# Rocket Design Help

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#### arobie

##### Guest
I would greatly appreciate any help I get. I'm trying to learn this stuff, and the best way for me to learn is to actually try, so I did.<br /><br />Earlier today, I started to figure numbers for a suborbital hop of an apogee of about 316 km. I started with the gravity losses. I figured it would loose 1.98 km/s in grav loss using this formula: (Don't worry about all these numbers. They are mainly for my benefit...to review my work as I post it...and to have a place to see it all organized. Although if anyone wants to look it over, you're welcome too. <img src="/images/icons/smile.gif" />)<br /><br />dVgrav = g,eff * t,b * sin(phi)<br />g,eff = local gravity on way up<br />g,eff = (g + g,end of flight) / 2<br />g = 9.807<br />g,end of flight = Earth's GM/ Rb<br />Earth's GM = 398600.44 km^3 / s^2 <br />Rb = Re + altitude<br />Earth's Re = 6371 km <br />Altitude = 316 km<br />Rb = 6687 km<br />g, end of flight = 398600.44 / 6687 = 0.008914051<br />--g, eff = (9.807 + 8.914051) / 2 = 9.3605255<br /><br />t,b = engine burn time<br />t,b = sqrt ( ( 2 * y ) / a)<br />y = 316000 m<br />a = 14.11 m / s^2<br />t,b = sqrt ( (2 * 316000) / 14.11)<br />--t,b = 211.6 seconds<br /><br />Sin(phi)<br />phi = flight angle<br />Phi = 90 degrees<br />--Sin(90) = 1<br /><br />dV = 9.3605255 * 211.6 = 1980.68 m/s = 1.98 km/s<br /><br />I have two questions. First, what is and how do I find 'a'? I already feel foolish for that question. I'm just learning this, and looking over my work on paper, I have no idea how I got 'a'. All that I started out knowing for that portion of the equation was y (altitude m).<br /><br />Second, what is the next step to figuring out how much total dV I need to get to 316 km altitude? <br /><br />Total dV = dV + gravity losses + drag losses<br />I quess I have one more question, is the above correct?

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#### spacester

##### Guest
Hi Arobie,<br /><br />The 'a' is the acceleration of your rocket. You should not feel at all foolish - what you have come up against is a hint of the difficulty of launcher design. <br /><br />As the rocketship designer, it's up to you to decide how fast the rocket accelerates. F = m*a is simple enough in most cases but for a rocketship m = mass is constantly changing because you're throwing stuff out the back end.<br /><br />So 'a' is not constant but<br />t,b = sqrt ( ( 2 * y ) / a)<br />is from the basic equation of motion, usually stated as<br />y = 1/2 * a * t^2<br />and it assumes a is constant.<br /><br />What you have to do is decide how fast it will jump off the launch pad (initial a). This value can be found by deciding how much thrust and how much the total mass of the rocket will be (GLOW - Gross Lift Off Weight).<br /><br />Now if you were to do this exactly, you would then use the rocket equation to find the value of acceleration over the time of the burn, and then use calculus to find your actual gravity losses. Let's not go there just yet. <br /><br />What you can do, and did do with your example, is use an average acceleration. You chose 14.11 m/s^2. If you happened to choose "one g" as your liftoff acceleration, then the acceleration at the end of the burn would be the number that averages with "one g" to get 14.11<br /><br />(9.807 + x) / 2 = 14.11<br />x = 18.413<br /><br />So you started your launch at "one g" = 9.807 m/s<br />and ended the burn at 18.41 / 9.807 = "1.88 g"<br /><br />OK have I confused you yet?<br /><br />This is an excellent example of an "iterative design". You can't figure out everything at once. You have to pick some values, work through the equations to find other values, plug those new numbers in to compare with the values you chose in the first place, adjust them, crank the equations again and so on and so on. It's also known as the "trial and error" method, but I've always disliked that term.<br /><br />As to your second question, I'm at work an <div class="Discussion_UserSignature"> </div>

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#### najab

##### Guest
While we're playing with numbers, I'm trying to write a Java applet that will let you play with orbits. The idea is that you put in the initial altitude, flight path angle and velocity and it'll draw the resulting orbit. I can work out the apoapsis, periapsis and eccentricity no problems, but when it comes to working out the position of the body - no dice.<p>Kepler's equations confused me when I did them in Physics at Uni and they still confuse me now! Looking at this site helped a lot, but I'm stuck with equations 1.29 - 1.30 - I'm sure that he's using the same term for two different quantities somewhere here, but I haven't been able to figure out which.<p>Hints?</p></p>

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#### spacester

##### Guest
Yeah, orbits are one thing but figuring out where a body is along that orbit takes it to a new level.<br /><br />Excellent reference, been there many times. I see nothing wrong with 1.29 and 1.30, but have never used them.<br /><br />I use 1.31 and 1.32 (or equations derived from them) for most everything and use True Anomaly to keep track of where a body is on its orbit. I have subroutines that take care of the nasty calcs so I'd have to look at those.<br /><br />Gotta run, more later <div class="Discussion_UserSignature"> </div>

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#### arobie

##### Guest
<font color="yellow">"Now if you were to do this exactly, you would then use the rocket equation to find the value of acceleration over the time of the burn, and then use calculus to find your actual gravity losses. Let's not go there just yet. "</font><br /><br />Ok, I'll be back in two years after taking calculus. I'm taking Advanced Math next year and Calculus my senior year. <br /><br /><font color="yellow">"OK have I confused you yet?</font><br /><br />Actually....no. <img src="/images/icons/smile.gif" /><br /><br />Doing it simply, without calculus:<br /><br />'A' is average acceleration. It's decided by me. I also choose my liftoff acceleration, and it averaged with the acceleration at the end of the burn equals the average acceleration...'a'.<br /><br />I do plan to have zero velocity at 316 km. There will be no horizontal velocity at apogee.

