IMFD IMFD Why and How Come Tech Questions

[PaulG]

This isn't a post to question HOW to use IMFD. I can USE it, now, I want to UNDERSTAND interplanetary travel. I don't want to be a button pusher, I want to understand why and how things happen.

Please, read and comment on my understanding and answer my questions. I hope this will help others as well.

Ok, scenario. I want to go to Mars. I want to take a fuel efficient method. I realize that a tangential transfer is the most fuel efficient method because I'm ejecting/intercepting along the same vector as the planet (Q1. is that right?)

So, in IMFD, I use the tangential intercept program to get an idea of when I will eject. Now, I switch to target intercept mode because I want a little more control over what is going on. I set the ejection date to 55116.742 and the injection date to 55430.604...because that is what the tangential program suggested. Well, thats 314 days. I thought a tangential transfer was around 260 days for Mars (Q2. Why the 54 day difference).

Then, I notice something in the display. I'm in OffPlane mode. The EjA is 0.644 and InA is 2.417. Thats pretty close to tangential. I can play a little with the start and end dates, but really, these aren't moving too much.

Now, Off-Plane mode means I'm moving neither in the plane of Earth nor the plane of Mars, but rather a "new plane" subject to the start and end points that lies somewhere between the two planes.

If I switch to Two plane mode, that means I travel along the plane of Earth, and at the node between Earth's plane and Mars' plane, I perform a plane correction, and finish the trip along Mars' plane. Now, when I switch, I notice the EjA is 0.035 and InA is 0.005. (Q3. Why did these change and become close to 0?)

Now, Source plane means I travel along Earth's plane until I'm 90 degrees from the target point and then switch to Mars' plane. Target plane means I travel along Earth's plane until I'm 90 degrees from the source point, then switch to Mars' plane. (Q4. Why 90 degrees?). Also, the EjA and InA angles change to 0.035 and 1.827 respectively for Source plane and 1.602 and 0.005 respectively for Target plane. (Q5. Why this change?)

I switch back to Off Plane because the total velocity change is 5.719k. (Q6. Am I correct that this velocity change is the deltaV needed to extend my orbit around assuming I have left the SOI of Earth and that additional dV is still needed to escape Earth?)

Next, when I look at the display I notice something, In terms of my orbital path from source to target, I escape earth slightly before the Periapsis (around the Sun) and arrive at Mars quite a bit after the Apopsis (around the sun). This means that I'm traveling outward to ApD and meeting Mars coming back in. It doesn't appear that my path is crossing Mars' orbit. (Q7. I would have thought the most favored transfer would have caused me to meet Mars at the ApD, why not in this tangential case?).

If I change the start and end points so that they nearly line up with PeD and ApD points, I essentially eject 180 degrees from the injection point. (Q8. Am I correct in my understanding that the Off-Plane mode can't be easily calculated because at 180 degrees the shortest path is not defined?)

In setting the ejection and injection points 180 degrees apart, I have achieved a transfer time of 270 days. The source and target paths are basically the same since I'm 90 degress from both. My EjA and InA are 0.889 and 4.622 for the Source Plane trip and 1.898 and 4.307 for the Target Plane trip. (Q9. Why are these different?) (Q10. If a typical tangential transfer is about 260-270 days, why are the angles not 0?)

So, now I'm ready to plan my escape. I need to first, launch, and then escape the influence of Earth. Lets assume each case. First, Off Plane mode. The Surface Launch program (course mode) tells me I need to launch 90°, for an inclination equal to my latitude. (Q11. Why? If I'm doing an offplane transfer, why does my launch heading matter? Is it because 90° gives the full benefit of Earth's rotation and therefore minimal energy to orbit? Does the inclination not matter because I'm creating a new plane "off plane"?)

If I switch to Source Plane mode, I still launch to 90°. If I launch from a very high latitude, I also launch to 90, according to the launch program. If I switch to Target mode, then it no longer launches to 90, but to a different heading, I'm assuming one that would give an inclination similar to that of Mars, relative to the sun? (Q12. Why does this happen?)

