Open Rocket Math

[Lt72884]

Look - post number 4.

But the above documentation link, which is in the Help:About screen of OpenRocket is the same paper.

My wife and sons aren’t very good at online searches, either.

well all be damned. I take accountability for that one. I even replied to your post. I completely forgot you even said that. I work at a drug and alcohol rehab center and had to restrain some boys on sunday so i had completely spaced your reply off. that is my bad. Ill edit my post right now

 

[Charles_McG]

well all be damned. I take accountability for that one. I even replied to your post. I completely forgot you even said that. I work at a drug and alcohol rehab center and had to restrain some boys on sunday so i had completely spaced your reply off. that is my bad. Ill edit my post right now

You might think that crow goes better with honey - but don't put it on too thick.

Also, MEMO TO YOU: As a new forum member, you might not have noticed that TRF has a rather short edit window. Remember that once you post it, TRF owns it. They have policies to protect it.

 

[Buckeye]

... apogee, velocity, etc. ...

The flight trajectory is simply governed by F = ma. Solve for a, then integrate for velocity, and integrate again for displacement. Other necessary parameters like air density and drag are functions of altitude and velocity, so everything is coupled. This has to be solved numerically. 4th Order Runge Kutta is best for these ODEs.

 

[Alan15578]

Lol, i feel you on that. Though i did like my numerical methods class.

Ok, ill look at your hand book as well as search for RASP-93 on google and see what comes up. One thing i have had for a while is this PDF

http://www.rocketmime.com/rockets/RocketEquations.pdf
and i have this link as well

https://en.wikiversity.org/wiki/PlanetPhysics/Single_Stage_Rocket_Burnout_Height

Ah, the joy of closed form solutions. Your first reference looks like the basic Malewicki equations, e.g. Estes TR10 and CenturiTIR-100, and the second is drag free; good luck with that one.

I like the skillful use of extended Malewicki equations. I would like to refer you to my old R&D report, but it has been lost for decades. "Topics inAdvanced Model Rocketry" is probably the best reference for derivation of the Malewicki equations, and alot of other useful stuff. You should be able to get a copy through you university library via Inter-library Loan. I could lend you one of my copies, but they ar selling for over $500, and I do not even know you name, address, school, next of kin, etc. When you go through the derivation yourself, be sure to get the complete solution to handle the cases of low thrust, and when the rocket is descending. The Malewicki equations are for constant density, which is not good for flying to 10000 Ft. You should look for papers written by Tom Keuchler in the NAR Tech Reviews in the NARTS section of the NAR website. He has a closed form solution for an exponential atmosphere that may be of some interest to you.

I can't say exactly what models are used by OpenRocket, however, the main reference should be the US Air Force Stability and Control Data Compendium (DATCOM). If it is not in your Aero. E. library you might want to consider transferring before it is too late. Of course OpenRocket should also have some supplemental modeling demanded by sport modellers.

 

[cls]

"Advanced Topics in Model Rocketry" is probably the best reference for derivation of the Malewicki equations, and alot of other useful stuff. You should be able to get a copy through you university library via Inter-library Loan. I could lend you one of my copies, but they ar selling for over $500,

MIL-HDBK 762 has all of that, flthe PDF is free

Here's one link

http://everyspec.com/MIL-HDBK/MIL-HDBK-0700-0799/MIL_HDBK_762_1855/

 

[peeler5150]

ok, that makes sense. i just did a runge-kutta assignment in excell a couple of weeks ago. very interesting math

Looking for opinions on the attachment.
Help, please, with Barrowman’s equations.
On page 28, looking at the vertical center of pressure on a fin, how does he take into account for the shape of the fin?

DeltaXf is found by

Given a for instance of m = 2, A(root) = 6 and B (Tip)=4. Clipped delta or trapezoid design.

M(A+2B) (A*B)

------------ + ( 1/6 x -----------)
3(A+B) (A+B)

(That doesn't turn out very good here.)
If the wider part of the fin is at the top this makes sense. Clipped delta fin shape.
If the wider part of the fin is at the bottom it doesn’t.
If it is a trapezoid shape the vertical center of pressure would seem to be half-way down A.
Width does not matter on the center of pressure on the fin, but doesn’t width matter on the center of pressure over the whole rocket?
Open Rocket and Rocksim must do something similar but how they do it is hidden from the user.
I suggest we change the formula depending on the shape of the fin.
For square, rectangle or ellipse we just use 1/2 of A. A simple vertical center.
If we use 1/3 instead of 1/6 we get better answers for a clipped delta fin, wide part at the bottom.
M=0, B=6 the center of pressure is 3” from the top. Proper trapezoid or rectangle shape fin.
M=1, B=5 the center of pressure is 3.24” from the top.
M=2, B=4 the center of pressure is 3.46” from the top.
M=3, B=3 the center of pressure is 3.67” from the top.
M=4, B=2 the center of pressure is 3.84” from the top.
M=5, B=1 the center of pressure is 3.90” from the top.
M=6, B=0 the center of pressure is 4.0” from the top. A proper delta shape fin.

