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something strange about the apollo program...


JtPB

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as a part of my munar-mission planning, i wanted to know what was exactly the profile of the apollo missions to copy this(they planned the missions as fuel-cheap as possible, for sure. this will be a key sentence as you will see below).

i assumed that they were at the moon-orbit that took the little dV for landing as possible, which according to my calculations is where H, the height of the orbit above the surface, is 2 times the radius of the body, independent on the body's mass, without considering the surface-rotation speed.

but that wasnt the parking orbit in apollo program. the orbit was at 110 km above the surface, which is clearly isnt the diameter of the moon(2 times the radius).

can anyone here explain this?:huh:

//i think that problem is symmetical, means that the dV that was required for landing is equal to the dV for taking-off to that height of orbit...

Edited by JtPB
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You should probably show us your calculations. If I'm not horribly mistaken, for minimal landing (and return) dv, you orbit as low as you dare. It's where your total energy is minimal among non-crashing orbits, and also the spot where old Oberth says your dv to energy change conversion rate is optimal.

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What calculations have you done? How can the delta-V required to land from a low circular orbit be higher than from a high circular orbit (without an atmosphere)? That's like saying that it takes more energy to get to a LEO than a GEO.

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as a part of my munar-mission planning, i wanted to know what was exactly the profile of the apollo missions to copy this(they planned the missions as fuel-cheap as possible, for sure. this will be a key sentence as you will see below).

No, they didn't. Just look at the launch windows, they were set-up in a way so that the TLI also came with a plane-change. reason was: if the first TLI-oportunity was missed for whatever reason, they wanted to have a second oportinity (and the orbit-plane angle changed relative to Apollos orbitplane with each orbit). So NASA set the launch incliniation in a way that the 2nd TLI window was within the .6° maximum plane-change capability; but this made the plange-angle for the 1st TLI opportunity also non-zero).

Then we had the Free-return-strategy: This is not the most fuel-efficient way for reaching lunar-orbit (but has an enourmous gain in security, as a SPS failure was possible).

i assumed that they were at the moon-orbit that took the little dV for landing as possible, which according to my calculations is where H, the height of the orbit above the surface, is 2 times the radius of the body, independent on the body's mass, without considering the surface-rotation speed.

Why? I didn't do the math but generally speaking: a lower orbit means less dV for the lander. Thats also why they changed the landing-procedure with Apollo 14: They used the SPS-engine to lower the Orbit (to about 17 km before undocking the lander (after undocking, the CM returned this his higher parking Orbit).

but that wasnt the parking orbit in apollo program. the orbit was at 110 km above the surface, which is clearly isnt the diameter of the moon(2 times the radius).

can anyone here explain this?:huh:

Not me - but on the one hand I'm not really sure if your calculations are correct, and on the other hand: The flight-time restrains along with the free-return trajectory put an narrow margin for where the perilune could possibly be. As far as I know, there were two other options: on with a flight time greater than 80 hours and one with a flight-time <60 hours, the later beeing inefficient as hell.

fly safe!

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You should probably show us your calculations. If I'm not horribly mistaken, for minimal landing (and return) dv, you orbit as low as you dare. It's where your total energy is minimal among non-crashing orbits, and also the spot where old Oberth says your dv to energy change conversion rate is optimal.
What calculations have you done? How can the delta-V required to land from a low circular orbit be higher than from a high circular orbit (without an atmosphere)? That's like saying that it takes more energy to get to a LEO than a GEO.
Maybe the maps showed a mountain in the way or something? Or you've did a mistake in your calculations.
No, they didn't. Just look at the launch windows, they were set-up in a way so that the TLI also came with a plane-change. reason was: if the first TLI-oportunity was missed for whatever reason, they wanted to have a second oportinity (and the orbit-plane angle changed relative to Apollos orbitplane with each orbit). So NASA set the launch incliniation in a way that the 2nd TLI window was within the .6° maximum plane-change capability; but this made the plange-angle for the 1st TLI opportunity also non-zero).

Then we had the Free-return-strategy: This is not the most fuel-efficient way for reaching lunar-orbit (but has an enourmous gain in security, as a SPS failure was possible).

Why? I didn't do the math but generally speaking: a lower orbit means less dV for the lander. Thats also why they changed the landing-procedure with Apollo 14: They used the SPS-engine to lower the Orbit (to about 17 km before undocking the lander (after undocking, the CM returned this his higher parking Orbit).

