M1.46? Hmm.
I am about to shoot myself in the foot, I'm sure, but here goes nothing.
Assuming a track of 195 degrees towards Indonesia, surely the speed required in order to make the "horizontal" vector equal 900 Kt would be about 3,477kt?
Turning the problem into a right angle triangle, with the hypotenuse being your track to Indonesia (15 degrees left of South), and the short ("opposite") side being 900kt:
Speed = 900/Sin 15 ?
Edited to add: Of course you could also be on a heading of due south (towards DPS), in which case the speed required would be infinite..
Go ahead: shoot, flame, educate!
Last edited by OK4Wire; 29th Aug 2014 at 05:34.