That seems like a lot to have to add on to the driveshaft. Using Pythagorean theorum, which states that (a) squared + (b) squared = (c) squared, when you extend the (a) side of the triangle, that would equate to height, by six inches, this would only increase the length of the (c) side of the triangle by a half inch or so. I had mine lengthened 1 1/2 inches, but this was mostly on the spline part for added travel. Something tells me that your driveshaft was too short to begin with, like maybe it should have gone to a 2dr sport. I can't give you all the numbers because I don't recall at what angle the driveshaft is in its stock location, but I had my driveshaft lengthened from 48 to 49 1/2 inches in length. This would mean that if we assume that the orginal driveshaft angle was zero or 180 degrees, that would make the horizontal length (side b) from the transfer case to the rear differential the same as the length of side (c), because side (a) is effectively zero inches in length. Therefor, 48 squared(side b) + zero squared (side a) equals 48 squared (side c), or c=b. If you increase the length of side (a) to 6 inches, assuming a six inch lift, this changes the formula to 48 squared plus 6 squared = c squared or 2304 plus 36 =c squared, or 2340 = c squared, or c= 48.3735464897912979141996793818226, an increase of merely 0.3735464897912979141996793818226. These numbers would be a little different if the original angle of the driveshaft is not zero degrees, which I am sure it is not, I just used a nice round number. The variance would not be large, however, as you can see from the difference of the first equation, where c=48, to the second, where c=48.3735464897912979141996793818226.
