[1.2] Galileo's Planet Pack (development thread) [v0.9]

[Jiraiyah]

5 minutes ago, OhioBob said:

Part of it could be that they're figuring the transfer from a lower orbit.  You used 80 km, but they could have used 70 km (the absolute minimum).  In that case I compute 1111 m/s, so I still don't get 1115 m/s.  Also in older versions of KSP, Kerbin's sidereal period was 6 hours (rather than it's current solar day), which made the stationary altitude a little higher.  But even taking that into account I still compute a value of only 1111.5 m/s.  I don't know where 1115 m/s comes from.

oh well, although the maps say they are calculating the low kerbin orbit as 80 Km, but i think nothing beats the math, so i assume we are the correct people here, meh, even with all the difference, 16 m/s as total is nothing i would be concerned about simply because well, as you mentioned, i would over reengineer my rocets at least with extra 300 tp 400 m/s simply because there is 75% chance i would screw up but anyways, review that file i sent and see if there is anything that would need fix if not, then there are few more formulas i may add here and there and it will get finished

for time being, good night

Edited October 31, 2016 by Jiraiyah

[OhioBob]

1 hour ago, JadeOfMaar said:

@OhioBob I would go with median, but any wise player would know to overengineer (even more?) a bit, just in case. As for my dV measurements I pad the values I get by just a little... Then again I'm not using career mode to test things.

I think what I'll use is a median launch window.  What I mean by that is that I'll compute the lowest Δv for several successive launch windows, and then take the median of those values.  For instance, launch windows for transfers from Gael to Tellumo occur about once every two years.  I'll determine what is the lowest transfer Δv is for each of those launch windows.  I'll then accumulate data for about 10-20 launch windows (spanning 20-40 years) and take the median of those numbers.  This way the Δv map will show a "typical" value if you launch at the optimum time during a launch window.  Some launch windows will be a little more and some will be a little less.  If you decide to launch at a sub-optimum time (i.e. days before or after the optimum launch window), then clearly the Δv requirement could be much higher.

How does that sound?  Anyone disagree?  I think this makes more sense than just using the absolute minimum Δv that might occur just once every few decades.  What good does that really do for you when you're planning a mission?

In some cases there could be very high Δv requirements even at a launch window.  I know that in the stock game some of the transfers to Moho can be ridiculous.  If that should occur here, I'll probably throw out the really bad ones and just use the transfers that are reasonable (assuming that no one will choose to launch during one of those really bad transfers).
 

Edited October 31, 2016 by OhioBob

[Galileo]

4 minutes ago, OhioBob said:

I think what I'll use a launch window median.  What I mean by that is that I'll compute the lowest Δv for several successive launch windows, and then take the median of those values.  For instance, launch windows for transfers from Gael to Tellumo occur about once every two years.  I'll determine what is the lowest transfer Δv is for each of those launch windows.  I'll then accumulate data for about 10-20 launch windows (spanning 20-40 years) and take the median of those numbers.  This way the Δv map will show a "typical" value if you launch at the optimum time during a launch window.  Some launch windows will be a little more and some will be a little less.  I you decide to launch at a sub-optimum time (i.e. days before or after the optimum launch window), then clearly the Δv requirement could be much higher.

How does that sound?  Anyone disagree?  I think this makes more sense than just using the absolute minimum Δv that might occur just once every few decades.  What good does that really do for you when you're planning a mission?

In some cases there could be very high Δv requirements even at a launch window.  I know that in the stock game some of the transfers to Moho can be ridiculous.  If that should occur here, I'll probably throw out the really bad ones and just use the transfers that are reasonable (assuming that no one will choose to launch during one of those really bad transfers).
 

makes sense to me!

[JadeOfMaar]

I agree fully as well. @OhioBob The Moho reference further explains to me why Moho is so much more dreadful, going by word of mouth, than I've yet experienced. I've laid out all the branches and colors, and most of the little circles for the dV map...

9 minutes ago, OhioBob said:

Δv

I see what you did there.

[OhioBob]

6 minutes ago, JadeOfMaar said:

I've laid out all the branches and colors, and most of the little circles for the dV map...

Sounds good.  I have no idea how long this is going to take me to complete.  There are a lot of numbers to compute and I've never done this before, so it could be awhile.

