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I'm getting interested in amateur rocketry. I'm realistic about it and understand that my skills and technology currently available won't let me construct anything beyond very simple designs, with glued paper hull and improvised nozzles, or at most with a hull from a pipe and water nozzle used fit to it.

What performance can I expect? How large such a rocket should be? Should I ever try build something longer than a couple of decimeters using such a primitive technology?

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  • $\begingroup$ Highly related question $\endgroup$ – Jack Jun 15 '18 at 11:10
  • $\begingroup$ @jack They are talking about quite advanced designs whereas I'll be happy if it climbs a few hundred feet $\endgroup$ – olegst Jun 15 '18 at 11:32
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Try a basic calculation using the Rocket Equation and an Isp of 110 seconds:

$$ \Delta v = I_\text{sp} \cdot g_0 \ln \frac {m_0} {m_f} $$

$\Delta v\ $ is the maximum change of velocity of the vehicle (with no external forces acting).
$m_0$ is the initial total mass, including propellant, also known as wet mass.
$m_f$ is the final total mass without propellant, also known as dry mass.

m0 and mf depend on your design. The formula doesn't take wind resistance and gravity into account so it's optimistic, but it'll give a ballpark figure for the final speed, and from there you can calculate the altitude it'll reach.

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  • $\begingroup$ So ignoring air resistance you're looking at >1000 m/s. That's very much supersonic. That in turn means you certainly can't ignore air resistance. $\endgroup$ – MSalters Jun 15 '18 at 14:14
  • $\begingroup$ what mass fraction did you use? Sugar rockets don't usually get a 95% mass fraction... $\endgroup$ – Hobbes Jun 15 '18 at 14:15
  • $\begingroup$ ln(m0/mf)=1, per engineering standards. Also Isp=100 and g=10. Now ln(m0/mf) means m0 = e* mf, which means the fuel fraction is somewhere near 2/3. With 2 digits precision, it drops to 60% fuel fraction. $\endgroup$ – MSalters Jun 15 '18 at 14:23
  • $\begingroup$ I've just done a simulation in OpenRocket, for an average model engine fitting in 15 by 3 cm thick cardboard casing it gave me 0.94M speed, 900 m apogee. Something is wrong here, isn't it? $\endgroup$ – olegst Jun 15 '18 at 14:52
  • $\begingroup$ Drag will be more of a factor in these small rockets than in large ones due to the square/cube problem (lots of surface area compared to internal volume/thrust). $\endgroup$ – Hobbes Jun 15 '18 at 15:00

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