I've been asked to find the maximum climb angle of a propeller driven aircraft from the graph of vertical velocity against airspeed:

graph of vertical velocity against airspeed

How can I do that?

  • 2
    $\begingroup$ Does this answer help you? $\endgroup$
    – Raffles
    Oct 26 '20 at 14:46
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    $\begingroup$ Can you post the graph? $\endgroup$ Oct 26 '20 at 14:50
  • $\begingroup$ I just noticed that your graph already includes the origin (0,0). Using the approach described in my answer, you can come up with the answer in less than a minute. No re-plotting is needed. $\endgroup$ Nov 1 '20 at 9:35
  • $\begingroup$ Tangent from 0, 0 to the curve yields point of max climb angle (maximum height for least distance traveled). Highest point on the curve yields maximum climb rate (obviously). $\endgroup$
    – Jan
    Nov 1 '20 at 12:38
  • $\begingroup$ Interesting airplane; looks like Vy is more than double Vx! (More characteristic of a jet with a high thrust loading -- see diagram on this answer -- aviation.stackexchange.com/a/22299/34686 ). Of course it's surely a made-up curve, but just saying... $\endgroup$ Nov 1 '20 at 15:50

All you need to do is to plot vertical versus horizontal speed using the same scale on both axes. Then the solution is the point with the steepest angle from the origin of the graph.

Of course, it helps to do this in metric units. And to add a description to the axes of a plot. The solution can then be directly read from the graph.


  • 1
    $\begingroup$ @quietflyer: Yes, this requires another arithmetical operation because airspeed is the vector sum of vertical and horizontal speed. Any spreadsheet will be able to do this. $\endgroup$ Nov 1 '20 at 8:20
  • $\begingroup$ "Of course, it helps to do this in metric units."-- why? $\endgroup$ Nov 1 '20 at 9:11
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    $\begingroup$ @quietflyer It does not matter as long as the units are the same. With metric units for speed (meters per second), this is the case. With imperial units, we typically use different units for horizontal speed (knots, i.e. nautical miles per hour) and vertical speed (feet per minute). $\endgroup$
    – Bianfable
    Nov 1 '20 at 9:31
  • $\begingroup$ This is a nice concise answer with a nice graphic. But unless the data is already loaded up into a spreadsheet ready for re-plotting to a new scale, there's a much faster way to solve the problem. $\endgroup$ Nov 1 '20 at 13:10

Of course when someone metricly assumes that one is both climbing and flying forward in m/s, that must be some airplane!

To truly be expert, one must confirm the units of climb and airspeed. Spreadsheets are indeed are helpful here, and may be programmed as follows:

  1. Same units of Velocity for climb and airspeed. (meters/second)

  2. Conversion of airspeed to horizontal velocity: Arc Sine ratio of Vv/Airspeed = climb angle Cosine of Climb Angle x Airspeed = Horizontal Velocity

For low angles of climb (non high performance aircraft) airspeed may be directly compared to vertical velocity, as the cosine of lower angles is very close to 1.

Plotting it as Peter Kampf did yields the airspeed of maximum climb angle. From the spreadsheet, this climb angle can be read.

  • $\begingroup$ Just as an aside, the vertical speed can't be higher than the airspeed, so it's not even theoretically possible that the units could be the same on both axes of the graph in the original question-- but if the lower axis were horizontal speed it would be possible, and yes that would be quite some performance! $\endgroup$ Nov 1 '20 at 15:19

Assuming no wind--

The steepest climb angle occurs when the ratio of vertical speed to horizontal speed is maximized.

In many aircraft the difference between horizontal speed and airspeed will be trivial. But even in high-performance aircraft capable of steep climb angles where this is no longer the case, it can be geometrically shown that the steepest climb angle also occurs when the ratio of vertical speed to airspeed is maximized. Both approaches are minimizing the same angle on the right triangle comprised of the vertical speed, horizontal speed, and airspeed vectors. And both approaches are maximizing the same angle on the right triangle comprised of the vertical speed, horizontal speed, and airspeed vectors. Therefore both approaches must have the same solution: when the ratio of vertical speed to airspeed is maximized, then the ratio of vertical speed to horizontal speed is also maximized. So your graph of vertical speed versus airspeed will work just fine for finding the maximum climb angle.

One way to find the maximum ratio of y to x on any graph of y versus x is to extend the axes of the graph to include the origin (0,0) and just run a line from the origin to any point on the curve and find the point on the curve where the slope of this line is the steepest, just as was done in these related answers: What is the typical climb angle (versus the ground) of a single engine piston plane? , Does an airfoil drag coefficient takes parasite drag into account? . Note that the line from the origin will have the steepest slope when it is tangent to the curve on the graph. Note also that the units of the graph need not be the same on each axis for this method to work. So no re-plotting is needed, just get out your ruler and start drawing. Since your graph already shows the origin (0,0), you are almost done already!

Once you know the data point where the climb angle is maximized, you can find the actual climb angle with a bit of trigonometry involving the right triangle comprised of the airspeed, horizontal speed, and vertical speed vectors. You don't have to know all three of these values-- any two are sufficient, and in this case you'll know the airspeed and the vertical speed. The climb angle will be the arcsine of (vertical speed / airspeed). Naturally, for this calculation you'll need to use the same units for both values-- a conversion may be necessary.

While you are at it, just the sake of curiosity you might want to also calculate the arctangent of (vertical speed / airspeed) for the same data point. That's the actual angle of the slope of the straight line you drew on the graph. If the two values are almost the same, that indicates that the airspeed and horizontal speed are so close as to be practically interchangeable, at least at that point on the performance curve. (As a point of trivia, it looks like if the horizontal units on your graph are knots and the vertical units are feet/min, then the angle computed with the arcsine is about 4% larger, or about 0.6 degrees larger, than the angle computed with the arctangent.)

Note that with a curve shaped like the one you've included in the question, the precision of your final answer is not going to be extremely sensitive to how well you are able to determine the exact point where the line from the origin is tangent to the graph. In other words if you vary the point of contact between the line and the curve slightly, the resulting climb angle will barely change at all. Note also that the data point you use for your trig calculation of climb angle can come from anywhere along the straight line you've drawn from the origin of the graph. You need not use the actual point where the straight line touches the curve. For example you can choose a point where the straight line crosses one of the airspeed or vertical speed indices, or where it crosses an intersection of both airspeed and vertical speed indices.

Or, if you've re-plotted your graph with the same scale on each axis, and you aren't worried about the difference between airspeed and horizontal speed, you can forgo the trig calculation and just use a protractor to measure the angle between the tangent line and the x axis of your graph of vertical speed versus airspeed. It's going to be harder to get a precise answer this way, and it may end up taking you longer. Since you've already shown us a perfectly good graph of vertical speed versus airspeed, you might as well use it via the method described here. You should be able to come up with the answer in less time than it took you to read this!

  • $\begingroup$ Don't let the length of this answer fool you, it is actually the fastest path to a solution, starting from what we're given to work with in the original question. $\endgroup$ Nov 1 '20 at 10:39

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