[Galileo]

1 minute ago, OhioBob said:

Sounds good.  I have no idea how long this is going to take me to complete.  There are a lot of numbers to compute and I've never done this before, so it could be awhile.

we have nothing but time

[OhioBob]

@JadeOfMaar, I'm planning to use this map as a guideline.  If you set up your map the same way, then hopefully my numbers will slot right in once I have them.

[Poodmund]

With respect to dV maps, I've actually always preferred reading it from an infographic listed like this: http://i.imgur.com/Vi8H41I.png Just food for thought.

2 hours ago, JadeOfMaar said:

I'll keep that in mind. It'll be quite welcome. I haven't opened KSPedia in a long time so I can't remember how those things look to readily decide if I want to make the GPP wiki stockalike. But I have decided I may likely omit the skybox I'm using in my wiki images as it may make the overall look of the thing too eccentric.

On that subject, @jandcando released v4 of the stock planet wiki. They have a cool sidebar now. http://imgur.com/a/H3RdP

For reference, here are a couple of shots of the ones I did bundled with OPM mimicking the stock layout:

Spoiler

 

Edited October 31, 2016 by Poodmund

[JadeOfMaar]

@Poodmund Aww. Slate's story sounds pretty similar to Thalia's. For her I wrote that she may have had an atmosphere but it was quite combustible and an impact ignited, seriously, ignited it.

As for your wiki entries, they're fabulous.

Blank map with aerobrake and jets-usable markers.

 

Edited November 1, 2016 by JadeOfMaar
Added SOI ranges and low orbit values

[OhioBob]

23 minutes ago, Poodmund said:

With respect to dV maps, I've actually always preferred reading it from an infographic listed like this: http://i.imgur.com/Vi8H41I.png Just food for thought.

I like that, it has a lot of useful information in it.  Some of it looks a lot like my personal notes.  We might be able to do something like that, but we're already set up for a more traditional Δv map, so I think we should go ahead and do that first.  We can always consider doing it the other way as well.
 

14 minutes ago, JadeOfMaar said:

Blank map with aerobrake and jets-usable markers.

Looks good.

[OhioBob]

4 hours ago, Jiraiyah said:

hmm, corrected that but still some what off ! (only 17 m/s but still why?)

Just thought of something.  If you want to make your orbit geostationary (rather than just geosynchronous), it will require a plain change.  For Kerbin the plain change is negligible because KSC is located almost on the equator, but on Gael KSC is 8.51 degrees off the equator.  This means that your parking orbit will be inclined at least 8.51 degrees.  Your Δv of 1098.92 m/s will get you into a geosynchronous orbit, but you'll still be inclined and will wander a little above and below the equator.  To be perfectly geostationary you need to remove the inclination so that you orbit above the equator.  If you add a 8.51 degree plane change into your second burn, that brings the total Δv up to 1113.58 m/s.  If you perform the plane change as a separate third burn, this adds 149.85 m/s and brings the total to 1248.77 m/s.

(edit)  Also note that 2863.3528 km is the GSO altitude using Gael's rounded off sidereal period.  If you use the exact sidereal period, the GSO altitude computes to 2863.3329 km.

Edited November 1, 2016 by OhioBob

[JadeOfMaar]

Updated image and previous post. I forgot Eta. But now all that's missing is numbers. The complete visual should help @OhioBob @Jiraiyah and @Galileo.

[Jiraiyah]

6 hours ago, OhioBob said:

Just thought of something.  If you want to make your orbit geostationary (rather than just geosynchronous), it will require a plain change.  For Kerbin the plain change is negligible because KSC is located almost on the equator, but on Gael KSC is 8.51 degrees off the equator.  This means that your parking orbit will be inclined at least 8.51 degrees.  Your Δv of 1098.92 m/s will get you into a geosynchronous orbit, but you'll still be inclined and will wander a little above and below the equator.  To be perfectly geostationary you need to remove the inclination so that you orbit above the equator.  If you add a 8.51 degree plane change into your second burn, that brings the total Δv up to 1113.58 m/s.  If you perform the plane change as a separate third burn, this adds 149.85 m/s and brings the total to 1248.77 m/s.

(edit)  Also note that 2863.3528 km is the GSO altitude using Gael's rounded off sidereal period.  If you use the exact sidereal period, the GSO altitude computes to 2863.3329 km.

