convert newton-seconds force to grams-force =advanced=

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shockwaveriderz

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Ok I'm stumped... I know that to covert newton-seconds force to lb-sec force I divide NS/4.5

So lets say I have an A3-4T motor

lets say the average thrust is 3 N-s. If I divide that by 4.5 I get .44 lb-sec force.. to get oz-sec force I suppose I would divide 0.44/16 ?

or so how do i convert # N-s force to grams-force?
 
1) Newton-seconds is not a force. It is force times time.

2) The Newton is a composite unit. It is the same as kilogram - meter/ second squared. (Remember that famous equation F = ma?)

So if you want to convert Newton - seconds to millinewton - seconds ( gram - meter/ second squared, multiply by 1,000. Or a better way to think about it is that you multiply by one in the form of:

1,000 millinewtons/ 1 Newton


One of the things that was drilled into me during enginerring thermodynamics and then fluid mechanics was unit analysis. It stuck. :)
 
Originally posted by shockwaveriderz
Ok I'm stumped... I know that to covert newton-seconds force to lb-sec force I divide NS/4.5

So lets say I have an A3-4T motor

lets say the average thrust is 3 N-s. If I divide that by 4.5 I get .44 lb-sec force.. to get oz-sec force I suppose I would divide 0.44/16 ?

or so how do i convert # N-s force to grams-force?
Let's back up a little first. Thrust is in Newtons, not N-s.

Now, to get oz of thrust, or oz-sec of total impulse, you *multiply* lb (or lb-sec) by 16, not divide, since there are 16 oz in 1 lb.

And then, since there are 28.34952 gm in 1 oz, you multiply your oz or oz-sec number by 28.34952 to get gm force of thrust or gm-sec of total impulse.

Or more simply, multiply N or N-s by 100 to get gm of force or gm-sec of impulse. That works out the same as the steps I just gave, but without having to work from metric to English units and then back again.
 
Dave's answer is excellent and unit conversions should always include dimensional analysis. Force is not impulse (force x time)nor is the weight (a force) the same as mass. However - just to confuse what would otherwise be simple - engineers use mass-lbs and force-kilograms:mad:

To get from 0.44 lbs to oz, use the fact that 16 oz = 1 lb to convert as follows:

0.44 lb (16 oz/1 lb) = 7.04 oz

When converting, put what you want in the numerator (oz) and what you want to eliminate in the demominator (lb).
 
Good explanations...I just want to add one more point.

The A3 has an average thrust of 3 Newtons (force)....don't confuse this with the Impulse (N-s). The letter designation gives you the impulse "class" of the motor. "A" motors are between 1.26 and 2.5 Newton-seconds.

The Impulse depends a lot on the burn time of the motor. Take the Estes D12 and the E9 for example. The D12 actually has more thrust but burns shorter than the E9. The longer burn of the E9 makes it and "E".

Hope that helps,
John
 
Using grams as a unit of force is wrong - its a mass unit.

1000 grams = 1 kg

Weight of 1 kg = (9.8 m/s^2)(1 kg)= 9.8 N (~ 10 N)

Weight of 1 gram = (9.8 m/s^2) (.001 kg) = .0098 N (~0.01 N)
Weight of 1 gram = 1 force-gram

so 100 force-grams ~ 1 N (actually 102 force-grams = 1 N).

Sorry to be picky, but incorrect units caused the $125,000,000 Mars Climate Orbiter to crash in 1999 and is a pet peeve of mine.

John
 
so basically, I should just use netwons as the force unit?

for example, as I said, I have 3 forces that I will be calculating :

Drag force + Lift Force + thrust force = 0.

so I can use netwons as the units for all 3 forces?

you guys with actual scientific and engineering educations and backgrounds are a great resource to somebody like me that was never trained or educated as either. thanks
 
Originally posted by shockwaveriderz
so basically, I should just use netwons as the force unit?

for example, as I said, I have 3 forces that I will be calculating :

Drag force + Lift Force + thrust force = 0.

so I can use netwons as the units for all 3 forces?
Shockie, Yes, you can use newtons for all three forces (provided they either are supplied that way or you convert them from whatever other units of force into newtons.

But your above equation sums the forces to 0. That implies that the acceleration (F=ma) will be zero as well - ie, constant velocity. About the only times in rocketry when the velocity is constant is (a) when it's sitting on the pad (v=0) or (b) when it's under chute, falling at a constant rate of descent (eg, 20 feet/sec).

...somebody like me that was never trained or educated as either...
I'll file that under "yet another reason E-Town was better than North" :)

BR,
Doug
 
doug: had to introduce etown/north hardin into the equation huh?

for those that have no idea what I am talking about... seems doug man graduated from etown high in 73 as I was graduating from north hardin 12 mile sup the road... and I did take intro to chem/physics as a freshman where i got an A... but that was like 36 years ago.....ouch.... there was a fierce rivalry between etwon and north hardin back then...

anyway there seems to be a 3rd case where all the forces must equal 0, ie boost gliders during the boost phase...

if Me = FeXe <--thrust x moment arm length

if Mw = Fw x moment arm length where Fw - wing force
if Mt = Ft x moment arm length. where Ft = tail force
if Mp = Fp x moment arm length where Fp is pod force

then Mtotal = Me+Mw+Mt+Mp = 0 for the glider to more or less go straight up...

In reality equlibrium is probably NOT really possible but is at a small enough magnitude such that it becomes a side force and the BG boost is a shallow arc........

during the coast phase Mtotal = Mw+Mt+Mp = 0

when a BG boost straight up(or in a shallow arc), the wing lift,stab lift, motor thrust and Pop Drag forces must be in balance. During motor thrusting, tehrefor, the BG probably is in an equilibroiu condition with the wing ooperating at a near-zero lift or slightly negative anagle of attack to produce minmal lift during boost. After motor burnout, the BG is again probably at an equlibium condition with the wing at a slightly higher angle of attack where its lift force offsets the pod drag/stabilzer lift forces.

this is why you see BG's go into opposite shallow arcs during boost and coast phase. (the infamous S curve) The closer you are to the equlibrium condition ( ie forces and moments balance) the less will be the arc , ie the S curve will be more flat therefore the trajectory is straighter which is good. You want to keep the BG during both boost and cast with a small trajectory "cone'" , preferbaly -5 to say +5 as lift and drag are both reduced by boosting and coasting at smaller anagles of attack to true vertical.
 
Originally posted by jwagner61
Sorry to be picky, but incorrect units caused the $125,000,000 Mars Climate Orbiter to crash in 1999 and is a pet peeve of mine.

John

I thought that was just the excuse to the press when the aliens shot it down ;)
 
For small estes motors...go to their web site and look under motors they have a chart with all their motors already converted...
 
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