Not me - but on the one hand I'm not really sure if your calculations are correct, and on the other hand: The flight-time restrains along with the free-return trajectory put an narrow margin for where the perilune could possibly be. As far as I know, there were two other options: on with a flight time greater than 80 hours and one with a flight-time <60 hours, the later beeing inefficient as hell.

fly safe!

well i thought you woould guess the calculation procedure because its pretty simple: generaly minimizing the dV(H) function which is consist of the orbit velocity part and the velocity the lander will reach when it hit the surface when the orbit's velocity is after zeroing. surface speed is not included in the formula cuze simplicity reasons.(the surface velocity on the moon is negligible anyway, 4 m/s approx)

anyway, thats what i did:

(when 'u' is the gravitional parameter GM of the body, 'R' is the radius(perfect sphere assumed) of the body, and 'h' is the orbit's height)

orbit velocity: V = sqrt(u/(R+h))

lander's impact velocity: Vi = sqrt(2u(1/R - 1/(R+h))

adding them together is: VL = sqrt(u/(R+h)) + sqrt(2u(1/R - 1/(R+h))

we can extract the u parameter which what will lead us to this:

(u^2)(sqrt(1/(R+h)) + sqrt(2(1/R - 1/(R+h)))

we will derive this thing, but before lets do some things to simplify the buisness:

1/(R+h) = x, 1/R = t

leaving us with:

sqrt(x) + sqrt(2)sqrt(t - x) //u^2 is omitted because the derivation that will be afterwards

derivation(d/dx):

1/2sqrt(x) - sqrt(2)/2sqrt(t-x)

transffering members & multiplaying by common denominator:

2sqrt(t-x) = 2sqrt(2)sqrt(x)

reduction by 2:

sqrt(t-x) = sqrt(2)sqrt(x)

square raise:

t-x = 2x -> t = 3x

remember that t = 1/R and x = 1/(R+h), and the above equation becomes

1/R = 3/(R+h)

multiply by common denominator again:

R+h = 3R -> h = 2R

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Care to explain your equation for the lander's impact velocity, it doesn't make any sense to me.

Either way, at PE, your lander's velocity relative to the surface (ignoring surface rotation) will be at a minimum the orbital velocity for a circular orbit with a height of zero above the surface. The higher and higher the Apoapsis is, the higher its velocity at perapsis is.

Your lander will have the greatest dV requirements the more it deviates from a circular orbit of height zero.

Edited by KerikBalm
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Your calculation describes the following landing profile:

- Stop the orbital velocity completely.

- Drop straight to the ground.

- Stop exactly at ground level.

But consider the following landing profile:

- Burn retrograde until the periapsis is at ground level.

- Coast to periapsis (ground level).

- stop exactly at periapsis (ground level).

The second landing profile is more efficient, you will need much less deltaV.

If you optimize the second landing profile, you see that the lowest possible orbit is the best.

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The real Moon's gravity is quite irregular, and the computers of the time couldn't fully simulate this, so the mission had to be prepared to make corrections. This quite possibly factored into the choice of orbits. Not a problem in KSP.

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Care to explain your equation for the lander's impact velocity, it doesn't make any sense to me.

Either way, at PE, your lander's velocity relative to the surface (ignoring surface rotation) will be at a minimum the orbital velocity for a circular orbit with a height of zero above the surface. The higher and higher the Apoapsis is, the higher its velocity at perapsis is.

Your lander will have the greatest dV requirements the more it deviates from a circular orbit of height zero.

the lander's impact velocity formula expands the case when an object is falling at constant field of gravity, which is constant acceleration. at the lander's case, you just calculate the work of the gravitional force along the height of the falling(energy of the lander at the beginning of falling, energy of the lander at surface) and the velocity at impact is calculated according to E/W = (mv^2)/2

Your calculation describes the following landing profile:

- Stop the orbital velocity completely.

- Drop straight to the ground.

- Stop exactly at ground level.

But consider the following landing profile:

- Burn retrograde until the periapsis is at ground level.

- Coast to periapsis (ground level).

- stop exactly at periapsis (ground level).

The second landing profile is more efficient, you will need much less deltaV.

If you optimize the second landing profile, you see that the lowest possible orbit is the best.

of course, i knew that the lander cant braking its speed immediately(iassumed this only when it braking the orbit velocity, which is completly worse assumtion, but when it goes downwards it must slow down it's speed sometime, and i suppose that this dV is anyway equal to the impact velocity. or i really dont know how they calculated all those things...