O~K what is the math to calculate that inclination change?

Hmm what did i screw?

as you see the gravitational parameter for the gael's sidereal calculation is from Ciro, and i used the exact numbers you gave few post before for Gael's SMA, the result? funny !

Even with using the calculator for getting the precise parameter for Ciro, I get this :

Edited November 1, 2016 by Jiraiyah

[OhioBob]

1 hour ago, Jiraiyah said:

O~K what is the math to calculate that inclination change?

The equation to compute the inclination change as a separate burn is,

Δv = 2*v*sin(θ/2)

where v is the orbital velocity and θ is the plane change angle.

For example, at geosynchronous distance, v = 1009.81 m/s, so for a 8.51^o plane change we have

Δv = 2*1009.81*sin(8.51/2) = 149.85 m/s.

If the plane change is combined with an altitude change, then the equation is

Δv = SQRT[ v_i² + v_f² - 2*v_i*v_f*cos(θ) ]

where v_i is the initial velocity and v_f is the final velocity.

If we are transferring from a 80 km orbit to a geosynchronous orbit, then v_i is the apoapsis velocity of the transfer orbit, i.e. v_i = 578.54 m/s.  So we have,

Δv = SQRT[ 578.54² + 1009.81² - 2*578.54*1009.81*cos(8.51) ] = 445.93 m/s.

Without the plane change the Δv is simply, 1009.81 - 578.54 = 431.27 m/s.

 

1 hour ago, Jiraiyah said:

Hmm what did i screw?

What you computed is Gael's sidereal orbit period (426 days), not its rotation period.  Normally the sidereal rotation period is given, i.e. it is something measured, not computed.  (We computed it for the moons because they are tidally locked, i.e. rotation period = orbital period.)  From the planet's sidereal rotation period and its sidereal orbit period, we calculate the length of its solar day using the following formula

length of solar day = (sidereal orbit period * sidereal rotation period) / (sidereal orbit period - sidereal rotation period)

where we just have to make sure that everything is in the same units.

For example, Icarus has an orbital period of 319.5 hours and rotation period of 213 hours, therefore its solar day is,

length of solar day = (319.5*213) / (319.5-213) = 639 hours.

Gael is a little different in that we're given it solar day (6 hours exactly) rather than is sidereal rotation period.  We therefore must compute its sidereal rotation period using a rearranged version of the above equation,

Sidereal rotation period = (sidereal orbit period * length of solar day) / (sidereal orbit period + length of solar day)

We have,

Sidereal rotation period = (2556*6) / (2556+6) = 5.98594848 hours.

 

1 hour ago, Jiraiyah said:

Even with using the calculator for getting the precise parameter for Ciro, I get this :

Ciro's radius is 70980 km, not 70.98 km.  This makes its gravitational parameter 1.2751483209192E+18 m³/s².

 

Edited November 1, 2016 by OhioBob

[Jiraiyah]

9 hours ago, OhioBob said:

The equation to compute the inclination change as a separate burn is,

Δv = 2*v*sin(θ/2)

where v is the orbital velocity and θ is the plane change angle.

For example, at geosynchronous distance, v = 1009.81 m/s, so for a 8.51^o plane change we have

Δv = 2*1009.81*sin(8.51/2) = 149.85 m/s.

If the plane change is combined with an altitude change, then the equation is

Δv = SQRT[ v_i² + v_f² - 2*v_i*v_f*cos(θ) ]

where v_i is the initial velocity and v_f is the final velocity.

If we are transferring from a 80 km orbit to a geosynchronous orbit, then v_i is the apoapsis velocity of the transfer orbit, i.e. v_i = 578.54 m/s.  So we have,

Δv = SQRT[ 578.54² + 1009.81² - 2*578.54*1009.81*cos(8.51) ] = 445.93 m/s.

Without the plane change the Δv is simply, 1009.81 - 578.54 = 431.27 m/s.

 

What you computed is Gael's sidereal orbit period (426 days), not its rotation period.  Normally the sidereal rotation period is given, i.e. it is something measured, not computed.  (We computed it for the moons because they are tidally locked, i.e. rotation period = orbital period.)  From the planet's sidereal rotation period and its sidereal orbit period, we calculate the length of its solar day using the following formula

length of solar day = (sidereal orbit period * sidereal rotation period) / (sidereal orbit period - sidereal rotation period)

where we just have to make sure that everything is in the same units.

