Questions and posers.

The Rocketry Forum

Help Support The Rocketry Forum:

stantonjtroy

Well-Known Member
Joined
Jan 28, 2009
Messages
1,114
Reaction score
116
Location
Glen Burnie MD
Two questions for Hi Power push. Is there a rule of thumb for a minium thrust to weight ratio. I know the higher the better but I have a specific envelope I need to work in and a minium is needed.
Second, is the ratio figured using total motor impulse or average thrust? Any help is welcome.

Troy
 

troj

Wielder Of the Skillet Of Harsh Discipline, Potent
Joined
Jan 19, 2009
Messages
14,443
Reaction score
358
Two questions for Hi Power push. Is there a rule of thumb for a minium thrust to weight ratio. I know the higher the better but I have a specific envelope I need to work in and a minium is needed.
Second, is the ratio figured using total motor impulse or average thrust? Any help is welcome.
5:1 is the general guideline, as that will work in most weather conditions that are within the safety code for both NAR & TRA. (I'm assuming UKRA, and CAR, as well, although I don't know their codes as well)

Initial thrust is more important, as it's all about getting the rocket up to the velocity necessary for stable flight before it clears the rod/rail/tower.

This is why motors with a high initial spike can be useful for heavier rockets -- it gives you that initial push to get the rocket going fast enough to be stable, then drops back, to give a longer burn.

-Kevin
 

powderburner

Well-Known Member
Joined
Jan 18, 2009
Messages
7,399
Reaction score
6
I would have given you the same answer (5:1 initial thrust-to-weight ratio) that troj just did.

Have you run any sims? The rule of thumb for velocity at the end of the launcher (rod, rail, whatever) says your rocket needs to be moving around 40 fps for the fins to be aerodynamically effective. If you have a small stability margin you may want even higher speeds.

Remember you can also use a longer launch rail.
 

atticus

Old and in the way
Joined
Jan 18, 2009
Messages
302
Reaction score
0
From the NAR report 'Launching safely into the 21st century'

https://www.nar.org/pdf/launchsafe.pdf

section 7.4.3 states:

"Best practice would indicate that rockets to be guided by launch rods, rails, or towers until they have
attained a forward velocity of at least 4 times the velocity at which the wind is blowing or gusting at
the launch site (e.g. 36 m/sec if the maximum gust is 20 mph). Introducing a “4 x wind”
requirement will require many models to use 8-foot or 10-foot rails or towers in significant
crosswinds for all but the lightest rockets with high thrust/weight ratios."

Of course, this will require an accurate simulation to predict the rail/rod exit velocity.
 

stantonjtroy

Well-Known Member
Joined
Jan 28, 2009
Messages
1,114
Reaction score
116
Location
Glen Burnie MD
Thanks for the replys. I should have run a topic search first; I found the thread in the HP section. Thanks again.
 

hardinlw

Well-Known Member
Joined
Jan 19, 2009
Messages
372
Reaction score
1
I'll give you the math behind these rules of thumb.

The acceleration of the rocket is due to the unbalanced force acting on it. Thrust acts upward, and weight acts down, so it's T-W. (Drag is negligible at low speeds.) The acceleration in g's for a given T/W is T/W - 1, so the acceleration in ft/sec is

a = 32.2 * (T/W - 1)

The distance an object travels at constant acceleration is

d = 1/2 a t^2

That distance is the length of the launch rod, so we can solve for the time, t, to hit the end of the launch rod as

t = sqrt(2 L / a)

The velocity at time = t is given by

v = a t

So we can get the velocity at the end of the launch rod for any length of rod without a simulation. A thrust-to-weight of 5 gives 39 ft/sec at the end of a six foot launch rod (or rail).

