Impluse ?

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mkmilion

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I was wondering the relation of impluse change when clustering. For example: Does a cluster of 2 A8-3's become the impluse A16-3 or does just become a stronger thrust with the same impulse?
I've been told that clustering allows the same altitude with more power; is this true?
If anyone can help answer or clarify these questions I would appreciate it. Thanks.

Ed
 

Elapid

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of the particular force

Impulse is measured in Newton seconds, or thrust per time period.

take two 20 Newton second (Ns) motors and together they produce 40Ns

two motors push harder than one
:)
 

cls

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Ed, 2 * A8s becomes 1 * B16. add the letters to get total power, add the numbers to get impulse.

advanced math: how about 3 * D12-5? (Richter Recker) that makes an "E36-5". arguably it would be an "E-and-a-half 36": D+D=E; D+D+D=E 1/2.

although, practically we don't say the "half"

another one: my Skunkworks 10' Saturn V flew on a K445 and 4 * J210s. What motor does all that make? Lessee, J+J=K; 4 * Js makes an L; L + K makes an L and a half - still under an M so it's OK for me (level 2) to fly.

last one: Andy built an N1 model which flew with 24 C6-5s and 6 E9-6s. what is the total power and equivalent thrust?
 

MarkABrown

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Actually, I don't think that is correct. I don't think you can just add the motors and assume the impulse class jumps a level. What you need to know is the thrust of the motor in Newtons. Then you add the thrust of all motors in the cluster and see where this cumulative thrust falls on the impulse scale. A chart of impulses is available here:

https://www.tripoli.org/tmt/motor_classes.shtml
 

Rocketjunkie

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Originally posted by cls
Ed, 2 * A8s becomes 1 * B16. add the letters to get total power, add the numbers to get impulse.

advanced math: how about 3 * D12-5? (Richter Recker) that makes an "E36-5". arguably it would be an "E-and-a-half 36": D+D=E; D+D+D=E 1/2.
3 D12's is a F36. Mid range F, but F nonetheless.
4 D12's would be a full F48.

Each letter is a *range* of total impulse. Add the total impulse of all the motors and the total will fall into one of the ranges. That letter is the power to use even if the total just makes the range (such as 85 N-s is still a G motor).

3 small J's such as J350's (700 N-s each, total 2100 N-s) is a K but 3 larger J's such as J415's (1200 N-s each, total 3600 N-s) is a L.

The thrust is different, if all the motors are the same, just multiply the average thrust number by the number of motors. If the motors don't all have the same burn time, divide the total impulse by the burn time of the cluster (usually the longest burning motor or if staggered starting, the total burn time.

Example: 3 G80's. Total impulse 3*120=360 N-s (Barely an I, max H is 320 N-s) with 3*80 newtons average thrust Equilivent motor I240. 2 G80's and 1 G40 all started together. A G40 burns twice the 1.5 seconds of a G80 but still has 120 N-s total impulse. Total impulse is still 360 but the average is 360/3 second burn time of the G40. This is an I120. However, the thrust curve is more like the I240 for the first 1.5 seconds and then just the G40 is burning for the last 1.5 seconds.
Clusters containing different motors need to have the types of motors arranged in a symmetrical pattern around the rocket centerline.
 

DynaSoar

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Originally posted by MarkABrown
Actually, I don't think that is correct. I don't think you can just add the motors and assume the impulse class jumps a level.
Technically, no, the motor class is the same. However, clustering MP motors CAN put you into the HP range, where you have to have L1 to fly it.
 

Zonker

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Sorry....click "Cluster Totals" in the menu on the left


Zonker
 

bobkrech

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That's a useful calculator, however the conversion of the % impulse class equivalent motor is incorrect.

For example if you put in 3 G80s,

the correct total impulse (360 N-S) is calculated,

as it the total thrust (80.9 pounds =360 N-S/4.45 N-S/pound),

however it lists this combo as 56.25% I class which is incorrect.

The correct values are calculated as follows.

100% H = 320 N-S = 0% I and
100% I = 640 N-S so the % impulse class is correctly calculated as

(360-320)/(640-320) x 100% = 40/320 x 100% = 12.5% I

Bob Krech
 

Zonker

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Interesting. I never noticed the discrepancy. I've used it to calculate 4 x D12s. The calculator lists it correctly at 68 NS and 85% of an F. As far as I can figure, that is correct. I'm not sure why the calculation for 3 G80s is wrong....maybe the database is out. Anyway, I find it to be a quick guide none the less.

Zonker
 

bobkrech

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Zonker

The calculator is wrong for that combo also.

4 x D12 has a total impulse of 68 N-S, but the calculator is calculating the % impulse class incorrectly using the formula total impulse/maximum class impulse = 68/80 = 85%.

This formula is incorrect. The correct formula for % impulse class is

(total impulse - minimum class impulse)/(max class impulse - min class impulse)

For 4 x D12 the answer is

(68-40)/(80-40) x 100% = 28/40 x 100% = 70% F.

Bob Krech
 

Zonker

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Bobkrech,

Thanks for the correction...I see what you mean. This reinforces the importance of knowing what impulse class you're dealing with. As DynaSoar said earlier in the post, it is quite possible to end up with an HP total which puts you into a different set of rules/regulations.

Thanks,

Zonker
 
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