*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.