Iron Hill Project Thread

[TMac3000]

I think part of the problem is that what I showed in the pic isn't really one ship, but four ships docked together.

On the CTD, I don't understand either. :shrug: I'm having no problem using the Dragonfly.
Let's see what your UCGO lander looks like. We can use that if there are no compatibility issues

 

[sorindafabico]

@Bruce: IMFD calculates the best parking orbit equatorial inclination using the data in Course program. It doesn't matter if you launch (or insert) the ship in the parking orbit a few hours or months before the injection. Try it by yourself

I'll make some screenshots when I get back to home, maybe i'm not explaining well.

---------- Post added at 07:37 PM ---------- Previous post was at 07:32 PM ----------

PS: AFAIK, it's the inclination relative to the ecliptic that slowly changes when you're on LEO, not the equatorial inclination.

 

[BruceJohnJennerLawso]

@Bruce: IMFD calculates the best parking orbit equatorial inclination using the data in Course program. It doesn't matter if you launch (or insert) the ship in the parking orbit a few hours or months before the injection. Try it by yourself

I'll make some screenshots when I get back to home, maybe i'm not explaining well.

---------- Post added at 07:37 PM ---------- Previous post was at 07:32 PM ----------

PS: AFAIK, it's the inclination relative to the ecliptic that slowly changes when you're on LEO, not the equatorial inclination.

Ummm I really hate to argue about something like this, but Im sure orbital inclination is constant in both the ecliptic & equatorial. A spacecrafts orbit is determined by some atm drag, its velocity, & the gravitational pull of the earth, none of which (save maybe the drag a little) are dependent on the rotation of the earth as a rigid body. What I think youre saying is that either Discoveries current orbit, or the one IMFD is proposing (~41 degrees?) have a line of intersection with the ecliptic that almost perfectly matches earths prograde vector. since with any orbit, that can only happen twice a year (and having one of those times line up with our launch window in equatorial orbit would be a miracle), I would think youre referring to a slightly changed inclination orbit that IMFD likes the best. No issue there, but it probably wants to have us make a burn do get that inclination right? sometime over the weekend Im going to run a rough complete simulation of our mission in orbiter just to clear things up

 

[TMac3000]

Guys, I'm telling you, Discovery is already in a near-equatorial orbit. The plane-change dV from her current orbit is very small

Wait...we are talking about the trans-Mercury injection, right?

 

[sorindafabico]

I feel more comfortable explaining with pictures, since, even after two years playing Orbiter, I still do not fully understand how the interplanetary transfers do work, and my approach is empirical.

Here's a picture of a Shuttle-PB on the ground. Just to find the BLL with IMFD's Surface Launch screen. To my current understanding of IMFD, BLL equals the best inclination of the parking orbit for the transfer planned in Course program.

After that, I did put the Shuttle-PB in the parking orbit with scenario editor. First to the "strange" inclination indicated by IMFD (~49 degrees equatorial).

These are the delta-v predictions. The ejection costs less than 10.5 km/s. I don't know if it needs some correction after leaving Earth's SOI, I'm just working with orbital ejection here.

Then, I used scenario editor to put the ship in an equatorial orbit - like the actual orbit of Discovery:

The delta-v matches that of the Course program, but there's a strange increase of the dVp vector.

Finally, with the ship in ecliptic plane:

Even higher delta-v than that of the equatorial orbit.

I have to run a full simulation of the transfer to measure the MCCs, but, currently, the 49 degrees orbit seems the best option.

 

[BruceJohnJennerLawso]

I feel more comfortable explaining with pictures, since, even after two years playing Orbiter, I still do not fully understand how the interplanetary transfers do work, and my approach is empirical.

Here's a picture of a Shuttle-PB on the ground. Just to find the BLL with IMFD's Surface Launch screen. To my current understanding of IMFD, BLL equals the best inclination of the parking orbit for the transfer planned in Course program.

After that, I did put the Shuttle-PB in the parking orbit with scenario editor. First to the "strange" inclination indicated by IMFD (~49 degrees equatorial).

