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For a rocket of:

  • mass, $m$ = 25kg,
  • dry mass $m_f$ = 19 kg,
  • Thrust = 2.1 KN,
  • total powered burn time, $t_b$ = 3.5 s,
  • $I_{sp}$ = 129

I get powered burn height = 549 m and total time to max ht of 5506 m in 35 sec. Is this correct? Assuming no drag and vertical liftoff.

I am using:

$$h = {I_{sp} \cdot g \cdot t_b} \cdot (1-ln(m_o/m_f)/(m_o/m_f -1)) - 0.5\cdot g \cdot t^2$$

Also, to calculate the burn rate I divided fuel_wt by total burn time. Is that roughly ok?

Can anyone help with this please? If possible, the relations/formulae that are to be used. Anyone know where can I get a mentor for rocketry.

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    $\begingroup$ I don't see drag anywhere there. $\endgroup$ Sep 22 at 11:24
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    $\begingroup$ Hello SE, its assuming zero drag $\endgroup$ Sep 22 at 13:20
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    $\begingroup$ For Rocketry, you may need to say where you are, for MODEL rocketry look at linked orgs from en.wikipedia.org/wiki/Tripoli_Rocketry_Association, for going to space type rockets you'd be looking at a different sort of mentor. $\endgroup$ Sep 22 at 13:58
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    $\begingroup$ Try nakka-rocketry.net This guy is the grandmaster of amateur experimental rocketry, with a whole passel of advice, tools, spreadsheets, instruction videos, etc. It's aimed at the amateur that builds their own motors, plays around with different fuels and systems, but does not plan to launch anything fancier than a GPS altimeter and camera to several thousand feet. $\endgroup$ Sep 23 at 11:36
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    $\begingroup$ If 6 kg of fuel produce 4 KN of thrust for 3.5 s, the $I_{sp}$ must be $\frac{4000 ~\text{N} \cdot 3.5~\text{s}}{6~\text{kg}\cdot 9.8~\text{m/s$^2$}}=238~\text{s}$. Also, your formula gives the height at the end of the burn (assuming no drag and constant acceleration during the burn), not the maximal height. $\endgroup$
    – Litho
    Sep 24 at 11:32
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I don't think that equation is correct. I don't see the benefits of using $I_{sp}$ when you have the thrust, $F_{t}$. The 1D equation of motion (neglecting drag) is:

$a_{b}(t)=\frac{F_t}{m_0-\dot{m}t}-g$

$a_{c}(t) = -g$

Where $c$ and $b$ are the coast and boost phases of flight. Your burn rate, $\dot{m}$, assumption is good, and since it is a constant you can analytically integrate to find velocity and height (its not worth it beyond velocity though):

$v_{b}(t)=-\frac{F_t}{\dot{m}}\ln(m_0-\dot{m}t)-gt + C_{1}$

$v_{c}(t)=-g(t-t_{b})+v_{b}(t_{b})$

$h_{b}(t)=-\frac{1}{\dot{m}^2}*(-C_{1}\dot{m}(m_0-\dot{m}t)-F_{t}(\dot{m}\ln(m_0-\dot{m}t)-\dot{m}t\ln(m_0-\dot{m}t)-m_{0}+\dot{m}t)+g(\frac{(m_0-\dot{m}t)^2}{2}-m_{0}(m_0-\dot{m}t)))-C_{2}$

$h_{c}(t)=-\frac{g}{2}(t-t_{b})^2+v_{b}(t_{b})(t-t_{b})+h_{b}(t_{b})$

Numerical integration (Euler, 0.01s dt) was far simpler, quicker, and yielded results within ~1% of the analytical (and you need numerical if you were to account for aerodynamic drag):

  • powered burn height: 501 m
  • max height: 5145 m @ ~34s
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    $\begingroup$ Thank you @BrendanLuke15 for the detailed answer. That was helpful. $\endgroup$ Sep 26 at 11:03
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    $\begingroup$ Hello @BrendanLuke15, What would the equation look like if I add air Drag to the calculation. $\endgroup$ Oct 26 at 9:50

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