View Full Version : Islander Vx, Vy, Vxse, Vyse conundrum

Tinstaafl

2nd Jan 2004, 03:56

I've always understood that Vx & Vy and Vxse & Vyse meet at the absolute ceiling for AEO or 1 inop, respectively. So how is it that a single, 65kts, figure is given for those four speeds even though the aircraft still has a large performance reserve?

Was it a quick & easy way through the certification maze? Something else? If a means of simplifying certification, what are the true figures for those speeds (ISA SL)?

Damned if I can figure it out.

Alex Whittingham

2nd Jan 2004, 18:02

The conventional textbook explanations for VX and VY concentrate on jet performance. As you know the thrust and power available curves for prop aircraft conventionally have a different shape, with Thrust vs EAS decreasing from a maximum at zero EAS and with Power available vs TAS increasing initially then falling off.

Because of the different shape of the thrust curve VX on a prop is a low speed, usually less than VMD. Likewise the different shapes of the power available and power required curves show VY to be less than VMD, at or very close to VMP and quite possibly the same as VX, give or take half a knot.

The effect of increasing altitude is to leave the EAS/IAS of VX substantially unchanged because the thrust line moves down and the drag line stays in position. The effect of altitude on the prop power available curve is to move it to the right and down while the power required curve moves right and up. The exact effect of this on the EAS/IAS of VY depends on the exact shape of the power available curve but usually slightly reduces the EAS of VY.

At the absolute ceiling VX and VY must be coincident, because it is impossible to have, for instance, a climb angle without having a rate of climb. The fact that the Islander has them the same at all altitudes indicates a power available curve that is flat around VMP, and that VX and VY both lie so close to VMP as to be co-incident.

Here's one for you, Tinstaafl, as I know you enjoy the theory. I have found an explanation of power available curves on power producing engines from a very respectable source that conflicts with the AP3456 version.

Aircraft Performance by Martin E Eshelby of Cranfield University, (ISBN 0 340 75897 X) Page 59 shows power available constant with increasing EAS and consequently defines VY for a power producing aircraft as being exactly at VMP.

The difference between the two explanations lies in the shape of the power available curve and in the axes of the graphs, the RAF plot power against TAS and Eshelby plots power against EAS. I can't figure out why there is a difference.

Tinstaafl

2nd Jan 2004, 20:18

Ta Alex.

Don't have any graphs available so I'll have to think on it. My initial thoughts re the BN2 was that the curves were particularly 'steep' ie compressed in the x (speed) axis, resulting in Vx & Vy (AEO & 1 inop) to be very close, compared to other aircraft. Still seems odd to me though, especially considering the large performance reserve (AEO)...

For your text discrepancy, what speed range are those graphs for? Perhaps the curve's shapes are an artifact of allowing/not allowing for compressibility effects?

bookworm

2nd Jan 2004, 20:31

I've always understood that Vx & Vy and Vxse & Vyse meet at the absolute ceiling for AEO or 1 inop, respectively. So how is it that a single, 65kts, figure is given for those four speeds even though the aircraft still has a large performance reserve?

I agree Tinstaafl. I don't think it's possible. To see this, it needs a little algebra.

Let T(v) be the excess thust (i.e. thrust available - thrust required, not the usual convention for T).

For max rate of climb, T is maximum and so dT/dv = T'(v) = 0. So T'(Vx) = 0.

Now look at excess power, which is v*T(v). For max rate of climb, v*T is maximum and so d(v*T)/dv = 0

So v*T'(v) + T(v) = 0.

Under normal circumstances this is satisfied at v = -T(v)/T'(v), which gives Vy. But in this case we know T'(Vx) = 0. So it can only be satisfied at v = Vx if T(v) = 0 also. That means no excess thrust, i.e. you're at the absolute ceiling.

Note that I've made no assumptions about the shape of T(v), so this applies equally to props and jets.

If you prefer thinking about curves, note that Vy is the minimum the excess power curve, where it's flat. Thus if you go to Vy minus a little, the excess power changes hardly at all (2nd order), while the distance travelled decreases a little (1st order). Thus the angle of climb must increase.

