Does anyone know what the plane change would have been between Columbia and the ISS (ballpark, I know it moved during the mission)? I could figure out the RAAN of Columbia from launch time, but I don’t know what it was for the ISS at the time.

[Sunday evening update]

Here’s what I added as a footnote to the chapter, as a result of discussion here:

The difference in orbital planes at the time was about ninety degrees. A sixty-degree plane change requires as much velocity as it takes to get into orbit (ignoring atmospheric drag and gravity losses), and ninety would take about forty percent more than that (or about as much as it would take to escape from earth’s orbit from the earth’s surface, again, ignoring those factors), so even if there had been a full external tank attached to the Columbia in orbit and the main engines could have been restarted, it still wouldn’t have had nearly enough propellant to get to the ISS.

I worked it out at the time… it was about 90 degrees.

Nick, thanks for the link… wedge angle was 88.1 deg at STS-107 launch and 90.6 deg when Columbia was lost. That makes for an out-of-plane delta-V ranging from 10.7 to 10.9 km/s.

Orders of magnitude beyond the capability of the Orbiter I am sure.

Right, M. The shuttle OMS had a delta-V capability around 0.3 km/s, most of which was used for insertion and deorbit.

Apologies in advance if the following isn’t of any help.

ISS orbital inclination is the same now as then: 51.6 degrees.

Columbia’s final mission was not, of course, ISS, so it used a more effeciant lower inclination which allowed more payload. 28 degrees is Kennedy’s inclination, and many non-ISS missions used it (ever degree in inclination change away from 28 cost them in the neighborhood of 600 lbs, IIRC) but Columbia didn’t. It had an Israeli dust measurement instrument, and they wanted to be able to cover Israel, North Africa, and the Med, so that was the reason given they used 39 degrees for orbital inclination.

Purely coincidentally, Iran’s northernmost point is at 39.4 degrees north. 🙂

Now, as for plane change… I DON’T know how to calculate the plane change (or take precession into account, RAAN, etc) (also, Colombia’s orbit was below ISS, I think). If the orbital periods differ, then you’d get a changing relative plane angle over time.

I do know that that much Delta-V (to match orbits and rendevous with ISS) was far beyond the ability of the OMS and reaction-control systems combined, even if they ran the tanks dry. Columbia, with that heavy a payload (That mission would have been the heaviest re-entry, if I remember right) could not have reached ISS orbit from launch, let alone after attaining a different orbit.

Could Columbia have “reached” ISS? Technically, yes (with a huge caveat). She could have “reached” it via raising her apogee to ISS altitude at the right time, to achieve intersecting orbits. I think she had the Delta-v to do that much. The caveat is that once having done so, she’d have little to no Delta-V remaining, and the relative Delta between Columbia and ISS (due to the differing orbital planes) would have been several thousand miles per hour at “rendezvous” (a very non-survivable event).

Shortly after Colombia’s loss, I had many arguments with people who kept saying “they should have just gone to the space station” so I do remember some of the details.

Hrmmm, if it was about 90 degrees, wouldn’t that put the needed Delta-v at rather close to what it takes to get into orbit in the first place?

(ever degree in inclination change away from 28 cost them in the neighborhood of 600 lbs, IIRC)

Far more than that. About 430 fps delta-V per degree. OMS prop consumption was about 25 lb/fps, so around 10,000 lbs OMS per degree. Total OMS capacity was 25,000 lb, or about 2.5 deg plane change with full tanks (which was never the case, OMS-2 ate up a good chunk of that).

Purely coincidentally, Iran’s northernmost point is at 39.4 degrees north.

Correct, it was a coincidence.

I DON’T know how to calculate the plane change (or take precession into account, RAAN, etc)

The expression for wedge angle between the planes can be derived using the Spherical Law of Cosines:

The delta-V for a given wedge angle and orbital velocity v is:

dv = 2*v*sin(wedge/2)

Hrmmm, if it was about 90 degrees, wouldn’t that put the needed Delta-v at rather close to what it takes to get into orbit in the first place?

Rather more than that, actually. As I wrote earlier, it would have been between 10.7 and 10.9 km/s, depending on when it was performed. Orbital speed is around 7.7 km/s, but the actual delta-V needed to achieve orbit is around 9.3 km/s, accounting for gravity and drag losses during ascent.