These are the delta-v predictions. The ejection costs less than 10.5 km/s. I don't know if it needs some correction after leaving Earth's SOI, I'm just working with orbital ejection here.

Then, I used scenario editor to put the ship in an equatorial orbit - like the actual orbit of Discovery:

The delta-v matches that of the Course program, but there's a strange increase of the dVp vector.

Finally, with the ship in ecliptic plane:

Even higher delta-v than that of the equatorial orbit.

I have to run a full simulation of the transfer to measure the MCCs, but, currently, the 49 degrees orbit seems the best option.

Sigh :hmm:

I have not touched IMFD in my life, but my guess is that it attempts to plan the trajectory it thinks works best, and gives you burn params that work for that particular trajectory. I dont recognize almost any of the anocryms that IMFD is using to describe the flight in your screenshots, but I think somewhere in there, IMFD is working agianst itself slightly. I will try a bare-bones flight plan test in Orbiter this weekend, but heres all I can say I know absolutely so far:

-Off-plane and on-plane transfers both work, but anything off-plane has to be at the right angle to the ecliptic for it to work. For example, a polar orbit can work perfectly for a lunar transfer, only twice a month, 3 days before the moon intersects that orbits plane.
-IMFD may be trying to help correct for us, as Mercuries orbit does have a 7 degree inclination to the ecliptic. Our encounter must be out of plane with mercury, If we only rendezvous properly top down (ie looking at the solar system from above). Im just not sure why it wants to add 49 degrees equatorial, as that would at least be 26 degrees past the ecliptic, and 19 past the correction for mercuries orbit

I highly suggest you try the videnie tool off OHM to help get a better feel for orbital mechanics & inclination. Its a complicated thing to visualize, but once you get it, youll be able to plan any type of mission you want perfectly. Stick with it, and maybe consider switching to TransX :tiphat:

 

[sorindafabico]

I'll do a simulation to see how much delta-v is needed to reach Mercury with each parking orbit inclination.

NOTE: the 49 degrees value is not static. It changes if you input different data in Course window.

 

[BruceJohnJennerLawso]

I'll do a simulation to see how much delta-v is needed to reach Mercury with each parking orbit inclination.

NOTE: the 49 degrees value is not static. It changes if you input different data in Course window.

What data?

 

[sorindafabico]

TEj and TIn, i.e., when you leave Earth and when you reach Mercury.

 

[BruceJohnJennerLawso]

TEj and TIn, i.e., when you leave Earth and when you reach Mercury.

Ah, so IMFD will still try to produce an answer for you, even if its way off any good Hohman transfer windows. Are the tests you ran above set to our expected launch & arrival times with discovery?

 

[sorindafabico]

I was testing now, but I had a CTD some "days" after a MCC.

The ejection cost was ~10,000 m/s (low). In the worst case, I was expecting a huge MCC to adjust the trajectory, but the first was only ~8 m/s. IMFD's Map screen - which is FAR more precise than OrbitMFD - was indicating a PeA of ~55,000 km on Mercury after that. I was planning a second MCC of ~7 m/s that would push my PeA to ~7,000 km.

I can run a full simulation (damn CTD), but I doubt the MCCs will cost more than the difference to the equatorial or ecliptic escape maneuvre. The only thing I didn't take note was the arrival velocity - but I don't know if the inclination of tha parking orbit makes so much difference in the arrival velocity.

Yep, I didn't understand why this inclination seems more cost-effective, maybe something related to nodes.

---------- Post added at 07:17 PM ---------- Previous post was at 07:15 PM ----------

Ah, so IMFD will still try to produce an answer for you, even if its way off any good Hohman transfer windows. Are the tests you ran above set to our expected launch & arrival times with discovery?

TMac is planning TEj = 56280 and TIn = 56385. I'm using these dates.

 

[TMac3000]

Yikes:blink:
You guys are going to have to translate for me. This is getting a little over my head:lol:

 

[BruceJohnJennerLawso]

I was testing now, but I had a CTD some "days" after a MCC.

The ejection cost was ~10,000 m/s (low). In the worst case, I was expecting a huge MCC to adjust the trajectory, but the first was only ~8 m/s. IMFD's Map screen - which is FAR more precise than OrbitMFD - was indicating a PeA of ~55,000 km on Mercury after that. I was planning a second MCC of ~7 m/s that would push my PeA to ~7,000 km.