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of course, i knew that the lander cant braking its speed immediately(iassumed this only when it braking the orbit velocity, which is completly worse assumtion, but when it goes downwards it must slow down it's speed sometime, and i suppose that this dV is anyway equal to the impact velocity. or i really dont know how they calculated all those things...

It has nothing to do with "breaking speed immediately". You can do that in calculations and get valid results. My reply hat literally nothing to do with that. I was talking about something completely different.

You can do a more efficent landing, not by stoping your orbital velocity completely (So that your apoapsis is at the center of the moon) and then dropping down, but by stopping your orbital velocity just to the point were your apoapsis is at ground level.

Edited by N_las
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Z-Man had the right hunch: the extremum you found is actually a maximum. Here's a graph for KSP's Mun (R=200 km, μ=6.5138e10 m^3/s^2).

2Wyfp46.png

The black curve is the total delta-v spent with this landing strategy. It peaks at 400 km (2R, as you correctly found).

Here's the corresponding graph for the better landing strategy which has been mentioned in this thread: first lower the periapsis to the surface, coast to the surface (i.e., periapsis) and then kill all your velocity.

tqZVLb4.png

Note that both landing strategies assume that TWR is large enough that velocity changes are nearly instantaneous. If this isn't true, expect landing costs to be higher than those calculated here.

Doing a "vertical drop" from high altitude is a bad idea since the higher you start the more gravity accelerates you on your way down, which is extra speed you have to kill to land. And for small altitudes the vertical drop is still not the best idea because it's more efficient to kill your orbital speed (which you have to do at some point either way) when you're as low as possible in the gravitational field (yes, the good ol' Oberth effect).

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It has nothing to do with "breaking speed immediately". You can do that in calculations and get valid results. My reply hat literally nothing to do with that. I was talking about something completely different.

You can do a more efficent landing, not by stoping your orbital velocity completely (So that your apoapsis is at the center of the moon) and then dropping down, but by stopping your orbital velocity just to the point were your apoapsis is at ground level.

i thought that the dV is independent on the deorbit strategy you will choose and is determined by this formula. the "high drop" method is more simple, however. Im not expert enough to use the other one...

Z-Man had the right hunch: the extremum you found is actually a maximum. Here's a graph for KSP's Mun (R=200 km, μ=6.5138e10 m^3/s^2).

http://i.imgur.com/2Wyfp46.png

The black curve is the total delta-v spent with this landing strategy. It peaks at 400 km (2R, as you correctly found).

Here's the corresponding graph for the better landing strategy which has been mentioned in this thread: first lower the periapsis to the surface, coast to the surface (i.e., periapsis) and then kill all your velocity.

http://i.imgur.com/tqZVLb4.png

Note that both landing strategies assume that TWR is large enough that velocity changes are nearly instantaneous. If this isn't true, expect landing costs to be higher than those calculated here.

Doing a "vertical drop" from high altitude is a bad idea since the higher you start the more gravity accelerates you on your way down, which is extra speed you have to kill to land. And for small altitudes the vertical drop is still not the best idea because it's more efficient to kill your orbital speed (which you have to do at some point either way) when you're as low as possible in the gravitational field (yes, the good ol' Oberth effect).

can you define "large enough TWR"?

and i dont see the graphs

// i assumed that the dV is eqaul for both landing and taking-off with these methods to the same height of parking orbit. see the note i edited in the first post.

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i thought that the dV is independent on the deorbit strategy you will choose

I think I can see why you'd think that: what's independent of the landing strategy is the required change in orbital energy. You start in orbit with an energy E0, and after landing your energy will be E1, and the difference E0-E1 is completely independent of the landing trajectory (since gravity is a conservative force). But a difference in orbital energy is not directly equivalent to a difference in velocity (a delta-v)! The Oberth effect is a great example of this non-equivalence: by burning deeper in a gravitational field, you produce a larger change in orbital energy with the same delta-v.

Said plainly: while the required change in orbital energy is indeed independent of landing strategy, the delta-v used isn't, and delta-v is directly related to fuel usage so that's the quantity that matters.

can you define "large enough TWR"?

Technically, both strategies assume velocity changes are exactly instantaneous, which means they assume infinite TWR. In practice, how large a TWR can effectively be considered "large enough" for purposes of these calculations will depend on the specifics of the landing profile and so it's hard to tell.

One thing that can be said though is that the impact of low TWR will be worse for the "vertical drop", since you'll be thrusting vertically (which is inefficient, since part of your thrust is used to fight gravity instead of decreasing your speed).

and i dont see the graphs

Weird. Did you click on the "Spoiler" links?

