Calculating Delta V Question

[The Flying Kerbal]

I wonder could someone clear up a query I've had for quite some time now concerning calculating Delta V?

I do like to try to do at least some of the maths when putting together a rocket, relying on Kerbal Engineer is very handy, but I do like to know how it does the calculations.  I know the equation for calculating the Dv on Kerbin is Isp*9.81*ln(Mfuelled/Mempty).  My head is telling me that to figure out the Dv required to return from - let's say - the Mun to Kerbin, this equation needs to be adjusted by replacing the 9.81 with the moon's gravity of 1.63.  However somewhere, possibly a Youtube video, someone says this isn't right, 9.81 is the number needed to calculate the Dv on all the bodies in the Kerbol system, and the fact it just happens to be the same as Kerbin's gravity is a happy coincidence.

Is this right?

Thanks everyone.

[FleshJeb]

That's good intuition but it's not correct. The 9.81 (known as g₀) is simply a conversion factor, and remains the same for all conditions/places.

The physical factor that drives the equation is v_e (effective exhaust velocity). The underlying concepts of momentum and energy rely on this (mv and 0.5mv²).

I_sp = v_e / g₀. It is represented in seconds so that different people working in both Imperial and metric units can use the same exact number for efficiency, and all the conversions to their native units happen on the right side of that equation. 9.81m/s^2 was picked as the conversion because it is the conversion from weight to mass on Earth and makes other calcs easier.

[IncongruousGoat]

To add to @FleshJeb's explanation, dividing exhaust velocity by g₀ gives you specific impulse in seconds, which happen to be the only unit that's common between Imperial and Metric. Specific impulse is, more than anything, a convention that allowed American and German engineers in the 50's to talk to each other about efficiency without having to do lots of unit conversion. Sure, it allows you to do some nifty things with thrust per weight flow of propellant, but that only works on Earth (or Kerbin).

[GoSlash27]

g0 in the Tsiolkovsky rocket equation is always 9.81 m/sec^2, no matter where you are.
And to clarify something else, you would not use this equation to determine the DV required for a maneuver (such as a return from the Mun to Kerbin). It is only used to determine the DV your rocket is capable of generating.

 Those calculations use the vis-viva equation and hyperbolic excess velocity.

Best,
-Slashy

Edited March 26, 2018 by GoSlash27

[The Flying Kerbal]

Thanks for clarifying that for me guys, very much appreciated.

1 hour ago, GoSlash27 said:

g0 in the Tsiolkovsky rocket equation is always 9.81 m/sec^2, no matter where you are.
And to clarify something else, you would not use this equation to determine the DV required for a maneuver (such as a return from the Mun to Kerbin). It is only used to determine the DV your rocket is capable of generating.

 Those calculations use the vis-viva equation and hyperbolic excess velocity.

Best,
-Slashy

Yes indeed, but I try and figure out in advance how much Dv each stage of a vessel will need to complete its specific task, while still in the VAB.  This means I don't send way too much fuel in the stage that will be used to get the vessel back to Kerbin, saving on weight.

[GoSlash27]

1 hour ago, The Flying Kerbal said:

Yes indeed, but I try and figure out in advance how much Dv each stage of a vessel will need to complete its specific task, while still in the VAB.  This means I don't send way too much fuel in the stage that will be used to get the vessel back to Kerbin, saving on weight.

A kerbonaut after my own heart

 

[OhioBob]

Also note that the actual value of g₀, and the one used by KSP, is 9.80665 m/s².  Though for most calculations 9.81 is close enough.
 

Edited March 26, 2018 by OhioBob

[The Flying Kerbal]

Thanks for all the great answers guys, huge help as always!

On 27/03/2018 at 12:09 AM, OhioBob said:

Also note that the actual value of g₀, and the one used by KSP, is 9.80665 m/s².  Though for most calculations 9.81 is close enough.
 

I didn't know that, good to know.

[OhioBob]

1 hour ago, The Flying Kerbal said:

I didn't know that, good to know.

I don't know how long you've been playing KSP, but there was a time when they used the rounded off value of 9.81.  And in rocket calculations they were actually using 9.82, at least that's what people said.  But they cleaned that up with the release of version 1.2.

I don't know if you've ever noticed, but take a look at Kerbin's information display in the Tracking Station.  It shows its ASL gravity as 1.00034 g.  Why such an odd number?  The reason is because back when Squad defined g₀ = 9.81 m/s², Kerbin had a surface gravity of exactly 1g.  But by making g₀ = 9.80665 m/s², had they left Kerbin's surface gravity at 1g, this would have changed its gravitational parameter.  And that would have changed familiar numbers that players had already gotten use to, like geostationary altitude.  So instead they left Kerbin's surface gravity at 9.81 m/s², which in standard gravities is, 9.81 / 9.80665 = 1.000341605 g.  They did the same thing with the other planets as well.
 

