Vx versus Vy

JDP

Well-Known Member
Okay, so i call myself looking for this question in past posts and searching for it but i can't seem to find anything on this. Why is there a difference between Vx and Vy. Before someone starts in on a textbook definition of best angle versus best rate I know that. I want to know is how an instructor would EXPLAIN it to a new private student. Thanks sorry for the beginner question
 
This would probably be hit or miss with most students, but I like to compare it to someone running up a wheel chair ramp versus someone walking up some stairs. If that even makes any freaking sense. To be honest, I only just now thought about this as a possible explaination after reading your thread. :D
 
If you make a graph of rate of climb vs. airspeed (with airspeed as the independant variable on the x-axis and rate of climb as the dependant variable on the y-axis), you will get an inverted "U" shape. If you draw a line from the origin of the axes (0,0) so that it will just touch the curve at ONE point, that is Vx. This way of showing it using basic geometry shows that Vx is of course this speed because it is the speed where the ratio of vertical speed to horizontal speed is the greatest. By the nature of the shape of that graph (an inverted "U"), Vx is then less than Vy (which would be the maximum point of the "arch".

The main point:
The difference is that at Vx, you are getting the most feet of altitude for each foot over the ground. At Vy you are getting the most feet of altitude for each minute flown. It can be confusing thinking about the multiple variables in the problem. The reason you don't want to climb at Vy if you need to clear an obstacle is because sure you will get higher sooner, but you will also have covered much more distance because you are going faster. At Vx, the rate of climb is less than Vy (which can be confusing because then why do you use that speed to clear an obstacle?), but you are moving more slowly forward, so at a given point (like the end of the runway), you will be at a higher altitude than if you climbed at Vy and were at that same point.

The reason that the graph of Rate of climb vs. airspeed is the shape it is has to do with lots of variables such as the difference between the drag curve (aka power required) and the power curve (aka power available).
 
The main point:
The difference is that at Vx, you are getting the most feet of altitude for each foot over the ground. At Vy you are getting the most feet of altitude for each minute flown. It can be confusing thinking about the multiple variables in the problem. The reason you don't want to climb at Vy if you need to clear an obstacle is because sure you will get higher sooner, but you will also have covered much more distance because you are going faster. At Vx, the rate of climb is less than Vy (which can be confusing because then why do you use that speed to clear an obstacle?), but you are moving more slowly forward, so at a given point (like the end of the runway), you will be at a higher altitude than if you climbed at Vy and were at that same point.

:yeahthat:

This is exactly how I explain it to my students. Makes sense to most of them.
 
Think about it in helicopter terms: Hovering slowly up at 10 feet per minute to clear a tree in a wooded area. He's going up 10 fpm, and forward 1 fpm, so after going forward 10 feet he has gone up 100 feet and can clear the tree.
 
The measurement of each is different, one is time, one is distance. That might help.
 
The measurement of each is different, one is time, one is distance. That might help.


:yeahthat:

That's probably the easiest way to explain it to a student. Units...

Vx = Best Feet/Feet
Vy = Best Feet/Min

I usually explain it that way. Basically I would tell a student you would use Vx to get above an obstacle in as short of a distance as possible, and use Vy to get to altitude ASAP.
 
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