The only thing I can think of is that Vx has to be greater than a minimum value for other reasons. For example, is the "book" Vx required to be a certain percentage above the stall?

Alex Whittingham

2nd Jan 2004, 20:49

Why is T a maximum for the max rate of climb? A maximum value of T as you have defined it gives max angle of climb, not rate.

I'll try and post the graphs. The speed ranges look comparable.

from AP3456:

http://www.bristol.gs/download_files/ap3456.jpg

from Eshelby:

http://www.bristol.gs/download_files/eshelby3.jpg

I've put the whole page up so you can see his generic formula for rate of climb.

Oh b*gger! and I can't edit or delete it. I'll try again:

AP3456

http://www.bristol.gs/download_files/ap3456.jpg

Eshelby

http://www.bristol.gs/download_files/ap3456.jpg

I can't get the images to post on here. I've put them on our forum here (http://www.topclasstraining.co.uk/ubbthreads/showflat.php?Cat=&Board=performance&Number=1396&page=0&view=collapsed&sb=5&o=&fpart=&vc=1&PHPSESSID=)

bookworm

2nd Jan 2004, 22:00

Why is T a maximum for the max rate of climb? A maximum value of T as you have defined it gives max angle of climb, not rate.

Yes sorry, typo. Should be:

For max angle of climb, T is maximum and so dT/dv = T'(v) = 0. So T'(Vx) = 0.

Tinstaafl

3rd Jan 2004, 06:31

Had aquick look at those graphs, Alex. I'll study them tomorrow. The most obvious thing to me is that one is depicting power required/available, the other is showing thrust.

I'd expect a difference since P=TxTAS ie not the same as thrust. Am I still missing something?

Wha??? We can't edit our posts after 10 mins?? What a pain!

Oops, Alex. Just noticed the third graph... :O

I wonder where the straight line depicting power available is from as well. Maybe he's normalised the graph's y axis to force a straight line depiction? That would help show where max. excess power is obtained by hiding the power available vs speed variable.

testing.....

How interesting. My additional post got added to my previous one. Wonder what will happen to this one?

Now I know. It got included in the initial post too. The timer is a bit out though. Tried the 'Edit' option after two minutes & still got the 'No editing after 10 mins' message

Keith.Williams.

3rd Jan 2004, 21:57

I do not have an answer to your question Alex, but the more I look at it, the more curious it all becomes.

If you look at Figure 3 in the AP3456 you will see that piston best ROC speed is slightly more than Vmp. But if you turn the page to Figure 4 you see two different situations. The low altitude curves show best ROC at a speed lower than Vmp, but the high altitude curves show it at slightly higher than Vmp. I suspect that this is simply an error and the author was concentrating on showing how the speeds vary with altitude, rather than showing their relationship with Vmp.

Eshelby's Figure 3.18 is curious in that it shows the thrust available curves as being convex to the origin. This is contrary to those in most other texts (every one that I have seen). Most texts show the thrust curve as being concave to the origin at the low end of the speed range and becoming slightly convex to it only at the high speed end. Also if we take his thrust against EAS curves from figure 3.18 and multiply them by increasing values of EAS (to get thrust power) we will not get the straight lines for thrust power against EAS, shown in figure 3.19.

If as Tinstaafl suggests, Eshelby has normalised his thrust power curves in Figure 3.19, then surely he should have also done this to the drag power curve, so that their relationship was not distorted. He does not appear to have done this, and perhaps this is the cause of his curious result of having best ROC at Vmp.

I suspect that like the author of the AP3456, he was concentrating on showing one aspect of the subject (Excess power = thrust power - drag power) and in so doing, inadvertently introduced an incorrect relationship between best ROC speed and Vmp.

Figure 3.19 is also curious in that the author has used the term "Maximum excess drag power" to represent thrust power minus drag power. Surely this should be "Maximum excess thrust power".

Hmmmmmm.....curiouser and curiouser.