 )t +(10843.5)t +(9654.2)t +(−5580.25)t +(1013.13)t +(−36.5455)t +(1.0471)t5 9 5 8 5 7 6 5 4 3 2
The graph of variable mass vs. time (using a 0.01 second time step) can be seen below in
Figure 12.
13. F − g Δmg − m)g vΣ h = mr − Fd + = ( mr + Δ − b = mdt
dv
To plot the response for Case II of the rocket flight, the equation found for thrust, T(t)
(seen in the preanalysis section), was input into a Taylor Series based on the force
equations above.The equations for both the linear and aerodynamic drag models are
seen below.
Linear drag, for time 0 seconds to 0.55 seconds
≈ v )Δt vi+1 i + ( m
T(t)
− g − m
bvi
For time after 0.55 seconds
≈ v − )Δt vi+1 i + ( g − m
bvi
Aerodynamic drag, for time 0 seconds to 0.55 seconds
≈ v )Δtvi+1 i + ( m
T(t)
− g − 2m
ρC Av ²D i
For time after 0.55 seconds
≈ v − )Δt vi+1 i + ( g − 2m
ρC Av ²D i
To plot the response for Case III of the rocket flight, the equation found for thrust, T(t)
(seen in the preanalysis section), is input into the Taylor Series based on the force
equations above. Additionally, the mass of just the rocket, , plus the variable massmr
m(t) is input into the equation in place of the constant mass, m. The equations for both
the linear and aerodynamic drag models are seen below.
Linear drag, for time 0 seconds to 0.55 seconds:
≈ v )Δtvi+1 i + ( T(t)
m +m(t)r
− g −
bvi
m +m(t)r
For time after 0.55 seconds, m(t) = 0:
≈ v − )Δtvi+1 i + ( g − mr
bvi
Aerodynamic drag, for time 0 seconds to 0.55 seconds:
≈ v )Δtvi+1 i + ( T(t)
m +m(t)r
− g − 2mr
ρC Av ²D i
For time after 0.55 seconds, m(t) = 0:
≈ v − )Δtvi+1 i + ( g − 2mr
ρC Av ²D i
Solutions for Unknown: