Vy vs Best Energy Airspeed

[Octavius]

Very nice explanation Badboy!  

So, by diving below the critical altitude for the merge and crossing through it as you zoom post-merge, you're in effect maximizing your energy state by remaining at the critical alt for the longest period of time... a sort of average half below the crit alt and half above the crit alt... is that correct?  Also, the critical alt is the most efficient altitude for fuel consumption/cruising speed/energy efficiency (in level flight) for said aircraft?

Diving/climbing through this critical altitude will give you more overall energy when compared to simply climbing (without the dive).

"Almost" like a slingshot... ?  Or a roller coaster with an energy deposit at the bottom.  (weird analogies that only make sense in my head )

[Scherf]

Every time I figure I'm simply going to quit surfing the boards, something like this comes along.

Didnt realise the Energy = Speed ^ 2 + Alt thing, damn that really puts things in perspective. I've been frikkin clawing for altitude...

Now that I see the diagram, I recognise that hook maneuver too. I get the impression jugs are really good at that type of thing, they don't turn, but they sure as hell can corner.

Thanks all,

Scherf

[HoHun]

Hi Badboy,

>Yep, that’s what classical climb analysis predicts, and that is what HiTech is referring to, but it has long been understood that the classical climb analysis has an embedded inconsistency.

I'm aware of that, but I'm not sure it makes any difference for the climb speed.

I just ran some numbers: Climbing at optimum steady-state climb speed, my WW2 fighter of choice needs 301 s to 4400 m. Optimum climb speed varies by +/-5 km/h indicated air speed in the interval (dropping from sea level to altitude). The energy needed for accelerating to the lower indicated but higher true climb airspeed I fail to include in my calculation is equivalent to 56 m altitude, which is equivalent of another 4 s of climbing. (That's the classical error.)

However, the climb speeds themselves don't seem affected by that error - on the contrary, if I use a constant climb speed the energy gap widens (insignificantly).

Altering speed from the steady-state climb speed only seems to give an advantage if you go slower, else the energy gap increases. However, I can gain only 2 s of the abovementioned 4 s because I have to fly at stall speed at least, and that will give me a 60 s steady-state climb time penalty so it's not really an improvement.

The document you mentioned may have some significance, but I've got to admit that I don't get it :-)

Regards,

Henning (HoHun)

[Badboy]

Originally posted by HoHun
Hi Badboy,

The document you mentioned may have some significance, but I've got to admit that I don't get it :-)

Regards,

Henning (HoHun)

Hi HoHun

Firstly, let me apologize for the following verbosity… I know I could explain this to guys like yourself and HiTech much more briefly than I’m going to, but I’m going to take the opportunity to get back to basics in the hope I can also open the subject to some of our less technically minded friends. Here goes…

It seems to me as though there are two different discussions here, one regarding best climb rate, and the other about best energy transfer rate, and they are absolutely not the same thing. Let me see if I can explain it a little better. Being in the best possible situation for an engagement, isn’t just about getting as high as possible in the shortest time, if it was, we could simply use our maximum climb rate to get as high as possible as quickly as possible and the problem would be solved. However, there are four vitally important energy related characteristics that need to be optimized to give a pilot the best possible chance of winning a fight, of which the climb rate solution only nets two of them, so in effect, solving the best climb rate problem only gets the job half done. The theory for finding the schedule for best energy transfer depends on optimizing the Es and Ps at the same time, so I’d better explain those terms:

Specific Energy:
Total energy by itself is not an accurate measure of maneuverability because of the inertia associated with weight. A B-17 weighing 40,000lbs may have more energy than a P-51 at 10,000lbs but the P-51 is clearly more maneuverable. A more accurate indication of maneuverability is the total energy divided by the gross weight to determine the energy per pound of weight, this is called the Specific Energy and is given by the expression Es = V^2/2g + h where V is the speed, g is gravity and h is the altitude, and is measured in units of length (feet say). Notice that the velocity is a significant part of this expression, having the best Es is not just about getting as much altitude as possible in the shortest time, it is about getting the best combination of speed and altitude. But that’s not all because getting to a fight with more altitude than your opponent isn’t any good if you don’t have the maneuverability to capitalize on it, which leads us to:

Specific Excess Power
The rate of change of energy with respect to time is called power, but once again, our B-17 with its four magnificent engines has more power than our P-51, but it still isn’t more maneuverable, so once again we are primarily interested in specific power, which is power divided by weight. Of course we are also only interested in the useful power, free to be used to maneuver, and this is called the excess power, that power left over once drag has been overcome. So we arrive at the specific excess power, or Ps and it is given by the expression Ps = V/g dV/dt + dh/dt  which is the velocity divided by gravity, times the rate of change of velocity with time, plus the rate of change of altitude with time.

