mardlamock

Hello everyone, i ve been trying to get the max height for my rocket taking drag into account (irl), variations in the angle of the rocket with respect to the airflow and a whole bunch of other things and i ended up with several differential equations. Im not going to explain much of what the equations look like themselves but just trying to get a general idea. The way i ve been solving them so far is by simply using excel spreadsheets where for a small time interval i treat all of the functions of time as a constant and then do an approximation as if the acceleration was constant. This could sort of be expressed like so, from t0 to t1:a=r(t1)-n(x'(t0))*(m(x(t0)). Then i get the displacement in t1: 1/2*(r(t1)-n(x'(t1))*(m(x(t0)))*t1^2, and velocity: t1*(r(t1)-n(x'(t1))*(m(x(t0))) From t1 to t2 acceleration would be: r(t2)-n(x'(t1))*m(x(t1)). As you can see, to solve for t=n i am using previous values from t=n-k, now would there be a way of expressing this whole thing with maybe sigma notation and then take the limit as t1-t2 approaches zero and the ammount of small t s i need approaches infinity. Sorry if you dont understand anything of what im saying, im not all that good at explaining myself and do not know all that much about these topics. Thank you very much!

The universe is everything that exists, therefore if it is outside the universe then it doesnt even exist.

No one gonna answer?

Hello, I ve been doing a bit more on those functions that express the maximum horizontal distance a rocket could travel without considering the sphereness of the earth and the atomsphere. Basically instead of looking for the rocket to stay at a same height through the time the fuel lasts I want to accelerate in a 45 degree angle as much as I can. In order to do so the rocket needs shed a bit of mass if the vertical component of its force at a 45 degree angle is smaller than the mass of the rocket times 9.81. So I ended up with this: http://latex.codecogs.com/gif.latex?%5CTheta%20%28t%29%3D%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20sin%5E-1%28%28m-r*t%29*g/f%29%20%26%20if%5C%2C%20t%3C%28-f/sqrt%282%29+g*m%29/%28g*r%29%5C%5C45%5E%7B%5Ccirc%20%7D%20%26%20if%5C%2C%20t%3E%28-f/sqrt%282%29+g*m%29/%28g*r%29%20%5Cend%7Bmatrix%7D%5Cright Now, the horizontal acceleration as a function of t is: http://latex.codecogs.com/gif.latex?a_%7Bh%7D%28t%29%3D%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%5Cfrac%7Bcos%28sin%5E-1%28%28m-r*t%29*g/f%29%29%29*f%7D%7B%28m-r*t%29%7D%20%26%20if%20%5C%2C%20t%3C%28-f/sqrt%282%29+g*m%29/%28g*r%29%5C%5C%20%5Cfrac%7B1/sqrt%282%29*f%7D%7B%28m-r*t%29%7D%26%20if%20%5C%2C%20t%3E%28-f/sqrt%282%29+g*m%29/%28g*r%29%20%5Cend%7Bmatrix%7D Where (-f/sqrt(2)+g*m)/(g*r) is equal to the time it takes for the rocket s force at 45 degrees to equal the weight of the rocket. Now, v(t) would be looking something like this right?:http://latex.codecogs.com/gif.latex?v_%7Bh%7D%28t%29%3D%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%5Cint_%7B0%7D%5E%7Bt1%7D%20%5Cfrac%7Bcos%28sin%5E-1%28%28m-r*t%29*g/f%29%29%29*f%7D%7B%28m-r*t%29%7D%20%26%20if%20%5C%2C%20t%5Cleq%20%28-f/sqrt%282%29+g*m%29/%28g*r%29%20%5C%5C%20%5Cint_%7B0%7D%5E%7B%28-f/sqrt%282%29+%28g*m%29/%28g*r%29%20%7D%20%5Cfrac%7Bcos%28sin%5E-1%28%28m-r*t%29*g/f%29%29%29*f%7D%7B%28m-r*t%29%7D+%5Cint_%7B%28-f/sqrt%282%29+g*m%29/%28g*r%29%20%7D%5E%7Bt2%7D%5Cfrac%7B1/sqrt%282%29*f%7D%7Bm-r*t%7D%26%20if%20%5C%2C%20t%20%3E%20%28-f/sqrt%282%29+g*m%29/%28g*r%29%20%5Cend%7Bmatrix%7D%5Cright. Now, wolfram can integrate that and actually has, the problem is that it is a ridiculously long function and im not even describing the parabolic motion it follows once the fuel has been depleted. Would there be a simpler way to do it? If there is anything I wrote in a weird way please tell me so I can explain it. Thanks!

I didint expect this to be so darn complicated, I somewhat understand it. It was just a question that came up to my mind when my math s teacher was explaining something about planes intersecting lol. A guy on IRC managed to explain it quite well to me. Thanks for the link man!

