Isp Calculation

[Ydoow]

I've been trying to learn Delta V and other calculations by reading older threads, and I got a little stumped on Isp

Here's what I'm reading and why it confuses me

From the VAB specs for the LV-909 engine:

Thrust = 50 kN at max throttle

Fuel flow = 2 litre per second at max throttle

Fuel density = (2.2 tonne/500 litre) = 4400 kg/m³

mass flow rate of fuel = 8.8 kg/s

I_sp = Thrust/(mass flow rate of fuel * g_kerbin) = 579.4 seconds

Effective exhaust velocity (V_e) = g_kerbin * I_sp = 5682 m/s

Very good, but the g value in ISP equations is always earth, 9.81. Doesn\'t matter if you\'re in space, on earth, on mars. Putting g(kerbin) is going to confuse people!

Then I read a bit on the wiki about g₀ (what they call g_kerbin)

Standard gravity, or standard acceleration due to free fall, usually denoted by g₀ or g_n, is the nominal acceleration of an object in a vacuum near the surface of the Earth. It is defined as precisely 9.80665 m/s2, or about 35.30394 (km/h)/s (≈32.174 ft/s2 or ≈21.937 mph/s)

To me, the wiki seems to be saying that these numbers only work if you're near the surface of Earth (or kerbin) and in a vacuum.

Further, I'm confused why TouhouTorpedo said that this number works no matter where you are; Earth, Mars, or space.

I understand that this value, g₀, is not G or g, but I don't understand why it remains constant and universal (literally according to TouhouTorpedo).

[semininja]

I think it's for purposes of comparison, that might just be how the unit is defined.

[Kosmo-not]

It's to convert it into a form that everyone can understand whether they are using imperial or metric. We use the surface gravity of our planet as a reference.

If you do the derivation of specific impulse, you will come up with the N*s/kg form.

[Ziff]

I guess I can take a crack at this. Someone correct me if I am wrong.

The units we use for specific impulse are artificial, we just arrive at them by using g in the equation. In reality, using g is basically a unit conversion for us to work with. Rockets work based on conservation of momentum, so the mass x velocity of the fuel burned off gives you the change in momentum of the rocket. Realize that it is the mass that is important. Because mass doesn't change regardless of where you are, it doesn't matter that we used g in the equation (it was just to give us that artificial number conversion anyway.) Specific impulse is just a measure of how efficient the engine is, or how much thrust per fuel unit consumed. Gravity doesn't effect that, what does however is atmospheric density, which is why engines have a lower Isp in thicker atmosphere.

If you notice, the g's cancel each other out, the specific impulse is the velocity of the fuel divided by g whereas the fuel rate is the mass of the fuel burned per second times g.

[Ziff]

It's to convert it into a form that everyone can understand whether they are using imperial or metric. We use the surface gravity of our planet as a reference.

If you do the derivation of specific impulse, you will come up with the N*s/kg form.

!! Damn, sometimes I am so stupid. I had no idea why we did that, but I understood it anyway. I swear I am half Kerbal sometimes.

[Ydoow]

Ah, okay this is making a bit more sense.

So really, the value of I_sp is arbitrary, it doesn't matter if I_sp is 5,000 or 5,000,000 as long as the units of that number is N*s/Kg

So g₀, as you said, is just to get those units lined up. Since I_sp can be an arbitrary number, and g₀ is only to give you the correct units, g₀ can also be arbitrary.

But as you said, we use earth's gravity simply because it's something we can relate to.

This makes more sense. Especially since I didn't see the connection between efficiency and gravity

Also, looking back at my whiteboard, arriving at Delta V makes a bit more sense; referring to calculating exhaust velocity

I_sp = F / m_flowrate * g₀

V_exhaust = I_sp * g₀

g₀ simply cancels out meaning gravity has no effect on exhaust thrust *phew*

So a quicker way about this would be...

V_exhaust = F/m_flowrate

right? My algebra can't be that bad, I just finished Calculus!

[MET?]

I might sound a little dumb here, but could you guys help me clear up some numbers?

So for g, thats a number we use when we are referring to standard gravity, and we can use this in all equations with g, and that number is approximately 9.81

However, that is different than G, which is the universal gravitational constant. When I see G in an equation, what number do I use for that?

Also: Kosmo-not

I was reading your interplanetary how to, and now that I've caught you on another thread, I was wondering...

In section 3, for planning our ejection angle, there are 4 formulas. Are these meant to be worked through one by one, so that after finishing the first, we get the variable we need for the second, and after finishing the second, we get the variable we need for the third, etc... until finally coming to the last part where we get our ejection angle?

Again, sorry if I sound ignorant, Im really trying to understand this...

Thanks!

[Kosmo-not]

gravitational constant: 6.67300 × 10^-11 m³ kg^-1 s^-2

^{If you have any questions for me, send me a PM.}

^{To answer your question about the interplanetary guide: yes, you assume correctly.}