This does not work for a swept fin or an off-center trapezoid. The point is we need to use something instead of 1/6.

 

[smstachwick]

Looking for opinions on the attachment.
Help, please, with Barrowman’s equations.
On page 28, looking at the vertical center of pressure on a fin, how does he take into account for the shape of the fin?

DeltaXf is found by

Given a for instance of m = 2, A(root) = 6 and B (Tip)=4. Clipped delta or trapezoid design.

M(A+2B) (A*B)

------------ + ( 1/6 x -----------)
3(A+B) (A+B)

(That doesn't turn out very good here.)
If the wider part of the fin is at the top this makes sense. Clipped delta fin shape.
If the wider part of the fin is at the bottom it doesn’t.
If it is a trapezoid shape the vertical center of pressure would seem to be half-way down A.
Width does not matter on the center of pressure on the fin, but doesn’t width matter on the center of pressure over the whole rocket?
Open Rocket and Rocksim must do something similar but how they do it is hidden from the user.
I suggest we change the formula depending on the shape of the fin.
For square, rectangle or ellipse we just use 1/2 of A. A simple vertical center.
If we use 1/3 instead of 1/6 we get better answers for a clipped delta fin, wide part at the bottom.
M=0, B=6 the center of pressure is 3” from the top. Proper trapezoid or rectangle shape fin.
M=1, B=5 the center of pressure is 3.24” from the top.
M=2, B=4 the center of pressure is 3.46” from the top.
M=3, B=3 the center of pressure is 3.67” from the top.
M=4, B=2 the center of pressure is 3.84” from the top.
M=5, B=1 the center of pressure is 3.90” from the top.
M=6, B=0 the center of pressure is 4.0” from the top. A proper delta shape fin.

This does not work for a swept fin or an off-center trapezoid. The point is we need to use something instead of 1/6.

So…I’m not really following the math presented this way, so I went right to the diagrams and pieced it together visually. I’ll try to explain here.





My guess is that the shape is accounted for through the interactions between XR, CT, CR, LF, and S. Positioning on the body tube is reflected by the XB value. For a forward-swept fin, I’d assume that XR would be negative. CT would be 0 for a delta or other triangular fin and a positive value for anything else.

For these trapezoidal fins, you can find LF by putting the fin on a grid, defining a right triangle with the hypotenuse extending from the middle of the root and tip chords, then calculating its length with the Pythagorean theorem. If you’re working with a triangular fin, do this with the middle of the root chord and the very tip of the fin. I can provide my own notes for this method if you wish, but it doesn’t require anything more difficult than comparatively simple algebra.

Are there shapes that this would not work for? I’d assume so, definitely if you have more than 4 points. But that’s one of the documented limitations of this method. Those shapes will have to be simplified.

Not sure how helpful that will be, but it makes sense to me. Definitely post any further questions, we have some real math sharks here.

 

[rockethead287]

So…I’m not really following the math presented this way, so I went right to the diagrams and pieced it together visually. I’ll try to explain here.

View attachment 542682



My guess is that the shape is accounted for through the interactions between XR, CT, CR, LF, and S. Positioning on the body tube is reflected by the XB value. For a forward-swept fin, I’d assume that XR would be negative. CT would be 0 for a delta or other triangular fin and a positive value for anything else.

For these trapezoidal fins, you can find LF by putting the fin on a grid, defining a right triangle with the hypotenuse extending from the middle of the root and tip chords, then calculating its length with the Pythagorean theorem. If you’re working with a triangular fin, do this with the middle of the root chord and the very tip of the fin. I can provide my own notes for this method if you wish, but it doesn’t require anything more difficult than comparatively simple algebra.

Are there shapes that this would not work for? I’d assume so, definitely if you have more than 4 points. But that’s one of the documented limitations of this method. Those shapes will have to be simplified.

Not sure how helpful that will be, but it makes sense to me. Definitely post any further questions, we have some real math sharks here.

damn my head get dizzy just by reading this ..

 

[smstachwick]

damn my head get dizzy just by reading this ..

It’s not quite as hairy as it looks. The operations are simple, there are just a lot of them. It’s also important to do them in the right order and pay attention to your groupings

You can find techniques online that allow you to pull square roots by hand, so with a little practice it’s entirely possible to work through the whole thing by hand in about 20 minutes.