Readings from Delta-V map don't match experiments.

[egoego]

I have problems understanding delta-V or delta-V maps.

The most recent delta-V map I could find (from this reddit post) sprung to my eye, because it says you only need 3200m/s to reach LKO. This might bee due to changes to lift and stuff in 1.0 (I used to play 0.90 and before.) But with atmospheric stuff being a bit more complex to calculate for me, I tried my luck with a Mun lander.

The map says from LMO (14km) down to Mun surface I would need 580m/s.

So I build a simple lander:

MK1 Pod : 0.84 tons

FL-T400 tank : 2.25 tons (2 tons fuel)

LV-909 Terrier engine : 0.5 tons (isp vac: 345s)

three LT-1 Legs

(about 3.7 tons total)

and put him into an 14km orbit around the Mun.

To calcualte how many delta-V it has I use the formula from the wiki (same as in this thread):

dV = Isp * gravity * ln( stageMass / (stageMass-fuelMass) )

The wiki says the gravity for the Mun is 1.63m/s².

That gives my lander about 437m/s delta-V and it is less than I expected.

According to the delta-V map I would need 580m/s, so I shouldn’t be able to get down to the Mun surface without crashing.

But I tested it and could land this thing with more then 50% fuel left on the Mun.

This lead to a few questions:

1. Did I read the map wrong?
   
2. If so, how should it be read?
   
3. Did I calculate the delta-V wrong? (Using Kerbins gravity constant gives a higher delta-V but I think I’m supposed to use the gravity constant of the object I’m orbiting.)
   
4. Less important: How are those delta-V maps created? Did the author use Kerbins gravity for all objects by accident?
   

EDIT:

After submitting this question I came up with the idea that I really should have used Kerbins gravity (9.81) for the calculation and that this was also used in creating the map. Physics class was a long time ago so a hint on why this could be true would be appreciated. At least it could explain why my numbers seem to be wrong.

Edited November 5, 2015 by egoego

[FancyMouse]

The gravity for calculating dV is always 9.81m/s^2, regardless of where you are. This is different from the TWR case.

Your edit is going to the right direction.

[BoilingOil]

I see a dv map here that look eerily similar. And THAT one is outdated!

Does that answer your question?

[Chaos_Klaus]

dV = Isp * gravity * ln( stageMass / (stageMass-fuelMass) )

It's not local gravity. It is g0. You always have to use 9.81m/s².

It's a little confusing, but this value is just there to do a unit conversion. The rocket equation actually goes:

delta v = ve * ln (m0/m1)

ve is the exhaust velocity given in m/s. Actually this value could be considered specific impulse. One explaination is that when german and american rocket scientists worked together they had to deal with different unit systems. So if everyone just used g0 in his prefered unit system and got the same value for ISP.

[GoSlash27]

egogo,

You've got it figured out. The map you're using is accurate. DV to orbit is a rough estimate for atmospheric bodies and calculated fairly precisely for airless bodies.

Having said that...

You will actually need more than 580m/sec DV to land on the Mun safely. Perhaps as much as twice that depending on your technique. Landings are highly- inefficient processes.

Best,

-Slashy

[Brainlord Mesomorph]

Those maps are minimum numbers for efficient burns. I blame my own sloppy piloting for burning way more fuel than that.

[KerikBalm]

345 *9.81 * ln(3.7/1.7) = 2,632 m/s dV....

Yea... that should get you down easily.

Half fuel remaining would be: 345*9.81*ln (2.7/1.7) = 1565 m/s

2632-1565 = 1066 m/s....

Well... I suppose you could have flown really inefficiently.

Its obviously not less than ~550 m/s... as that is the speed at which you orbit... If you are in very low orbit... like scraping the top of mountains low... check your surface velocity... on an airless body, that is approximately the minimum dV you need to land... and also the dV you need to get back to orbit + some extra due to gravity drag losses.

If you are orbiting in a low circular orbit at ~550 m/s, its obviously impossible to land (come to a stop, as you contact the ground), without a change of velocity of ~550 m/s

"a hint on why this could be true would be appreciated."

Its just a constant used to relate Isp to m/s... Your rocket's dV is going to be the same if its in orbit around the sun, jool, gilly, eve, or Kerbin... so why would you start changing the constants (like g0 ) so that you get different numbers? if you change the value of g0... then you need to change the way Isp is calculated....

If Isp was calculated usng Mun's gravity as a constant instead of 9.81 m/s... then the number you use for Isp would be 9.81/1.63 higher...

Isps are measured relative to Kerbin/Earth gravity/9.81m/s/s.... always use 9.81 m/s/s

You don't want to use a different Isp value for different celestial bodies (and indeed, gravity even varies with the distance to the body... so its completely impractical to use different values for g0.... its always 9.81... always)

Edited November 5, 2015 by KerikBalm

[radonek]

egogo,

You've got it figured out. The map you're using is accurate. DV to orbit is a rough estimate for atmospheric bodies and calculated fairly precisely for airless bodies.

Having said that...

You will actually need more than 580m/sec DV to land on the Mun safely. Perhaps as much as twice that depending on your technique. Landings are highly- inefficient processes.

Best,

-Slashy

No. Seriously. Sure, you need a bit more then what ideal "suicide burn with infinite TWR" numbers say, but if you need to double that, you are doing something very very wrong. Probably landing with TWR of 1.001?

[egoego]

Thanks all for your answers. I should have taken the time to read more into the toppic instead of just using the formula without knowing how/why it works.

[GoSlash27]

No. Seriously. Sure, you need a bit more then what ideal "suicide burn with infinite TWR" numbers say, but if you need to double that, you are doing something very very wrong. Probably landing with TWR of 1.001?

radonek,

A lot of people use a "stop 'n' drop" technique to land at a specific location. I've seen instances of this going upwards of 200% minimum. See kerikbalm's analysis above. My favorite technique for pinpoint landings exceeds 150% minimum. It's not "doing something wrong", it's simply a tradeoff.

If you want to land in a precise location, it's gonna cost you.

Best,

-Slashy

[Tatonf]

Consider using this delta-V page in the future for an always up to date map :

http://forum.kerbalspaceprogram.com/threads/96985

About how it is made, I think it's empirical.

For bodies without atmosphere, it's quite simple : just take the velocity at the indicated orbit, add a bit more for climbing up to it, and there you go.

For bodies with atmosphere, it's complicated. Considering you use atmospheric engines, do a nice gravity turn and have an aerodynamic rocket, you are almost 100% sure to go to orbit with the indicated delta-V (in vacuum). But some crazy design and ascent profile can do much better, I remember seeing domeone orbiting Kerbin with a bit les than 3000 m/s.

[GoSlash27]

tatonf,

DV to orbit is empirical for atmospheric bodies. Everything else is calculated and rounded to the nearest 10 m/sec.

For airless bodies it's the orbital velocity at sea level plus the DV required to Hohmann transfer to the specified low orbit altitude.

Best,

-Slashy