Not quite, you're drag is actually lower at Vx than at Vy. Aerodynamic efficiency is better (if by efficiency we refer to drag). Take a look this graph:

http://www.undaerospace.com/images/s...s/altitude.gif
Notice that Vy is part way up the power required (green) curve.

Take a look at the chart in

Blackhawk's PDF. Sorry, I couldn't find a replica on google to link. Looking at the first graph on page one you see that the tangent line to the rate of climb curve gives us our Vx speed. The top of the rate of climb curve is Vy.

To increase our angle we sacrifice some of our climb performance to allow us to fly at a slower speed. Flying at a slower speed gives us more time to gain altitude. So Vx is at a slower speed to increase time available, over a given distance, to climb. As you put it, altitude/distance.

An example, using the same first graph in the pdf mentioned above, (apologies for the numbers):

Aircraft 'A' flies at Vx = ~68 knots giving ~640 fpm

Aircraft 'B' flies at Vy = ~80 knots giving ~700 fpm

After flying for 1 minute:

Aircraft 'A' has covered 6886 feet in one minute, gaining 640 feet of altitude.

Aircraft 'B' has covered 8101 feet in one minute, gaining 700 feet of altitude.

Using some simple trig we find climb angles of each:

Inverse tangent (640/6886) gives us Aircraft 'A' at a climb angle of 5.3 degrees.

Inverse tangent (700/8101) gives us Aircraft 'B' at a climb angle of 4.9 degrees.

You may ignore the math used to get to these numbers, the conclusion to draw here is:

Aircraft 'A' was flying at Vx of 68 knots, rate of climb of 640 fpm, and a resultant climb angle of 5.3 degrees.

Aircraft 'B' was flying at Vy of 80, rate of climb 700 fpm, and a resultant climb angle of 4.9 degrees.

Math doesn't lie. Find this helpful or gibberish? TY blonde bombshell Melissa, right?