View Full Version : best angle, rate, or min manuevering speed

16th Sep 2010, 04:45
OK, company manual for 737ng mentions using one of these speeds on climb out if obstacle is a factor. Confused?! Surely there is just one speed, and in my mind wouldn't that be best angle of climb?

16th Sep 2010, 05:17
How far out is the obstacle?
In addition, overspeed takeoffs (IE:improved climb procedure) can also be used.

16th Sep 2010, 07:51
flyb - cannot answer unless you quote your manual text!

16th Sep 2010, 08:08
Hi flybubba,

Most of my previous aircraft have specified those speeds like this:

If you imagine a drag (Y) curve plotted against speed (X), you get a cup shaped graph. "U" shaped.
(1) Min drag speed is where the tangent is horizontal.
(2) Best rate is where the tangent goes through the origin of the graph.

(1) That is where you'll have maximum excess thrust at minimum speed = Best Gradient. You'll be limited to maximum 15 degs bank because of your stall margin so Min Maneuver is normally + 10 kts on top.

(2) If ATC ask for "Best Rate" - then the most sensible speed for economy and ROC is at that point.

16th Sep 2010, 13:54
After looking at manual again, the wording includes phrases such as "ATC constraints, etc". Makes more sense now; they are talking about not only obstacle constraints, for which best angle would apply.

16th Sep 2010, 14:06
Generally speaking

Maximum Angle of Climb - in this climb regime the airplane the greatest altitude gain in the shortest possible horizontal distance flown => best climb gradient. (speed Vx is the maximum gradient of climb)

Maximum Rate of Climb - in this regime the airplane achieve the greatest altitude gain in the shortest possible time. (speed Vy is the maximum rate of climb)



Minimum maneuvering speed required a high AOA => higher drag and the vertical component of this drag reduces the total lift => the over all climb performance are reduced.

So for obstacle I'll suggest Vx!