Author Topic: Basic Rocket Science Q & A  (Read 502363 times)

Offline R7

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Re: Basic Rocket Science Q & A
« Reply #760 on: 03/04/2013 11:21 am »
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Quote
The galactic year, also known as a cosmic year, is the duration of time required for the Solar System to orbit once around the center of the Milky Way Galaxy.[1] Estimates of the length of one orbit range from 225 to 250 million "terrestrial" years.[2] According to NASA, the Solar System is traveling at an average speed of 828,000 km/h (230 km/s) or 514,000 mph (143 mi/s),[3] which is about one 1300th of the speed of light.
AD·ASTRA·ASTRORVM·GRATIA

Offline spacecane

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Re: Basic Rocket Science Q & A
« Reply #761 on: 03/06/2013 12:42 pm »
What is the tremendous speed relative to?  If all the Galaxies are rotating "in sync" then maybe we aren't moving at all???  Hmmmm...

It's interesting to think about.  We are certainly moving at a pretty good clip relative to a given point on the sun's surface!

Offline QuantumG

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Re: Basic Rocket Science Q & A
« Reply #762 on: 05/01/2013 09:01 pm »
Great answer.

Welcome to the forum!
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Offline stevetaylor

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Re: Basic Rocket Science Q & A
« Reply #763 on: 05/11/2013 03:14 pm »
a Question?
is the recovery time from a prolonged stay in zero-g affected by recovering in a reduced gravitational environment, say one-third-g?

Offline Jim

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Re: Basic Rocket Science Q & A
« Reply #764 on: 05/11/2013 04:22 pm »
a Question?
is the recovery time from a prolonged stay in zero-g affected by recovering in a reduced gravitational environment, say one-third-g?

Unknown

Offline scienceguy

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Re: Basic Rocket Science Q & A
« Reply #765 on: 05/15/2013 03:34 am »
When something has a specific impulse (Isp) of 5000 s, what does that mean? What lasts for 5000 s?
e^(pi*i) = -1

Offline QuantumG

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Re: Basic Rocket Science Q & A
« Reply #766 on: 05/15/2013 03:59 am »
When something has a specific impulse (Isp) of 5000 s, what does that mean? What lasts for 5000 s?

Heh. I like the phrasing.

Specific impulse is a measure of engine efficiency. Like most measurements of efficiency, that means it is something divided by something. For a rocket, we're concerned with how much thrust it produces for each unit of propellant, per second. That is, we want to divide the thrust by the change in propellant per second, which is often called "mdot".

Say we take the thrust and divide it by the mdot, what do we get? In the civilized world we measure thrust in newtons and we measure mdot in kg/s. We prefer to say newtons, because Newton was awesome, but they are just kg*m/s^2. If we divide kg*m/s^2 by kg/s, we get m/s (it's the same as multiplying kg*m/s^2 by s/kg, see how the kg and one of the s's cancel?) which is a measurement of velocity. What's this velocity measuring? Newton's third law tells us: it's measuring the exhaust velocity of the rocket!

Hang on, we're almost there. In many parts of the world, rocket scientists and engineers don't bother with isp.. they just talk about the exhaust velocity. For various reasons, not the least of which that smaller numbers are easier to remember, it's convenient to divide the exhaust velocity by g, the acceleration due to gravity at the Earth's surface. Like many measurements, this is entirely arbitrary.

So, we have m/s and we divide by m/s^2, what do we get? Well, the m and one of the s's cancel, so you get just seconds! That's why isp is measured in seconds.

To answer your specific question, an isp of 5000 s is telling us that the exhaust velocity of the rocket is 49033.25 m/s, which is very fast indeed. If we can put 1 kg of propellant through the rocket per second, it will produce 49033.25 newtons of thrust!
« Last Edit: 05/15/2013 04:04 am by QuantumG »
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Offline strangequark

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Re: Basic Rocket Science Q & A
« Reply #767 on: 05/15/2013 04:46 am »
When something has a specific impulse (Isp) of 5000 s, what does that mean? What lasts for 5000 s?

QuantumG is correct, but if you're looking for a physical meaning to latch onto, it is the number of seconds that 1 pound of propellant can be used to produce 1 pound of thrust.

Offline Jim

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Re: Basic Rocket Science Q & A
« Reply #768 on: 05/15/2013 06:53 am »
[quote author=QuantumG link=topic=13543.msg1052304#msg1052304 date=136859038

Say we take the thrust and divide it by the mdot, what do we get? In the civilized world we measure thrust in newtons and we measure mdot in kg/s. We prefer to say newtons, because Newton was awesome, but they are just kg*m/s^2. If we divide kg*m/s^2 by kg/s, we get m/s (it's the same as multiplying kg*m/s^2 by s/kg, see how the kg and one of the s's cancel?) which is a measurement of velocity. What's this velocity measuring? Newton's third law tells us: it's measuring the exhaust velocity of the rocket!