Now, where did that four-to-one ratio come from? I turns out that all thin airfoils stall between 12 and 15 degrees angle of attack. The angle of attack that the fins see while the rocket is on the launch rod is the inverse tangent of the wind speed divided by the speed of the rocket. Four-to-one is roughly 15 degrees. Basically, what is being said is that the speed you need to have when you hit the end of the launch rod depends on highest wind speed you might encounter coming off the launch rod. If the rocket leaves the rod with less than 4x the wind speed, it is likely that the fins will be stalled and the rocket will be unstable. This is not a desirable condition. ;-)
 

Graham Orr

Well-Known Member
Joined
Mar 14, 2009
Messages
297
Reaction score
0
On the old TRF I posted the results of a few multi-degree of freedom simulations I did in Matlab for weathercocking prediction based upon wind conditions and rocket caliber. The fact of the matter is that lower F/M flights can be done safely in very still conditions. I wish I still had the figure to show this (I don't have the time to redo it right now), but if anyone has Matlab and would like to run the code, it is posted below:

////////////////////////////

function void = weatherCock(Vw,F,B,M,Xcg,L,D,c)
N=5000;
T=.01;
rho=1.2; %air density kg/m^3
D=.01; %rocket diameter
I = M*Xcg^3/(3*L)+M*(L-Xcg)^3/(3*L); % moment of inertia
As=L*D;
Ac=pi*(D/2)^2;

theta=zeros(N,1);
Vx=zeros(N,1);
Vy=zeros(N,1);
Sx=zeros(N,1);
Sy=zeros(N,1);
a=zeros(N,1);
a(1:B/T)=F/M;

clf
Vy(1)=10; %velocity when leaves rail (m/s)
n=N;
for i=2:N
Vy(i)=Vy(i-1)+a(i)*T*cos(theta(i-1))-9.8*T-T*(As*sin(theta(i-1))+Ac*cos(theta(i-1)))*(Vy(i-1)^2+Vx(i-1)^2)*rho*.75/(2*M);
Vx(i)=Vx(i-1)+a(i)*T*sin(theta(i-1));
dthetadt = .75*rho*As*c*D*((Vw+Vx(i)*cos(theta(i-1)))^2-(Vy(i)*sin(theta(i-1)))^2)/(2*I);
theta(i)=theta(i-1)+dthetadt*T;
Sx(i)=Sx(i-1)+Vx(i)*T;
Sy(i)=Sy(i-1)+Vy(i)*T;
if Vy(i)<0
n=i;
break
end
end
subplot(2,1,1)
plot(0:T:T*(n-1),180/pi*theta(1:n))
subplot(2,1,2)
plot(Sx(1:n),Sy(1:n))
 

MKP

Well-Known Member
Joined
Jan 18, 2009
Messages
200
Reaction score
0
I think my head just exploded.
 

cjl

Well-Known Member
Joined
Jan 18, 2009
Messages
12,549
Reaction score
11
Now, where did that four-to-one ratio come from? I turns out that all thin airfoils stall between 12 and 15 degrees angle of attack. The angle of attack that the fins see while the rocket is on the launch rod is the inverse tangent of the wind speed divided by the speed of the rocket. Four-to-one is roughly 15 degrees. Basically, what is being said is that the speed you need to have when you hit the end of the launch rod depends on highest wind speed you might encounter coming off the launch rod. If the rocket leaves the rod with less than 4x the wind speed, it is likely that the fins will be stalled and the rocket will be unstable. This is not a desirable condition. ;-)
The problem with this statement is that not all airfoils stall at that angle of attack. Specifically, a highly swept delta can go well past 20 degrees, and in several cases can exceed 30 degrees without stalling. Conveniently enough, this also happens to be the shape of the fins on many rockets.

For an example of this, look at the CL vs alpha plots for the SR-71, shown here: https://www.sr-71.org/blackbird/manual/6/6-5.php

Note that the slope of the curve is still positive at 25 degrees. There are still good reasons that you wouldn't want to be launching in conditions when the wind angle at rod exit is 25 degrees (such as the severe weathercocking that would occur), but the fins are not necessarily going to stall, due to the high sweep angle and short span. Of course, this only applies to rockets with a high sweep angle and low aspect ratio, but I think I can safely say that this is true of many HPR rockets.
 
Last edited:
Top