By solutions i mean points where f is undefined

I dont mean a function that has an equal amount of solutions and impossible solutions, that is easy. Ie sqrt x. its got infinite solutions on the positive side and infinite impossible solutions on the negative side (disregarding imaginary numbers). I kind of tried to explain it here http://imgur.com/L8NeyQr

Hello everyone, I was just thinking about the possible solutions to a certain function with or without complex numbers and I just wanted to know if there could be any way to express a function that between any two values has as many solutions as impossible solutions. And that if for example i was to take another two random values within the original ones i would also have equal number of solutions and impossible solutions (i really dont know what is the opossite of solution in math). What do you think, is it possible to have something like that?

so it would be fair to say that a(t) sqrt(1-(m0-mr*t)*9.81/f)^2)*f/(m0-mr*t)

The horizontal acceleration with respect to t is =cos(sin^-1((m0-mr*t)*9.81/f)))*f/(m0-mr*t). First i need to find the indefinite integral and then the definite integral right? So i can get the f out of the way by factoring it out, then i have no idea what to do. How can i do integration by substitution here? Im sorry for being so dumb, I never had anyone explain calculus irl for me so all i know is from the interwebz and it is not much. Thanks!

I just confirmed that it is right by looking at the graph slowly and replacing by t=0 so now all i have to do is integrate that (UGH)

That makes total sense tavert. I dont really know much about trig functions, seriously i just dont understand them. I can do some operations with trig identities and such and sort of understand where they come from but how the trig functions are graphed and how radians work just eludes me. Thanks, i will look into it when. But is my reasoning right, If it is right i know i can then integrate it just fine.

I was asking if the function i proposed was right:theta(t)=sin^-1((m0-mr*t)*9.81/f)). It gives me some weird numbers when i graph it, ie, the angle is like 1 degree which is kind of odd.

I expressed myself wrong, what im trying to find is the angle of the rocket as a function of time. I want the rocket to fly like a plane, that is exactly what i meant by saying that i wanted 0 vertical acceleration. You can think of it as if it was a missile launched parallel to the ground and i want it to remain at the exact same height through the time the engines are activated. By solving for this angle then i can calculate what the horizontal part of the force vector would be with respect to time, if I do the definite integral for that formula then i can calculate the maximum horizontal speed for my rocket and also the maximum horizontal displacement for the time the engines run. That is what i am looking for, the maximum horizontal displacement without atmosphere for the time the engines run.

Hello everyone, as you may have already noticed im building a small rocket and I want to know what would be the furthest distance my rocket could go only hoirzontally, without any vertical acceleration. I started off by saying that the force exerts a certain force f which is a constant, the mass of the rocket as a function of t is m0-mr*t where m0 is the initial mass, mr is the mass flow rate, and t is time). So what I thought was that the best idea would be to say that the vertical component of the force f would at all times have to be equal to the mass of the rocket times g. This could be written as: sin (theta)= m*g/f. as i said before was m0-mr*t, so: sin(theta)= (m0-mr*t)*g/f. In order to find the angle I apply the inverse sine function, this gives us: theta(t)=sin^-1((m0-mr*t)*9.81/f)). Is this correct, because when i plug it into a wolfram to graph it it gives me some weird angle and i dont really understand what im doing wrong. I know that the optimal angle would be 45 degrees but that also means the rocket would displace upwards which i am not trying to analyze here (i want to think of my rocket as if it was to be used like an rpg or something like that). Thanks!

You can get kno3 saying it is for aquaponics or fish related stuff (i bought it from a guy that does that), but i dont know how it is in germany

Im building my own rocket in argentina without any permits whatsoever, no one has told me anything and the final rocket i plan to build will weigh 4kg. You can easily do static tests in a city, just get to a rooftop where no one cares. As far as the fuel, you can use potassium nitrate and sugar aka Rocket candy, and go prett damn high. I got 2kg of KNO3 (you can also get it from stump remover) for only 15 bucks in my country, you can make a decent concrete nozzle. Let me send you a few vids. http://www.youtube.com/watch?v=KWeFe5REzhs (notice the buildings in the background), http://www.youtube.com/watch?v=qnNXqx6GVho.

Oh woaw, thanks man! I hadnt noticed the wrong sign nor thought about where i wanted the integral from. Thank you very much! Do you know what i should read to derive the equations for eliptical orbits and such? I learnt some calculus on my own (two years until i learn it at school), but im not smart enough to find out the laws of motion for an object in a eliptical or hyperbolic orbit. Thanks!

Hello everyone, I would like some help trying to derive the rocke equation. I tried to do it my way, so sorry for the mistakes. Ok, so I started off by saying that f=ma, then a=f/m, so the acceleration at any given time is = F/(m0-mr*t) , m0 being the initial mass, mr being the mass flow rate. Velocity is the integral of acceleration therefore if i integrate f/(m0-mr*t) I should get the velocity as a function of time. I did the integration by substitution part and i finally got that v(t)=(f/mr)*ln(m0-mr*t), which is wrong because the rocket equation says it is (f/mr)*ln(m0/mr*t). What did i do wrong?

Governments and monopolies. The selfishness of human being and lack of critical thought in certain societies. The fear of the unknown, the fear of change. Those are the biggest problem humanity faces.

http://imgur.com/jGwCShA, on their website they have this, so this i assume means they re fully capable of doing this and cannot deny our request. http://www.spacex.com/falcon9

I think this is a must, and also something like KAS and maybe some FAR in it as well

Damned Robotics

Thanks guys i now understand it!

Oh i see! Thanks, now i get it.