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#### mrmorris

##### Guest
<font color="yellow">"There will be no horizontal velocity at apogee."</font><br /><br />Hmmm -- nitpicking, but would be more accurate to say that there will be 'negligible' horizontal velocity.

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#### arobie

##### Guest
mrmorris,<br /><br />True, true....I almost said something in my post about if the pilot was good enough...superhuman...that the horizontal velocity would be none. I didn't put it though...this is only hypothetical number games.<br /><br />Now if I was flying like I do in Orbiter, well then my horizontal velocity might end up a couple of ten's of meters per second either way. That's not alot, I've gotten pretty good at it. <img src="/images/icons/smile.gif" /> <br /><br />Suborbital hops are my favorite flights...although I did manage to dock with the ISS just two nights ago. First time ever. <img src="/images/icons/smile.gif" /> <br /><br />The funny thing is...after spending some three hours rendevous and docking, I let it run all night and day the next day. When I came home from school, I went to undock. As soon as I undocked, thats when all the warning lights went off and my ship started beeping urgently at me. My main computer (on the ship) told me my crew was dead. It turned out I had forgotten to close the damn airlock hatches before undocking. DOH! <br /><br />Things learned from that mission: <br />-The crew does not like open space.<br />-Undocking before closing the airlocks exposes them to open space.<br />-Always check to make sure the airlocks are closed before undocking!

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#### spacester

##### Guest
The formula to find the equivalent deltaV corresponding to the change in potential energy going from a planet's surface at radius R to radius (R + elevation) is:<br /><br />dV = sqrt((G * M)*((1 / R) - (1 / (R + elevation))) <div class="Discussion_UserSignature"> </div>

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#### arobie

##### Guest
dV = sqrt((G * M)*(1 / R) - (1 / (R + elevation)))<br /><br />What is G, and what is M?<br /><br />I would assume 'M' would be mass, but I thought 'm' stood for mass.

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#### najab

##### Guest
G is the Universal Gravitational Constant (6.67300 × 10<sup>-11</sup> m<sup>3</sup> kg<sup>-1</sup> s<sup>-2</sup>), and M is the mass of the earth.

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#### drwayne

##### Guest
G is, I believe, the gravitation constant, and M is the mass of the Earth. (I believe, but I am in the middle of something)<br /><br />Wayne <div class="Discussion_UserSignature"> <p>"1) Give no quarter; 2) Take no prisoners; 3) Sink everything."  Admiral Jackie Fisher</p> </div>

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#### drwayne

##### Guest
You beat me Naja (what else is new).<br /><br />Wayne <div class="Discussion_UserSignature"> <p>"1) Give no quarter; 2) Take no prisoners; 3) Sink everything."  Admiral Jackie Fisher</p> </div>

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#### arobie

##### Guest
Universal Gravitational Constant = (6.67300 × 10 ^ -11 m3 kg-1 s-2)<br /><br />Wow. The number is almost as big as it's title...and Universal Gravitational Constant is no short title.<br /><br />So I take 6.67300 × 10^-11 and plug it into the equation?

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Yup.

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#### drwayne

##### Guest
Eventually, you will run across a simple equation for the gravitational attraction between two masses:<br /><br />F = G * m1 * m2 / r^2<br /><br />where m1 and m2 are the masses of the point masses, and r is the distance between them...<br /><br />Trivia, I know...<br /><br />Wayne <div class="Discussion_UserSignature"> <p>"1) Give no quarter; 2) Take no prisoners; 3) Sink everything."  Admiral Jackie Fisher</p> </div>

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#### arobie

##### Guest
<font color="yellow">As to your second question, I'm at work and my reference on that is at home so later for that. Do you want to have zero velocity when you get to 316 km altitude?</font><br /><br />So what if I happened to want to have a horizontal velocity of 7.72 km/s instead of my simple 0 km/s? (Of course my flight angle will change from its steep 90 degrees to make this possible.)

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#### arobie

##### Guest
drwayne,<br /><br />Cool...nice to know trivia. <img src="/images/icons/smile.gif" />

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#### spacester

##### Guest
Um, is 7.72 km/s the orbital velocity for a circular orbit at 316 km altitude? If yes, then you find the orbital energy and forget about the potential energy of 316 km elevation. The PE is accounted for in the vis-viva equation, which is the basis for most all this stuff.<br /><br />dV to orbit =<br />Orbital energy<br />Plus Gravity losses<br />Plus Drag losses<br />Minus elevation advantage compared to sea level<br />Minus latitude advantage due to the Earth's spin<br /><br />A 316 km straight up flight is not an orbital flight so vis-viva does not apply, you use ballistics equations (Newton's equations of motion) <div class="Discussion_UserSignature"> </div>

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#### arobie

##### Guest
Yes, it is. I got it from this cool little website to find orbital velocity. <br /><br />Ok, thank you. I did a google on orbital energy and found what I needed. I'll learn that and then put it too work along with the rocket equation to study the logistics of reproping at different altitudes of Martian orbits. <img src="/images/icons/laugh.gif" />

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