So, putting all this together, we need to find a way of optimizing four terms all at once, they are the four terms that comprise the Es and Ps expressions above. We need to find the best possible combination of velocity, rate of change of velocity, altitude and rate of change of altitude, all at the same time! It is the fighter pilots equivalent of nirvana, the mother of all sweet spots!

By climbing to an engagement at the best climb rate, you are only optimizing two of those terms, the altitude and the rate of change of altitude, but that’s only half the story… it simply won’t cut it. If you ignore the other two terms, you may arrive at the fight with a little more altitude, and a better climb rate, but you simply won’t be in the best energy situation for the ensuing engagement, you can only find that sweet spot by optimizing the Es and Ps together and those considerations push up the airspeed beyond that required for the best climb rate alone… That’s what I’ve tried to illustrate in the diagram I posted, I have never been talking about just getting as high as possible as quickly as possible, even though that can be a simple and effective strategy to use in the arena if you have a plane that can do it really well (the G10 springs to mind) but in terms of correct air combat, it simply isn’t the right thing to do, as almost any current fighter pilot will confirm.

Hope that helps…

Badboy

[HoHun]

Hi Badboy,

I think your E diagram is quite sensible - for going from A to D most quickly, A->B->(just short of) C->D looks like the optimum.

However, numerically I'd say that your speed B->C is much too high as it would be far above Vy for a typical WW2 propeller fighter. Your preference of speed for best energy transfer seems to confirm that you've selected a different speed on purpose. (A->B would be almost invisible on my variant of the chart :-)

At Vy, I've got Psmax for 1 G - it's the optimum speed to maximize energy and has a defined technical meaning.

Is there such a defined meaning for the speed of best energy transfer as well? It seems to me that you're thinking of a concept of a similar nature as corner speed ...

Regards,

Henning (HoHun)

[Crumpp]

Great Post, Badboy!

That goes right along with what some of the veterans have been telling me.  They did not use the "best climb rate".

No disrepect to them is intended in this statement but many of them do not know very much about aerodynamics.  They were combat pilots and not engineers.

The general rule was to climb with enough speed to do a loop over the top if you had too.  Not very scientific but seems to go right along with the gist of your post.

Crumpp

[JG14_Josf]

Drag, Lift, gravity, and thrust contribute to change where one plane is relative to another plane in time.

Drag resists movement.

Gravity accelerates objects in one direction.

Thrust and lift force the plane in specific directions.

Drag is minimized when the plane is unloaded.

So, if the plane spends a minimum time in an unloaded state where the force of drag is minimized such as an unloaded dive followed by a hard turn to an unloaded climb then will the plane gain an overall energy advantage over a plane flying at a constant best climb speed in the same amount of time?

Plane A is the diving and climbing plane.

Plane B is the constant best climb speed plane.

Both plane A and plane B have the same amount of gravity forced upon them except I think the plane spending more time higher has a slight advantage.

Both planes have a similar amount of thrust from the engine during the time period except some differences associated with efficienies at different altitudes and speeds.

Plane B has a consistant force of lift and drag applied except this relationship does change some with the change of altitude.

Plane A spends an amount of time where drag is minimum and an amount of time where the force of drag is high.

So, does Plane A gain more or less total energy compared to plane B?

I think this is a very interesting question because certain planes inevitably were better at pushing the numbers in favor of gaining energy during unloaded flight, better in dives and zooms than other planes and these particular attributes contributed to enable one plane an advantage over another plane when utilizing dives and zooms in close in, one on one dog fights.

Also, Plane B goes from the starting point to the ending point in a straight line and therefore goes the shortest distance from the starting point to the ending point.

Plane A turns during the flight time and therefore has to go a longer distance if the ending point is common to the ending point of plane B.

[muerto]

Great posts Badboy.  This explains a lot of the questions I had.  I knew Vy seemed sort of right, but I also knew entering a fight a Vy is a bad idea.

Is there any kind of rule of thumb one can use to determine the speed that can put you in the sweet spot?

Is there a correlation to Vy (a percentage of Vy perhaps) or anything simple like that, or does it depend on too many factors to have a general rule of thumb to shoot for?

Crumpp said:

The general rule was to climb with enough speed to do a loop over the top if you had too. Not very scientific but seems to go right along with the gist of your post.

and this goes along with what I was trying to say before (what I was trying to interpret from Shaw's book and my own experience in AH) that you need to be at manuvering speed at the engagment, not at Vy.
I think Shaw calls it vertical manuvering speed, then says something like (its been a while since I read it) whenever possible go to best energy speed (when the bogey is not in guns range or not in a position for a guns solution) and climb, but be able to have vertical manuevering speed while closely engaged.