Thats it for now, there will be more. I hope that respones will be lively and educational. If you are reading this line, then I really do appreciate the time you took to read everything everything above and even more grateful for any answers.

Thank you,
Paul

 

[tblaxland]

(Q1. is that right?)

Sort of. There are other more efficient transfers available (like weak-stability boundary transfers) but they take more time and are quite difficult to plan in Orbiter.

(Q2. Why the 54 day difference).

You may be targeting an intercept near Mars aphelion and/or leaving near Earth's perihelion. Mars' position with respect to the ecliptic may also be a factor (see below for more info).

(Q3. Why did these change and become close to 0?)

I would expect InA to go close to zero if you are ejecting into a plane aligned with the Earth's orbit. The low EjA shows you have a good eject time.

Now, Source plane means I travel along Earth's plane until I'm 90 degrees from the target point and then switch to Mars' plane. Target plane means I travel along Earth's plane until I'm 90 degrees from the source point, then switch to Mars' plane. (Q4. Why 90 degrees?).

A question for the author I guess. It always seemed a rather arbitrary number to me.

Also, the EjA and InA angles change to 0.035 and 1.827 respectively for Source plane and 1.602 and 0.005 respectively for Target plane. (Q5. Why this change?)

These seem opposite to what I would expect, I would expect InA=1.827 for Target plane and InA=0.005 for Source plane. Are you sure you copied the data correctly?

I switch back to Off Plane because the total velocity change is 5.719k. (Q6. Am I correct that this velocity change is the deltaV needed to extend my orbit around assuming I have left the SOI of Earth and that additional dV is still needed to escape Earth?)

It is the dV you would need to give Earth if you wanted to send it to Mars. The orbit eject program helps you get you ejection burn from your Earth orbit so the final trajectory is aligned with this transfer orbit.

(Q7. I would have thought the most favored transfer would have caused me to meet Mars at the ApD, why not in this tangential case?).

The most effecient transfers to Mars will intersect Mars when it is ascending or descending through the plane of the ecliptic. It will not always be possible to get your line of apsides aligned with Mars' line of nodes. A pork chop plot will help you get close.

(Q8. Am I correct in my understanding that the Off-Plane mode can't be easily calculated because at 180 degrees the shortest path is not defined?)

The transfer orbit size is well defined but the inclination of it is not. At that point you may be better off noting down the eject and intercept times and switching to the dV and map programs.

(Q9. Why are these different?)(Q10. If a typical tangential transfer is about 260-270 days, why are the angles not 0?)

I'm not sure why they are different. What does Two Plane show for these eject and intercept times (should be close to zero).

(Q11. Why? If I'm doing an offplane transfer, why does my launch heading matter? Is it because 90° gives the full benefit of Earth's rotation and therefore minimal energy to orbit? Does the inclination not matter because I'm creating a new plane "off plane"?)

You want your parking orbit to be closely aligned to your transfer orbit, ie, inclination to the tranfer to be as low as possible and the LAN aligned with the LAN of the transfer orbit.

If I switch to Source Plane mode, I still launch to 90°. If I launch from a very high latitude, I also launch to 90, according to the launch program. If I switch to Target mode, then it no longer launches to 90, but to a different heading, I'm assuming one that would give an inclination similar to that of Mars, relative to the sun? (Q12. Why does this happen?)

Say your transfer orbit (whichever one) has an inclination to the ecliptic of 4 deg. The highest latitude you can reach that inclination from is dependant on your transfer orbit LAN, but best case is 23.45+4 = 27.45 deg, and worst case is 23.45-4 = 19.45 (23.45 is the angle of Earth's equator to the ecliptic). If I get time, I'll try and do a drawing to show why but maybe the numbers will help you visualise it for now.

If you are reading this line, then I really do appreciate the time you took to read everything everything above and even more grateful for any answers.

Your welcome. It was a well thought out, well researched post.

 

[atuhalpa]

Q4 reply: All orbital planes intersect the center of the Sun. Since a Hohmann transfer is an elipse with the source point and the target point 180 degrees apart, if you have to change planes due to a difference in inclination the planes intersect with each other 90 degrees from the source and 90 degrees from the target. does that clarify it? ( I hope I understand the concept clearly, so I don't confuse you even more...)