Hang on, we're almost there. In many parts of the world, rocket scientists and engineers don't bother with isp.. they just talk about the exhaust velocity. For various reasons, not the least of which that smaller numbers are easier to remember, it's convenient to divide the exhaust velocity by g, the acceleration due to gravity at the Earth's surface. Like many measurements, this is entirely arbitrary.
[/quote]

In the uncivilized world, we measure thrust in lb and we measure mdot in lb/s and when you divide them, you get s

too late to continue.  Will reconcile lb and g tomorrow day.
« Last Edit: 05/15/2013 06:58 am by Jim »

Offline QuantumG

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Re: Basic Rocket Science Q & A
« Reply #769 on: 05/15/2013 07:20 am »
lbf and lb are not the same thing, and yes, we know you're not civilized ;)
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Offline Jim

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Re: Basic Rocket Science Q & A
« Reply #770 on: 05/15/2013 12:42 pm »
lbf and lb are not the same thing, and yes, we know you're not civilized ;)


That is why I was going to reconcile lb an g

Offline scienceguy

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Re: Basic Rocket Science Q & A
« Reply #771 on: 05/15/2013 05:14 pm »
Thanks for the responses. I knew it had something to do with fuel, but I was trained in metric and I knew we wouldn't be measuring thrust in kilograms.
e^(pi*i) = -1

Offline ugordan

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Re: Basic Rocket Science Q & A
« Reply #772 on: 05/15/2013 07:58 pm »
I knew it had something to do with fuel, but I was trained in metric and I knew we wouldn't be measuring thrust in kilograms.

Didn't the Russians use kilogram-force for a long time for their engine specs?

Offline strangequark

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Re: Basic Rocket Science Q & A
« Reply #773 on: 05/15/2013 08:22 pm »
I knew it had something to do with fuel, but I was trained in metric and I knew we wouldn't be measuring thrust in kilograms.

Didn't the Russians use kilogram-force for a long time for their engine specs?

Yes. The Soviets used metric, but not SI. They still used seconds for Isp, and pressures were something weird too (kgf/cm^2 maybe).

Offline baldusi

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Re: Basic Rocket Science Q & A
« Reply #774 on: 05/16/2013 04:41 pm »
I knew it had something to do with fuel, but I was trained in metric and I knew we wouldn't be measuring thrust in kilograms.

Didn't the Russians use kilogram-force for a long time for their engine specs?

Yes. The Soviets used metric, but not SI. They still used seconds for Isp, and pressures were something weird too (kgf/cm^2 maybe).
Actually, all the metric pressure units are intimately related. And the SI are the culprits of the pressure measure being so diverse (even as metric).
The correct SI unit is the Pascal (N/m²). Which is very nice as a definition of two units, but it's simply "too small" for normal use.
The traditional measure was the Bar, which is 10N/cm², but also 1.013atmospheres. Thus, for second order approach to atm is an excellent unit and one very easy to relate to. But since 10N/cm² is not unit/unit, SI decided standardize on Pa, but then 1atm=0.1013MPa, a lot of unnecessary numbers.
Now enter the kgf/cm², which is 0.97atm, 0.98Bar and gives you the thrust straight to tonnes force, which is what you usually want to know the T/W at liftoff. Also, when working with pneumatic systems, it gives you your thrust in kg, with is extremely useful when you use kg as the unit of weight. Once you are in orbit, N is the correct unit for acceleration. And for high isp engines like ion drive, N is a more comfortable unit to work with. But those only work in orbit, where you can forget about gravity losses.
Thus, kgf/cm² or Bar would be the units that you'd actually use for pressure systems if the SI snobs hadn't polluted the measures because Bar was not "beautiful enough".

Offline Proponent

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Re: Basic Rocket Science Q & A
« Reply #775 on: 05/17/2013 11:51 am »
To nit-pick, the SI unit of pressure is the pascal, not the Pascal.  The SI symbols for units derived from people's names are capitalized, but not the units themselves.