(ever degree in inclination change away from 28 cost them in the neighborhood of 600 lbs, IIRC)

Far more than that. About 430 fps delta-V per degree. OMS prop consumption was about 25 lb/fps, so around 10,000 lbs OMS per degree. Total OMS capacity was 25,000 lb, or about 2.5 deg plane change with full tanks (which was never the case, OMS-2 ate up a good chunk of that).

Clarification: I realized after posting you were probably referring to the penalty at launch for increased inclination, where I was referring to the penalty for performing an in-orbit plane change. The launch penalty is closer to your number.

Rand, the Sunday evening update looks good to me.

Egad! All along I just assumed it wasn’t possible due to the 12 degree difference in inclination, assuming their ascending nodes aligned.

You can get the historical data from Celestrak: http://celestrak.com/NORAD/archives/

I worked it out at the time… it was about 90 degrees.

Nick, thanks for the link… wedge angle was 88.1 deg at STS-107 launch and 90.6 deg when Columbia was lost. That makes for an out-of-plane delta-V ranging from 10.7 to 10.9 km/s.

Orders of magnitude beyond the capability of the Orbiter I am sure.

Right, M. The shuttle OMS had a delta-V capability around 0.3 km/s, most of which was used for insertion and deorbit.

Apologies in advance if the following isn’t of any help.

ISS orbital inclination is the same now as then: 51.6 degrees.

Columbia’s final mission was not, of course, ISS, so it used a more effeciant lower inclination which allowed more payload. 28 degrees is Kennedy’s inclination, and many non-ISS missions used it (ever degree in inclination change away from 28 cost them in the neighborhood of 600 lbs, IIRC) but Columbia didn’t. It had an Israeli dust measurement instrument, and they wanted to be able to cover Israel, North Africa, and the Med, so that was the reason given they used 39 degrees for orbital inclination.

Purely coincidentally, Iran’s northernmost point is at 39.4 degrees north. 🙂

Now, as for plane change… I DON’T know how to calculate the plane change (or take precession into account, RAAN, etc) (also, Colombia’s orbit was below ISS, I think). If the orbital periods differ, then you’d get a changing relative plane angle over time.

I do know that that much Delta-V (to match orbits and rendevous with ISS) was far beyond the ability of the OMS and reaction-control systems combined, even if they ran the tanks dry. Columbia, with that heavy a payload (That mission would have been the heaviest re-entry, if I remember right) could not have reached ISS orbit from launch, let alone after attaining a different orbit.

Could Columbia have “reached” ISS? Technically, yes (with a huge caveat). She could have “reached” it via raising her apogee to ISS altitude at the right time, to achieve intersecting orbits. I think she had the Delta-v to do that much. The caveat is that once having done so, she’d have little to no Delta-V remaining, and the relative Delta between Columbia and ISS (due to the differing orbital planes) would have been several thousand miles per hour at “rendezvous” (a very non-survivable event).

Shortly after Colombia’s loss, I had many arguments with people who kept saying “they should have just gone to the space station” so I do remember some of the details.

Hrmmm, if it was about 90 degrees, wouldn’t that put the needed Delta-v at rather close to what it takes to get into orbit in the first place?

Far more than that. About 430 fps delta-V per degree. OMS prop consumption was about 25 lb/fps, so around 10,000 lbs OMS per degree. Total OMS capacity was 25,000 lb, or about 2.5 deg plane change with full tanks (which was never the case, OMS-2 ate up a good chunk of that).

Correct, it was a coincidence.

The expression for wedge angle between the planes can be derived using the Spherical Law of Cosines:

cos(wedge) = cos(i1)*cos(i2) + sin(i1)*sin(i2)*cos(raan2 – raan1)

The delta-V for a given wedge angle and orbital velocity v is:

dv = 2*v*sin(wedge/2)

Rather more than that, actually. As I wrote earlier, it would have been between 10.7 and 10.9 km/s, depending on when it was performed. Orbital speed is around 7.7 km/s, but the actual delta-V needed to achieve orbit is around 9.3 km/s, accounting for gravity and drag losses during ascent.

Clarification: I realized after posting you were probably referring to the penalty

at launchfor increased inclination, where I was referring to the penalty for performing anin-orbitplane change. The launch penalty is closer to your number.Rand, the Sunday evening update looks good to me.

Egad! All along I just assumed it wasn’t possible due to the 12 degree difference in inclination, assuming their ascending nodes aligned.