I can run a full simulation (damn CTD), but I doubt the MCCs will cost more than the difference to the equatorial or ecliptic escape maneuvre. The only thing I didn't take note was the arrival velocity - but I don't know if the inclination of tha parking orbit makes so much difference in the arrival velocity.

Yep, I didn't understand why this inclination seems more cost-effective, maybe something related to nodes.

---------- Post added at 07:17 PM ---------- Previous post was at 07:15 PM ----------



TMac is planning TEj = 56280 and TIn = 56385. I'm using these dates.

Ah anocrym overload :lol:

10 Kps is reasonably good, but optimal DV would be about 9.5 Kps. The extra 500 m/s probably is unavoidable, angle change to ensure we intersect mercuries orbit, which has an ecl inc of 7 degrees, at the right point. It is fairly hard to explain why that inclination works, but I imagine it is right.

 

[TMac3000]

When you first pull up a target planet in the IFMD Course program, the data you see is not a good Hohmann window--if you try to launch with it, you will end up lost in space. That's why you have to lock TOF, and adjust the arrival date to minimize the total dV

Then the inclination you get should make sense

 

[BruceJohnJennerLawso]

When you first pull up a target planet in the IFMD Course program, the data you see is not a good Hohmann window--if you try to launch with it, you will end up lost in space. That's why you have to lock TOF, and adjust the arrival date to minimize the total dV

Then the inclination you get should make sense

now thats better! :lol:

And I thought Transx was bad when I was learning it.

By the way, Im going to try to post my utility satelitte here this evening.

 

[sorindafabico]

When you first pull up a target planet in the IFMD Course program, the data you see is not a good Hohmann window--if you try to launch with it, you will end up lost in space. That's why you have to lock TOF, and adjust the arrival date to minimize the total dV

Then the inclination you get should make sense

I'm using the dates planned for Discovery: Dec 19 (MJD 56280) for TEj and Apr 03 (MJD 56385) for TIn.

Use scenario editor to test by yourself: move in time to a few hours before TEj, put the ship in 49.13 degrees equatorial and input the data in IMFD

---------- Post added at 09:45 PM ---------- Previous post was at 08:24 PM ----------

Done.

Here's the delta-v budget:

From Earth escape to Mercury SOI (trans-hermian injection + MCCs): 10205 m/s. MCCs are less than 1% of the cost.

Mercury orbit insertion: 24806 m/s (TOUGH! It was my first time on Mercury). 70x70 km barely polar orbit.

Total delta-v: 35011 m/s.

 

[TMac3000]

It's interesting to note that HOI takes about twice as long as THI. It's a combination of picking up a lot of speed as you fall toward the Sun, and the wide difference between Earth's mass and Mercury's. We're already going fast just to leave Earth, then we pick up a buttload of velocity during the trip, and then we have to brake into Mercury's pea-sized gravity well.

When you combine all those factors, the insertion burn at Mercury is longer than a launch burn to Saturn!

 

[sorindafabico]

This was the biggest cause of our maps of Mercury were half empty until 2011.

The orbit insertion is hard. You have a lot of burn time into a very very small window. My ship was a million kilometers away just 24 hours before PeT!

 

[TMac3000]

Total delta-v: 35011 m/s

And that's just one way. A little bit less to come back, since Mercury is small and Earth is big. Total trip dV: I'd say around 60 km/s.

And THAT'S not counting the dV needed for the MESSENGER rendezvous we will perform some time in the second week after arrival

---------- Post added at 08:17 PM ---------- Previous post was at 08:15 PM ----------

Oh, and it will cost more fuel than you might think to get 35 km/s of dV, because I don't think you included the mass of the cargo and crew

 

[sorindafabico]

I used a shuttle-pb just to measure the delta-v from escape to orbit insertion. The Sh-PB doesn't have delta-v for a round trip.

Anyway, there's a significant cost reduction with this strange 49 degrees parking orbit (~3,000 m/s). I would like to know why...