[GoSlash27]

For almost all cases, the discrepancy between g0 and literal sea level gravity amounts to an error of 3 hundredths of 1 percent; inconsequential. The only case where that's significant is calculating orbital period vs. radius, and that uses 9.81 m/sec. I just use 9.81 m/sec for all calculations and don't worry about the tiny error.

Best,
-Slashy

[OhioBob]

2 hours ago, GoSlash27 said:

For almost all cases, the discrepancy between g0 and literal sea level gravity amounts to an error of 3 hundredths of 1 percent; inconsequential. The only case where that's significant is calculating orbital period vs. radius, and that uses 9.81 m/sec. I just use 9.81 m/sec for all calculations and don't worry about the tiny error.

I agree that for something like calculating the delta-v of a rocket, 9.81 is plenty close enough.  But I don't agree with that part about orbital period vs. radius using 9.81 m/s².  Orbital period calculations don't even use g₀, they use the gravitational parameter, μ.  While the value of μ can be defined directly in a celestial body's config, most often it's computed from given values of surface gravity and radius.  We use the formula μ = gr², where g is in m/s² and r is in meters.  In KSP, g is given in units of standard gravities, so we must convert.  We have, g(m/s²) = g(gees) * g₀, where g₀ = 9.80665 m/s².  I haven't found any instances where KSP uses a different value of g₀.

Of course I agree that on Kerbin g = 9.81 m/s², but g is not g₀.  g₀ is a universal constant having the value 9.80665 m/s², while g depends on where you are.  But since gravitational parameter uses g and not g₀, for Kerbin μ = 9.81*600000^2 = 3.5316E+12 m³/s².

My point in bringing up the exact value of g₀ is if somebody is doing something like writing an Excel spreadsheet to do the calculations, then why not just use the exact value.  You type it in once and you're done.  On the other hand, if you're punching the number into a calculator each time, then the shortcut of using 9.81 is good enough.

 

 

[DunaManiac]

There is a website called strout.net that calculates delta v for you. I used it for many successful missions. (link here --->) http://www.strout.net/info/science/delta-v/intro.html

[Laie]

On 3/26/2018 at 6:29 PM, The Flying Kerbal said:

My head is telling me that [...] replacing the 9.81 with the moon's gravity of 1.63. 

I see how you got there, but sorry, no.

The g in the equation is only needed because you start with ISP measured in "seconds". Quote Wikipedia: "I_sp in seconds is the amount of time a rocket engine can generate thrust, given a quantity of propellant whose weight is equal to the engine's thrust."

See how it says"weight" rather than "mass"? You need gravity to get from one to the other, and whoever came up with the idea of expressing ISP in seconds silently assumed earth standard gravity (as usual when confusing weight and mass, it goes without saying, doesn't it? Though my physics teachers would blow their top when they caught me doing this). Anyway, that's how it got in there, and why it remains 1g wherever you are.

When multiplying ISP in seconds with 1g, you get exhaust velocity in "meters per second", which is what actually goes into the rocket equation: start mass, end mass, and the velocity at which the difference is expelled.

Edited March 31, 2018 by Laie
rephrasing

[DoToH]

 

43 minutes ago, Laie said:

When multiplying ISP in seconds with 1g, you get exhaust velocity in "meters per second", which is what actually goes into the rocket equation: start mass, end mass, and the velocity at which the difference is expelled.

This ^^

[OhioBob]

5 hours ago, Laie said:

When multiplying ISP in seconds with 1g, you get exhaust velocity in "meters per second", which is what actually goes into the rocket equation: start mass, end mass, and the velocity at which the difference is expelled.

Tsiolkovsky's rocket equation,

where V_e is the exhaust gas velocity.  But what we use is actually the effective exhaust gas velocity.

The thrust equation is, 

F = q * V_e + (P_e - P_a) * A_e

where q is the mass flow rate, V_e is the actual exhaust gas velocity, P_e is exhaust pressure at the nozzle exit, P_a is the ambient air pressure, and A_e is the area of the nozzle exit.  The term q * V_e is called the momentum thrust, and (P_e - P_a) * A_e  is called the pressure thrust.  Pressure thrust is the result of unbalanced pressures forces at the nozzle exit.  For a nozzle that is correctly adapted to its operating environment, the pressure thrust is zero, or near zero.

To simplify things we introduce the concept of effective exhaust gas velocity, C, which combines the effect of both momentum and pressure thrust into a single parameter,

F = q * C

The equation for specific impulse is,

I_SP = F / (q * g₀)

which we rearrange as follows,

F = I_SP * q * g₀

So we now have two equations for thrust that we can set equal to one another,

q * C = I_SP * q * g₀

dividing through by q, we have

C = I_SP * g₀

So that's the long explanation for why we use I_SP * g₀ when we solve the rocket equation.
 

Edited March 30, 2018 by OhioBob

[The Flying Kerbal]

On 30/03/2018 at 12:42 PM, DunaManiac said:

There is a website called strout.net that calculates delta v for you. I used it for many successful missions. (link here --->) http://www.strout.net/info/science/delta-v/intro.html

Good one mate!