[JG14_Josf]

One thing that becomes obvious when thinking about this puzzle involves technique.

In order to illustrate this point please consider reading the following personal experience.

My brother allowed me to pilot his plane and he wanted to see if my pilot simulation experience helped. He challenged me to get the plane at a higher level speed. My efforts resulted in about 30knots less than he was able to get. I tended to overcontrol. The plane spent too much time flying sideways under my ham fisted piloting whereas my brother managed to keep the plane relatively arrow straight.

Application:

If Plane A were flown from point A at 10,000 feet to point B at 15,000 feet in a straight line that happened to be at a climb angle corresponding with the planes best climb speed then this plane would be converting the power of fuel into the power of altitude in the most efficient manner possible. As long as the pilot managed to keep the plane arrow straight without it yawing sideways, pitching forward, or back; then no other pilot could possibly get the same plane from point A to point B any faster; all else being equal.  

Correct?

What if pilot B is flying side by side with plane A at point A in the same plane at the same starting speed and:
The pilot of plane B pushes the stick forward hard enough to create lift with the elevator that causes drag and this control input rotates the plane's pitch nose low and even starts causing the main wing to create lift on the bottom surface again creating more drag. The ham fisted pilot then allows the plane to speed up but by now the plane is pitched into a very steep dive. He pulls back on the stick to make the dive less shallow, after all he is in a race with pilot A to point B, and point B is higher than the starting point at point A.  As soon as pilot B settles into a constant dive angle he starts pulling back on the stick to climb into a zoom climb by pointing his plane at point B somewhere ahead of plane A. Pilot B spends the rest of the time it takes to reach point B pitching his plane up, pulling gs, and creating drag.

Which plane reaches Point B first?

Which plane has more energy at the time the first plane reaches point B?


Looking at the problem from this angle is similar to looking at the race between my brother and me in his plane.
How can drag be minimized?

Back to pilot B starting at point A both side by side, both flying at the same speed and angle (best climb speed, best climb angle):

Pilot B pushes the stick forward only enough to unload as much G force as possible. Plane B is now on a ballistic trajectory determined by engine thrust, gravity and momentum. All forms of lift and therefore drag are canceled out to their minimums.

Plane B remains on this trajectory for just the right amount of time before pulling into a zoom climb at just the right time to, again, minimize drag during the vector change at the wings most efficient speed and corresponding g load. The vector change is made just so as to allow the pilot to return to an unloaded ballistic trajectory that will cause his plane to arrive at point B.

Now which plane arrives at point B first and which plane has more energy at the time both planes arrive at point B?

What if pilot A is using his rudder pedals as foot rests?

[Pyro]

But the original question was not in regard to entering combat, it was simply asking at whether the best speed to gain energy was Vy.

[muerto]

Pyro Said:

But the original question was not in regard to entering combat, it was simply asking at whether the best speed to gain energy was Vy.

True, true, true.

And I think Vy is the correct answer to the original question I asked.

It just so happens that, in my head, there was a lot more to the question than what I actaully asked.
Shame on all of you for not being mind readers.

But because of what has come out in this discussion I have gained a greater insight into the overall question I sought to have answered and I thank you all.

The only thing lacking now is knowing the acutal speeds for each aircraft (not Vy, but the best energy transfer speed to fit into Badboys diagrams) or knowledge of how to determine it purely from flying the aircraft.

That, and the ability to instantly judge a con's relative energy state, and the skills to perfectly execute the proper judgement with regards to the correct tactics to apply based on any given situation, and... etc, etc,...

[Badboy]

Originally posted by Pyro
But the original question was not in regard to entering combat, it was simply asking at whether the best speed to gain energy was Vy.

I thought the original question was about a comment Shaw made in his book about “maximising the rate of total energy gain rather than just the rate of climb” Shaw’s comment is in the context of air combat. I assumed that getting an insight into air combat was the real motivation behind the questions being asked... Just trying to be helpful here, didn't mean to hijack the thread.

Badboy

[Pyro]

It's certainly not a hijacking.  I just think the original question was just being interpreted differently by various parties which was causing some confusion.

[muerto]

I agree, no hijack at all.

I'm very glad you chimed in, Badboy, because without your post I would have recieved the answer to my initial quesiton, but I still would have had more unresolved questions about Shaw's mention of Best Energy" speed.

[HoHun]

Hi Badboy,

>Shaw’s comment is in the context of air combat.

After re-reading Shaw, I'd say the speed of best energy transfer is only relevant for supersonic jets that have a superior climb ability above Mach 1 than below in parts of their envelope. Wow! :-)

Regards,

Henning (HoHun)