For example, Icarus has an orbital period of 319.5 hours and rotation period of 213 hours, therefore its solar day is,

length of solar day = (319.5*213) / (319.5-213) = 639 hours.

Gael is a little different in that we're given it solar day (6 hours exactly) rather than is sidereal rotation period.  We therefore must compute its sidereal rotation period using a rearranged version of the above equation,

Sidereal rotation period = (sidereal orbit period * length of solar day) / (sidereal orbit period + length of solar day)

We have,

Sidereal rotation period = (2556*6) / (2556+6) = 5.98594848 hours.

 

Ciro's radius is 70980 km, not 70.98 km.  This makes its gravitational parameter 1.2751483209192E+18 m³/s².

 

I Think I finally nailed it down?

something else made me confused, isn't formula for orbital velocity is sqrt(gravitational parameter / radius of orbit) ?

that formula give V = 1009.81 for stationary orbit, but for 80 km parking orbit (with summing it to 600Km radius) gives 2278.93 m/s, so how did you calculate that 578.54 m/s  ?

using SQRT (mu / r) i calculated the dV by that Δv = SQRT[ v_i² + v_f² - 2*v_i*v_f*cos(θ) ] and look at the result :

it's way different than what u say simply because the V at 80 km origin (680km) is not what you say?

Edited November 1, 2016 by Jiraiyah

[OhioBob]

4 hours ago, Jiraiyah said:

that formula give V = 1009.81 for stationary orbit, but for 80 km parking orbit (with summing it to 600Km radius) gives 2278.93 m/s, so how did you calculate that 578.54 m/s  ?

578.54 m/s it the velocity at apoapsis of the 80 km x 2863 km transfer orbit.  The variables v_i and v_f are the velocities immediately before and immediately after the second burn, i.e. the circularization burn that takes place at apoapsis of the transfer orbit.  Surely you already know how to compute that velocity because one of your spreadsheets shows the Δv of the second burn as 431.268 m/s, i.e. 1009.81 - 578.54 = 431.27 m/s.  Just use those same two velocities to compute the Δv with plane change.  The velocity at 80 km is irrelevant in this calculation.

[Jiraiyah]

1 minute ago, OhioBob said:

578.54 m/s it the velocity at apoapsis of the 80 km x 2863 km transfer orbit.  The variables v_i and v_f are the velocities immediately before and immediately after the second burn, i.e. the circularization burn that takes place at apoapsis of the transfer orbit.  Surely you already know how to compute that velocity because one of your spreadsheets shows the Δv of the second burn as 431.268 m/s, i.e. 1009.81 - 578.54 = 431.27 m/s.  Just use those same two velocities to compute the Δv with plane change.  The velocity at 80 km is irrelevant in this calculation.

Oh i see, i got them mixed up dang it lol space math is hard

[OhioBob]

Just now, Jiraiyah said:

Oh i see, i got them mixed up dang it lol space math is hard

Well, it is rocket science after all.

[Jiraiyah]

32 minutes ago, OhioBob said:

Well, it is rocket science after all.

you are awesome, got it working

[OhioBob]

4 hours ago, Jiraiyah said:

you are awesome, got it working

I still see one problem.  The total Δv for the three-burn method is,

667.652 + 431.268 + 149.846 = 1248.766 m/s

That part you got right.  First burn inserts you into the transfer orbit, the second burn circularizes the orbit at geosynchronous distance, and the third burn changes the plane to 0^o inclination.

However, the total Δv for the two-burn method is,

667.652 + 445.933 = 1113.585 m/s

In this scenario the second burn is a combination maneuver that both circularizes the orbit and changes the plane.  What you did is, 667.652 + 431.268 + 445.933 = 1544.853 m/s, which is incorrect.

Combining a plane change with an altitude change can result in a significant savings in Δv; however, getting the timing and angles exactly right to perform a two-burn strategy is more challenging.  With the three-burn method you just have to make sure that you are over the correct part of the planet when you reach apoapsis.  You don't have to worry about where the nodes are.  You just perform the circularization burn at apoapsis, and then you can take your time and perform the plane change when you reach one of the nodes.  With the two-burn method you not only have to make sure that apoapsis occurs over the correct part of the planet, but apoapsis must also coincide with one of the two nodes.

