Forums › General Discussion › Club Membership Information › membership ?
- This topic has 11 replies, 9 voices, and was last updated 15 years, 6 months ago by
Warren B. Musselman.
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AuthorPosts
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May 11, 2010 at 10:48 pm #41008
Brad Walker
ParticipantOne more thing about membership. If I join NCR, is the membership based on a calendar year or does it run for a complete year from time of signing up.
Thanks.
May 12, 2010 at 2:17 pm #52687Ken Plattner
ParticipantIt runs for a complete year from when you join.
Ken
May 13, 2010 at 11:18 pm #52688SCOTT EVANS
Hey…. WHats that formula up there?
Am I supose to know that? 😕May 13, 2010 at 11:57 pm #52689Cineroc
Hey…. WHats that formula up there?
Am I supose to know that?only if you want to claim you’re a rocket scientist…
Roger
May 14, 2010 at 1:15 am #52690Mitch Smith
Unless my brain isn’t working correctly tonight…. The formula is for the acceleration of a rocket taking into account the rate at which the fuel is consumed along with the effects of atmospheric presure. Let us know how far I am off Ken.
May 14, 2010 at 4:55 am #52691Chris LaPanse
I don’t see an atmospheric pressure term. It’s just a basic form of the rocket equation.
(For that matter, it doesn’t even appear to have a gravity or drag term)
May 14, 2010 at 2:26 pm #52692Ken Plattner
ParticipantRight on, Chris. It’s the Ideal Rocket Equation – derived from the conservation of momentum of the rocket and fuel by Russian scientist Konstantin Tsiolkovsky in 1903 or so.
Simplistically, P=mV, so you can say that mv of the rocket = mv of the fuel. For the rocket, m is essentially constant, but the velocity changes. For the propellant, the mass changes, but the exhause velocity of the burning fuel is constant.
So,
m delta V = V delta m
mdv = vdm
Do the integration, etc…
May 14, 2010 at 2:31 pm #52693SCOTT EVANS
Right on, Chris. It’s the Ideal Rocket Equation – derived from the conservation of momentum of the rocket and fuel by Russian scientist Konstantin Tsiolkovsky in 1903 or so.
Simplistically, P=mV, so you can say that mv of the rocket = mv of the fuel. For the rocket, m is essentially constant, but the velocity changes. For the propellant, the mass changes, but the exhause velocity of the burning fuel is constant.
So,
m delta V = V delta m
mdv = vdm
Do the integration, etc…
duh……….Hey Joe, what size motor should I put in this one? ……just give me the biggest one that will fit! 😉
May 14, 2010 at 9:36 pm #52694Bruce R. Schaefer
Scott E, don’t question a guy with a degree in advanced mathematics. 🙄 He will differentially equate your derivative in no time flat, not to mention integrate your limits!
May 15, 2010 at 7:00 am #52695Ed Dawson
Thanks Scott. I have not laughed so hard in a while.
Hey Joe, what size motor should I put in this one? ……just give me the biggest one that will fit!
That’s what it all about, eh?
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