// i assumed that the dV is eqaul for both landing and taking-off with these methods to the same height of parking orbit. see the note i edited in the first post.

Sure, with perfect flying landing and takeoff costs are identical, but I'm not talking about takeoff here, just the landing part.

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i thought that the dV is independent on the deorbit strategy you will choose and is determined by this formula. the "high drop" method is more simple, however. Im not expert enough to use the other one...

It can be easily shown that the dV is not independent of the deorbit strategy. You just have to choose a ridiculous landing strategy to see that. We could imagine a "landing" strategy that starts with a transfer to Jool, followed by a short pause at KSC to get new snacks, and then we touch down on Muns surface. If dV would be independent of landing strategy, this would use just as little dV as a direct landing on Muns surface.

Or if we reverse the situation. We can show that the dV from surface to orbit is not a constant value, but is dependent on the start strategy. Gravity Turn is more dV efficient than simply burning up until you reach orbital height and then burning sideways until you reach orbital velocity.

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Technically, both strategies assume velocity changes are exactly instantaneous, which means they assume infinite TWR. In practice, how large a TWR can effectively be considered "large enough" for purposes of these calculations will depend on the specifics of the landing profile and so it's hard to tell.

One thing that can be said though is that the impact of low TWR will be worse for the "vertical drop", since you'll be thrusting vertically (which is inefficient, since part of your thrust is used to fight gravity instead of decreasing your speed).

is the LV-909 enough for the most of the cases?

Weird. Did you click on the "Spoiler" links?

yep. and the link of the image. just a blank page with (1x1) as a title

Sure, with perfect flying landing and takeoff costs are identical, but I'm not talking about takeoff here, just the landing part.

so what you are saying is that if i reverse the taking-off(which i of course do as an horizontal burn, mainly) ill get the correct landing procedure? but how do i verify the wanted landing site? what will i search on youtube to see a tutorial for this? is it like 'suicide burn' when you always point your ship at retrograde?

It can be easily shown that the dV is not independent of the deorbit strategy. You just have to choose a ridiculous landing strategy to see that. We could imagine a "landing" strategy that starts with a transfer to Jool, followed by a short pause at KSC to get new snacks, and then we touch down on Muns surface. If dV would be independent of landing strategy, this would use just as little dV as a direct landing on Muns surface.

Or if we reverse the situation. We can show that the dV from surface to orbit is not a constant value, but is dependent on the start strategy. Gravity Turn is more dV efficient than simply burning up until you reach orbital height and then burning sideways until you reach orbital velocity.

didnt understand your 'suggested' procedure... we talking about the methods that start on the parking orbit around the moon and do not goes out of it...

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didnt understand your 'suggested' procedure... we talking about the methods that start on the parking orbit around the moon and do not goes out of it...

If dV is independent of landing procedure, then it wouldn't matter if we leave the Moon or not. But even if you enforce the rule "don't leave the moon", we could invent a stupid landing procedure to show that dV isn't independent:

Imagine you would cancel all your orbital velocity, and then you drop to the surface. But you burn the whole time, so that you drop incredibly slow, and you reach the surface only after one million years of incredibly slow dropping. Clearly this landing procedure takes much more dV than other procedures, since it involves burning for millions of years. The landing procedure DOES matter, and the dV isn't independent.

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is the LV-909 enough for the most of the cases?

It depends on the mass of your ship and the number of LV-909s, as well as the balance between engine Isp and mass. It's complicated.

The bottomline is that it's all a matter of thrust-to-weight ratio: the higher it is, the closer you can get to the optimal landing trajectory (in terms of delta-v). While it's possible to land with a TWR less than one, it becomes rather expensive.

yep. and the link of the image. just a blank page with (1x1) as a title

Weird, they work fine on my browser. Here are direct links to the graphs:

Delta-v cost for "vertical drop" landing strategy

Delta-v cost for "lower periapsis" landing strategy

so what you are saying is that if i reverse the taking-off(which i of course do as an horizontal burn, mainly) ill get the correct landing procedure?

Exactly. The most efficient landing strategy is the same as the most efficient takeoff strategy, just in reverse: always try to burn as horizontal as possible. Doing that minimizes delta-v lost to fighting gravity ("gravity losses").

what will i search on youtube to see a tutorial for this? is it like 'suicide burn' when you always point your ship at retrograde?