Also, technically it is the parking orbit that has the 8.51^o inclination and the destination orbit that has the 0^o inclination.  This is because KSC in GPP is located at a latitude of 8.51^o N.  Unless you fly some sort of dogleg trajectory during ascent, it's impossible to launch into an orbit with an inclination less than the latitude of the launch site.  If you fly due east out of KSC, you'll end up in an orbit with an inclination of 8.51^o.  The plane change is to change the inclination from 8.51^o to zero, which is required for a geostationary orbit.

I also don't know why you are showing V_i = 149.846. From beginning to end, you use four different orbits in the three-burn method, and three orbits in the two-burn method.

Three-burn method
Orbit #1:  r = 680 km, v = 2278.93 m/s, i = 8.51^o.
Orbit #2:  r_p = 680 km, r_a = 3463.333 km, v_p = 2946.58 m/a, v_a = 578.54 m/s, i = 8.51^o.
Orbit #3:  r = 3463.333 km, v = 1009.81 m/s, i = 8.51^o.
Orbit #4:  r = 3463.333 km, v = 1009.81 m/s, i = 0^o.

Burn #1:  Δv = 2946.58 - 2278.93 = 667.65 m/s
Burn #2:  Δv = 1009.81 - 578.54 = 431.27 m/s
Burn #3:  Δv = 2*1009.81*sin(8.51/2) = 149.85 m/s

Two-burn method
Orbit #1:  r = 680 km, v = 2278.93 m/s, i = 8.51^o.
Orbit #2:  r_p = 680 km, r_a = 3463.333 km, v_p = 2946.58 m/a, v_a = 578.54 m/s, i = 8.51^o.
Orbit #3:  r = 3463.333 km, v = 1009.81 m/s, i = 0^o.

Burn #1:  Δv = 2946.58 - 2278.93 = 667.65 m/s
Burn #2:  Δv = SQRT[578.54²+1009.81²-2*578.54*1009.81*cos(8.51)] = 445.93 m/s

(edit)  Is anybody else seeing + 431.268 as a hyperlink?  I have no idea why it's doing that.  Weird.
 

Edited November 1, 2016 by OhioBob

[Galileo]

13 minutes ago, OhioBob said:

(edit)  Is anybody else seeing + 431.268 as a hyperlink?  I have no idea why it's doing that.  Weird.

nope looks normal to me

[OhioBob]

13 minutes ago, Galileo said:

nope looks normal to me

Umm, very strange.  On my screen I'm seeing the characters "+ 431.268" as a hyperlink (blue underlined).  Although I get the hyperlink icon when I hover the mouse pointer over it, it doesn't actually link to anything.  When I switch to edit mode I see it as regular black text.  I've tried re-typing the text, but every time I click save it turns back into a hyperlink.  Oh well, I guess I'll just live with it (even though it's bugging me).

(edit) In this post too I'm seeing it as a hyperlink.

Edited November 1, 2016 by OhioBob

[Galileo]

1 minute ago, OhioBob said:

Umm, very strange.  On my screen I'm seeing the characters "+ 431.268" as a hyperlink (blue underlined).  Although I get the hyperlink icon when I hover the mouse pointer over it, it doesn't actually link to anything.  When I switch to edit mode I see it as regular black text.  I've tried re-typing the text, but every time I click save it turns back into a hyperlink.  Oh well, I guess I'll just leave with it (even though it's bugging me).

you should just link it to something irrelevant would be hilarious

Edited November 1, 2016 by Galileo

[OhioBob]

6 minutes ago, Galileo said:

you should just link it to something irrelevant would be hilarious

You mean something like this + 431.268

[Jiraiyah]

19 minutes ago, OhioBob said:

You mean something like this + 431.268

lmao, do you know what was the promo for that link? LIFE trailer, completely about space !!!!

by the way, i can't say how much i am grateful for your help, take a look at this final chart i hope it's solid rocket science now

Now i can go to sleep (1 AM here)m i am planning on developing a mod doing all this crazy math very soon, that is if i can find a way to get my hands on planets, their needed info, the moons the same blah blah, anyone know the api for it?

Edit : replaced the cells, now better look and feel

Edited November 1, 2016 by Jiraiyah