# Pytels Dynamics 12.18: Rocket dynamics

Tags:
1. Jun 17, 2017

### Alexanddros81

1. The problem statement, all variables and given/known data
A rocket engine takes 8 seconds after firing to reach its full thrust. Assuming that
the rocket was fired at t=0, use the following time-elevation data to estimate the
velocity and acceleration (a) at t=0; and (b) at t=8s

2. Relevant equations

3. The attempt at a solution
I don't know which equation to use
Solution given is for a) v=-2.5m/s, a=0m/s2 (b) v=160m/s, a=33m/s2

#### Attached Files:

• ###### 12_18 Pytel.jpg
File size:
28.9 KB
Views:
31
2. Jun 17, 2017

### haruspex

This seems a rather strange question, and the answers even stranger.
It is nigh impossible to estimate the acceleration at t=0 from these data, but I would have thought it should be > 0. And I do not see how the velocity could be negative then. So I will violate the usual HW forum rules and lay out my solution.

I approached it by supposing the height is a polynomial function of time and looked for the highest degree. Seems to be about 3. So then I used an online tool to find the best cubic fit to the data. That doesn't quite work because it gives a nonzero height at t=0. I fixed that by dividing the heights by the times (leaving out the t=0 point) and finding the best quadratic fit, then multiplying back by time. This gives
y = 7.08E-001t3 + 2.177 t2 -2.303t.
That's a pretty good fit to the data, but it still gives a negative velocity at t=0. You could fix that by extending my procedure for solving the nonzero height, giving you a polynomial with only t2 and t3 terms.

Edit: I believe it makes sense to assume height and velocity are 0 at time 0. So divide by t2 and then look for a polynomial fit. I had to go up one more degree to get a good fit:
y = -4.09602831·10-2 t4 + 1.278815296 t3 - 9.82402941·10-2 t2

Last edited: Jun 17, 2017
3. Jun 17, 2017

### rude man

You could use the first three and last three data pairs to answer the question. But then I didn't get either given answer, missing the t=0 one big-time. But then, they did say "estimate"

You could also assume thrust F(t) = kt or F(z) = kz, fitting the data to get k (where z = elevation). But I wouldn't bother.

4. Jun 18, 2017

### Alexanddros81

I finally found in Pytel book what I was looking for. At the begining of the book at preface it states:
"An asterisk is also used to mark problems that require advanced reasoning. Topics and problems
associated with numerical methods are preceded by an icon representing a computer disk"
Having an image of a computer disk in front of a problem number means it is solved using numerical method.
Up to this problem sample problem 12.4 was used numerically using numerical differentiation with finite
differences which is summarized in Appendix E.

I don't know if this helps...

5. Jun 18, 2017

### Ray Vickson

One can also try a constrained least-squares (L2) fit and/or least-absolute-value (L1) cubic fit to the data. In both cases we seek a fit of the form $y = at + bt^2 + ct^3$ under the constraint $a \geq 0$. The L2 fit minimizes $\sum_{i=1}^9 (y_i - at_i - b t_i^2 - c t_i^3)^2$ (using a nonlinear optimization package to handle the constraint on $a$), while the L1 fit minimizes $\sum_{i=1}^9 |y_i - at_i - b t_i^2 - c t_i^3|$ (by converting it to a linear programming problem and solving using a computer package). The results are:
$$\begin{array}{rl} \text{L1 fit:} & y = 1.58 t^2 + 0.74 t^3 \\ \text{L2 fit:} & y = 1.4940 t^2 + 0.75573 t^3 \end{array}$$.
The initial velocities are both zero (obtained as output, rather than input), and the initial accelerations are 0.74 (m/s) and 0.75573 (m/s) are reasonably consistent.

I would much more trust such a "fitting" method than a "finite-difference" method in this case, because the question asks for a second-derivative estimate at an end-point and I do not believe finite-difference methods can be reliable in such cases.

However, as you have said already, there is no truly convincing way of estimating the figures the question seeks.

Last edited: Jun 18, 2017