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Rule of Thumb

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Heisenberg

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Which is a better rule of thumb for selecting a motor for a specific rocket, the 5 to 1 thrust to weight ratio, or the minimum launch rail/rod velocity? It appears to me the 5 to 1 rule of thumb is much more frequently cited. I've only seen the minimum velocity cited on a few places namely info-central. org and thurstcurve.org but they both use a different velocity for the minimum.

Brian
 

cjl

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5:1 is usually used, because it is easy to figure out on the spot. It's quite a bit harder to calculate the velocity at which something would leave the rod/rail on the spot if you need a quick estimation.
 

ben_ullman

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5:1 is usually used, because it is easy to figure out on the spot. It's quite a bit harder to calculate the velocity at which something would leave the rod/rail on the spot if you need a quick estimation.
but I bet you could do it ;)

Ben
 

WillMarchant

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As CJL mentions, the 5:1 is intended as a conservative rule to be used on the flying field. Calculating the rail departure velocity is a much safer approach. The fixed exit velocity numbers are the subject of quite a bit of discussion in the NAR's safety report. People are arguing that the number is really a function of the current windspeed at launch. You can find the "Launching safely into the 21st century" report at http://www.nar.org/pdf/launchsafe.pdf
 

MarkII

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Whatever looks best in the RockSim flight summary. ;)

MarkII
 

powderburner

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You can also improve the separation velocities at the end of the launch rod by using a longer launch rod.

You can go to many local hardware stores and find "round rod" in the bins with all-thread, angle stock, bar stock, etc, and 1/8th inch diam is usually available in lengths up to four feet. I don't happen to like using 1/8th that long (it gets kinda wobbly, especially for medium or heavy model rockets) so for four foot length I recommend stepping up to 3/16th or 1/4 inch diam.

You will need bigger launch lugs though-

You can also switch lugs for buttons, and launch from rails. These will give you whatever guided launch lengths of 8 or 10 feet or whatever you are willing to wrestle with (when they get too long, look for a rail launcher that comes apart in sections). Then you'll be all ready for mid power, and high power....
 

cjl

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As CJL mentions, the 5:1 is intended as a conservative rule to be used on the flying field. Calculating the rail departure velocity is a much safer approach. The fixed exit velocity numbers are the subject of quite a bit of discussion in the NAR's safety report. People are arguing that the number is really a function of the current windspeed at launch. You can find the "Launching safely into the 21st century" report at http://www.nar.org/pdf/launchsafe.pdf
True. It's also highly rocket dependent. A rocket with a large aerodynamic restoring moment and low inertia (in other words, one with large fins and a light weight) needs less speed to be stable off the rod than one with a small aerodynamic restoring moment and high inertia, such as one that is heavy with lots of noseweight and small fins. The low-inertia, high restoring force rocket will also be more susceptible to weathercocking.
 

highpowerrocket

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As mentioned the safest method is minimum launch rail velocity and it IS dependent of winds. Most simulation programs use a "small angle" assumption to simplify the stability equations. This assumption is usually valid for angles of attack less than 10 degrees or so. The angle of attack leaving the rail/rod is a direct relationship between rocket and wind velocity. The stronger the wind the higher the rocket velocity needs to be leaving the rail.
 

mikec

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It's quite a bit harder to calculate the velocity at which something would leave the rod/rail on the spot if you need a quick estimation.
Vrail = sqrt(2*(F/m)*l) where F is the initial thrust of the motor in newtons, m is the total mass of the rocket in kg, and l is the length of the rail in meters. The hardest part is knowing what the initial thrust of the motor is.

If you want to pick the minimum allowable V(rail), then you can solve for maximum allowable liftoff mass: 2*F*l/(Vmin^2). 15 m/s would be pretty conservative. So, as an example, the AT G79 with a maximum thrust of 94N, with a rail length of 1.5m, would have maximum liftoff mass of 1.25 kg.

It is simpler to use 5:1.
 

cjl

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Agreed. Besides, you also have to know what an acceptable rail velocity would be. I would tend to think that 15m/s is a bit on the high side for a calm day, and you could probably get away with 8-10m/s. On a windy day, you need more, but it's still rocket dependent. As you said, 5:1 is simpler.
 

bobkrech

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If you condense the LAUNCHSAFE rule of thumb to compensate for wind, you find that the recommended T/W is the wind speed in MPH with a minimum of 5:1

Many folks think that these suggestions are conservative in strong winds, but after seeing several model rockets cartwheel 2 to 4 time 50' above the launch rod in 15-20 mph winds, I'm a believer.

Bob
 

Graham Orr

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I'm reposting this from the old forum. It was a 3D (X,Y,pitch) numerical solution I did.


In reality, there really isn't a max lift-off weight... besides the upper end being your motor thrust level.

The rule of thumb of 4-5 G is there to ensure the rocket has adequate velocity in relation to the wind... This is with the assumption of something like a 6 ft launch rail.

I ran some simulations in Matlab to better demonstrate (see attached)

Sim details:
300 Ns motor, square thrust curve
1 kg rocket mass
2 m vertical launch rail
54 mm aiframe, Cd=0.4
Graph shows varying 2-10 G's lift-off acceleration

(sorry about all the lines being the same color.... see if you can figure out which is which)
I think there was like a 3-4 m/s wind in their too...

flight characteristics.jpg
 
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