Rod exit velocity?

Static stability of the model is a big factor. Mean Machines can take off slower than Fat Boys.

Jerry

I would venture a guess that wind speed plays a pivotable role.

shockie B)

Bob, I thought the minimum velocity was around 33 fps.

Time to upgrade to a coat hanger. ; )

Randy

I asked this question before and got some mixed answers - mostly relating to the wind because it definitely affects stability at the end of the rod. More wind means less stable at launch.

Short Technical Answer:

You must be certain that the actual air velocity (vector quantity - speed and direction) at launch rod exit is such that you do not stall the fins or cause the CP to drift farther forward than a margin to CG. This usually means that the angle of attack is less than 10 or 15 degrees (for stall), or that your stability is calculated based on the center of lateral area (CLA), ie., the cardboard cutout method. The CLA method takes into account the force generated by the tube, where the Barrowman type equations do not.

Rule of Thumb Answer:

I found a section on

that recommends 30 mph or about 44 fps.

Practical Answer:

Get a longer launch rod or use a higher average thrust motor. (This is the way I "solved" the problem of my Estes Phoenix. I moved to 1/4" lugs and launch off of the 5' rod of a Mantis. The Aerotech D15's and E18's also get it moving better than the Estes D12.)

By the way 30 fps might not be too slow, depending upon the rocket. You might however, want to defer that launch until the wind is very low.

Quick calc - 30 fps up rocket velocity, 6 fps wind velocity. sin AOA = V horizontal / V vertical = 6/30 = 0.2 AOA = arcsin (0.2) AOA = 11.5 degrees (marginal?) Airspeed = sqrt[(V horizontal)^2 + (V vertical)^2)] Airspeed = sqrt[(6^2) + (30^2)] = [36+900] Airspeed = sqrt[936] Airspeed = 30.6 fps (quick lesson on vector math, heh heh)

At 44fps (rule of thumb) 11.5 degrees AOA is at 8.8 fps. Anecdotal evidence will suggest that we can go much higher than 11.5 degrees, even in my own experience. Maybe 15 to 20 degrees is more suitable a rule, since 8.8 fps is only a 6 mph wind.

20 degrees AOA for a 30 fps launch rod exit is calculated like this: sin AOA = V horizontal / V vertical sin 20 = V horizontal / 30 0.342 = V horizontal / 30 V horizontal = 10.3 fps

and for 44 fps vertical = 15 fps or 10 mph horizontal.

Hmmmmmmmm.

I guess the real question is "how do you calculate the CP position at a given angle of attack?"

.....

Have a look at:

- Robert Galejs

snip

I think you are dead on Thomas.

My question is how is it that Homer didn't use a rod at all? He simply placed his AUK series on the ground, or did I miss something in the book?

Randy

I was reading over Rober Galejs paper and noticed that his angles were derived off of the tangent of the AOA and I used the sin. Roberts is correct, mine is not.

If your gonna do vector math, you need to know trig.

Redoing the calcs...

Quick calc - 30 fps up rocket velocity, 6 fps wind velocity. tan AOA = V horizontal / V vertical = 6/30 = 0.2 AOA = arctan (0.2) AOA = 11.3 degrees (was 11.5)

At 44fps (rule of thumb) 11.3 degrees AOA is at 8.8 fps. Anecdotal evidence will suggest that we can go much higher than 11.3 degrees, even in my own experience. Maybe 15 to 20 degrees is more suitable a rule, since 8.8 fps is only a 6 mph wind.

20 degrees AOA for a 30 fps launch rod exit is calculated like this: tan AOA = V horizontal / V vertical tan 20 = V horizontal / 30 0.364 = V horizontal / 30 V horizontal = 10.9 fps

and for 44 fps vertical = 16.0 fps or 10.9 mph horizontal. (was 15.0 and 10.0)