You're in luck: some time ago Kosmo-not posted a very nice video showing how to land using a "constant-altitude trajectory":

As you can see, it's not a vertical suicide burn. It's rather much like a "horizontal takeoff" in reverse. The fundamental idea is to control your vertical speed by pitching your spacecraft. At the start you'll need very little pitch to keep your descent speed low and thus will be thrusting almost horizontally, which means your thrust is used almost entirely to decelerate instead of fighting gravity.

but how do i verify the wanted landing site?

It's indeed hard to control your landing site since the landing trajectory is long and it's difficulty to predict exactly when to start the landing burn. So in practice one usually trades off a bit of efficiency for more control over the landing spot (by doing a slightly more "vertical" landing profile).

What I normally do is lower the periapsis so it's close to the surface and past my intended landing site. I figure out approximately how much time I need to burn to kill my orbital velocity and use that to determine when to start burning. When I'm close to the landing site, I then use pitch to fine tune where I will touch down. It's a matter of practice, and one can get fairly close to the figures quoted in delta-v maps.

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Also keep in mind that unless you want to visit an anomaly/easter egg site or perform a surface rendezvous, your landing precision isn't that important. Very few airless bodies actually have you passing over the surface at any significant speed (hello Tylo) in a low orbit, and even tiny biomes like Midland Craters or the like are many square kilometers across. Even in my most noobish of days I have never missed my intended target so far that I ended up in a different biome!

And if all else fails, you can still do a hybrid approach... go into a very low circular orbit (between 8 and 10 km). Then, instead of planning a maneuver node on the opposite side of the Mun that makes the periapsis touch the ground at your intended landing site, push the maneuver node closer to your intended target and increase the dV so that the periapsis disappears into the ground slightly behind your intended target. The further forward (closer to landing) you push the maneuver node, the more dV you will have to invest to achieve this trajectory, but also the steeper the descent curve will be. It's worth noting that this approach is often more useful in practical application than the idealized variant anyway, since the idealized variant ignores the fact that the Mun has topography and will sometimes result in plastering yourself into a mountain or crater rim while coasting to periapsis.

Since you're now just slightly overshooting, you just coast down and, when you're directly above where you want to be, just perform the old "kill horizontal velocity, drop straight down" maneuver. At this point you'll be only maybe a kilometer above the surface and moving very slowly already, so you won't really loose much in the way of efficiency.

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i have the trajectories-prediction mod, which tells me where will i impact by considering the rotation of the body.

also can i just burn in radial direction(away from the surface. yep i know this is bad idea for using in dV, but nonetheless) to land?

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Your calculation describes the following landing profile:

- Stop the orbital velocity completely.

- Drop straight to the ground.

- Stop exactly at ground level.

But consider the following landing profile:

- Burn retrograde until the periapsis is at ground level.

- Coast to periapsis (ground level).

- stop exactly at periapsis (ground level).

The second landing profile is more efficient, you will need much less deltaV.

If you optimize the second landing profile, you see that the lowest possible orbit is the best.

Second profile is, in fact, the optimal policy if you allow for infinite thrust. (It is a bit of work to derive.) Furthermore, if we now vary height as a parameter, we see that the first-order optimal condition has no real solution. That means that the optimal altitude for the orbit is at one of the boundaries, either surface elevation or +inf. Simple algebraic check confirms that you want your orbit as low as possible to minimize delta-V requirement. QED.

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the lander's impact velocity formula expands the case when an object is falling at constant field of gravity, which is constant acceleration.

This is only (approximately) true for very small changes in altitude. Otherwise, the acceleration due to gravity is not constant. (g® = GM/R^2.) I'm guessing you probably know this but just misspoke at the time. Either way, calculating impact velocity based on change in gravitational potential energy, which is what I think you did, is correct.

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i have the trajectories-prediction mod, which tells me where will i impact by considering the rotation of the body.

also can i just burn in radial direction(away from the surface. yep i know this is bad idea for using in dV, but nonetheless) to land?

Only if you're starting from a dead stop.

Landing from orbit is about neutralising two velocities: horizontal and vertical. You need 'em both near zero at touchdown.

Point retrograde, fire up the thrusters, and tweak your pitch up and down to control your vertical drop rate while you're zeroing out your horizontal velocity. Ideally, you want to spend as little as possible fuel on the vertical (because you'll just get it back from gravity if you kill it too soon) and focus as much as possible on the horizontal (once that's gone it's gone for good).

The shallower an angle that you come in at, the less fuel you waste fighting gravity. However, it takes a bit of finesse to keep things cheerful when you're 500m off the deck but still whizzing sideways at 500m/s. Especially when there are hills around.

Edited by Wanderfound
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