 

[jarmonik]

(Q3. Why did these change and become close to 0?)

Little background information at first, the angles EjA and InA are also containing off-plane components meaning that they are angles of vessel's velocity vector and planet's velocity vector in all three dimensions. In off-plane transfer mode you can't get neither of these angles to zero because of the off-plane components. If you leave a plane or you approach a planet from it's orbital plane in that case the off-plane component is zero. In "two-plane" transfer mode, off-plane components of departure and arrival are both zero, meaning that you can get the EjA and InA down to zero as well.

(Q6. Am I correct that this velocity change is the deltaV needed to extend my orbit around assuming I have left the SOI of Earth and that additional dV is still needed to escape Earth?)

The so called total delta velocity is a sum of outward velocity, plane change velocity and inward velocity where outward and inward velocities are not exactly a delta velocities. True delta velocity needed for a vessel can be calculated from: dV = sqrt(x^2 - 2*GM/r) - v, where x is outward or inward velocity, r is a radius of parking orbit and v is a vessel's velocity in the parking orbit. In other words you will get a "boost" from the planet's gravity and the true delta velocity needed for the vessel is a less than the outward velocity. This is well explained in Duncan's deepspace manual. Also, because the r is unknown to IMFD it can't give exact velocities.

(Q1. I realize that a tangential transfer is the most fuel efficient method because I'm ejecting/intercepting along the same vector as the planet. is that right?)

Perhaps from some point of view. But in generally, it should give pretty good deal but it's definately not the most efficient.

Before any calculations are made in Tangential Transfer program the transfer case is translated to a co-planar case, meaning that there are no plane changes (i.e. off-plane components) during initial calculations. The plane change is added afterwards. Due to "boost" effect and some other things it would be good to give some plane change velocity during departure and arrival rather than during the mid-course plane change.

(Q4. Why 90 degrees?)

That would minimize the magnitude of the mid-course plane change. This is explained in the BMW page 379.

(Q9. Why are these different?)

In source plane mode a transfer will begin with co-planar transfer leg in source planet's orbital plane and goes in off-plane after a mid-course plane change. Target plane mode will do it in opposite order.

(Q10. If a typical tangential transfer is about 260-270 days, why are the angles not 0?)

There are off-plane componets in source and target plane modes but you should be able to get one of these angles to zero.


I am going to clean up this mesh in IMFD 6 by replacing source, target and two plane modes with bi-elliptic mode. It would minimize the delta velocity in 4 to 6 dimensions depending from user's will to define times of departure or arrival. That would also requite to take other plans in account like launch and target approach plans.

 

[PaulG]

Thank you all for the responses. I'll have a new array tonight after I digest some of this. For now, just a couple quick questions:

Tblaxland: Thank you, I'll have to check on some of the questions you asked, I may have reversed some of the values, I was toggling between screens so I probably mentally transposed.

jarmonik: What is the BMW?

Thanks,
Paul
http://orbiter-forum.com/member.php?u=86

 

[tblaxland]

What is the BMW?

[ame="http://www.amazon.com/Fundamentals-Astrodynamics-Roger-R-Bate/dp/0486600610"]Amazon.com: Fundamentals of Astrodynamics: Roger R. Bate, Donald D. Mueller, Jerry E. White: Books[/ame]

 

[PaulG]

Hmm...interesting....well, I'm reading a book on methods for multivariate analysis right now...but for $15 and free super saving shipping, I'll have to add this one to my list.

 

[tblaxland]

Hmm...interesting....well, I'm reading a book on methods for multivariate analysis right now...but for $15 and free super saving shipping, I'll have to add this one to my list.

You can preview it on Google Books:
http://books.google.com.au/books?id...zrX_CA&sa=X&oi=book_result&ct=result&resnum=1

 

[ar81]

Here some concepts
[ame="http://www.orbithangar.com/searchid.php?ID=3403"]Tutorial: Concepts for Interplanetary MFD 4.2.1[/ame]

You may find answers there.