Offline baldusi

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Re: Basic Rocket Science Q & A
« Reply #776 on: 05/17/2013 03:39 pm »
To nit-pick, the SI unit of pressure is the pascal, not the Pascal.  The SI symbols for units derived from people's names are capitalized, but not the units themselves.
Yeah, another of those "logic" that come out of an international committee. The full name is not capitalized, to "avoid" confusion the the actual person (when the whole purpose of naming it after someone is to immortalize its name), but it's degrees Celsius (the "d" is not capitalized!). And, to add insult to injury, the symbol is capitalized (Pa). I love metric, but once you allow a bunch of bureaucrats to make decisions nothing practical can come out of it.
That's why the Russians use the metric units the way you actually work with them. When you want pressure and use US customary, you want the force to be in lbf. Same for metric, you usually need it in kgf. So, don't let the SI bureaucrats get in the way of using the most practical units for actual work.
Again, it does makes sense to use Pa for things like Martian atmosphere, and N for the thrust of an orbital vehicle. But plumbers think in terms of the force units of mass.

Offline crickmaster

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Re: Basic Rocket Science Q & A
« Reply #777 on: 05/17/2013 06:08 pm »
Got a question regarding interplanetary transfers.  I'm trying do a 2D simulation of a hohmann orbit between Earth and Mars. 

Suppose I have all the orbital elements of Earth and Mars at departure and arrival, and if I had the magnitudes of the velocities in orbits, how do I convert/transform the magnitude into it's xyz components?

I know we're supposed to use trig and some of the angles from the COE, but I'm not entirely sure how to go about it.  Anyone can link me/tell me what these transformations are?
« Last Edit: 05/18/2013 06:11 am by crickmaster »

Offline 93143

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Re: Basic Rocket Science Q & A
« Reply #778 on: 05/17/2013 06:59 pm »
But plumbers think in terms of the force units of mass.

Plumbers use the Hazen-Williams formula too.  What's your point?

Any experienced SI user can switch between Pa (or kPa, or MPa; there's a reason metric prefixes exist) and bar almost subconsciously.  He might have to whip out his calculator to switch between bar and atm, if the answer is particularly important...  As for N vs. kgf, the use of separate units for force and mass maintains generality and keeps you from doing stupid things.  So what if it's a little more work to multiply or divide by 9.80665 (or whatever the local value of g happens to be)?  Consistency of units is a valuable thing, as I've found in my CFD research.

Measuring rocket performance in s seems to be a sort of grandfathered artifact that persists partly because it gives you the same number in all measurement systems (and will until someone comes up with an alternative unit of time).  For actual calculations, everyone multiplies it by g.

...

I don't see how you can maintain that a unit is "too small" for a given application, when metric provides such a simple way to scale up or down in multiples of 1000...

I personally think it's great that metric produces integer powers of ten that correspond almost exactly to physical measures like the density of water (998.2 kg/m³ at 20°C) or the Earth's surface gravity (9.80665 N/kg ± 0.5% or so) and atmospheric pressure (101.325 kPa on a standard day at sea level).  It means you can do quick calculations really easily in your head, but when it comes time to do it accurately, it doesn't fool you into thinking you're dealing with a physical identity when it's really only a coincidence.  (Water is the worst for this, because at 4°C it really is almost exactly 1000 kg/m³; this is by design, but it's no longer the standard...)

...

Also, a bar is 100 kPa, or 0.986923 atm.  Your number was backwards.

...

I use bar all the time.  Then I move the decimal five places to the right, because my code is in straight MKS...

...on the other hand, cgs confuses the hell out of me (I don't actually know what an erg is off the top of my head).  Especially in electromagnetics.  I suppose it's mostly to do with what you learned first...
« Last Edit: 05/18/2013 07:14 am by 93143 »

Offline strangequark

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Re: Basic Rocket Science Q & A
« Reply #779 on: 05/20/2013 07:09 am »
To nit-pick, the SI unit of pressure is the pascal, not the Pascal.  The SI symbols for units derived from people's names are capitalized, but not the units themselves.
Yeah, another of those "logic" that come out of an international committee. The full name is not capitalized, to "avoid" confusion the the actual person (when the whole purpose of naming it after someone is to immortalize its name), but it's degrees Celsius (the "d" is not capitalized!). And, to add insult to injury, the symbol is capitalized (Pa). I love metric, but once you allow a bunch of bureaucrats to make decisions nothing practical can come out of it.
That's why the Russians use the metric units the way you actually work with them. When you want pressure and use US customary, you want the force to be in lbf. Same for metric, you usually need it in kgf. So, don't let the SI bureaucrats get in the way of using the most practical units for actual work.
Again, it does makes sense to use Pa for things like Martian atmosphere, and N for the thrust of an orbital vehicle. But plumbers think in terms of the force units of mass.

I'd like to add to 93143's post above by pointing out that using kgf/cm^2 throws a wrench into things when you're working with dynamic and stagnation pressure. You'd be stuck with cm/s for velocity, and hyls/cm^3 for density to keep a consistent set of units. I'll take SI please and thank you.
« Last Edit: 05/20/2013 07:17 am by strangequark »

 

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