Energy Height Questions?

[Stoney]

For anyone familiar, I was doing some reading about sustained turn rate, which happened to be in the same chapter as E-M Methods.  As some of the energy equations came before the sustained turn rate stuff, I figured I'd start at the beginning and work my way through. 

The concept of energy height intrigued me, mostly because we were having a discussion a couple of weeks ago about which aircraft retained energy best.  Energy height is basically the total energy of an aircraft (i.e. energy state) divided by the weight of the aircraft and corresponds to the altitude of the aircraft at zero velocity.  So, I began doing some simple excercises computing energy height for a number of in-game aircraft, at similar configurations, to see if there was any useful performance comparisons.

Energy Height Formula:

He = H + 1/2G X V^2

Where:

He = Energy height (in feet)
H = Altitude (in feet)
G = Load Factor (in Gs)
V = Velocity (in feet/sec)

One thing I noticed early was that regardless of the weight of an aircraft, two aircraft at the same altitude and velocity have an identical energy height.  Is this true, or did I do my math incorrectly?  Seems counter-intuitive that a fully loaded P-47 doing 300 mph TAS at Sea Level would have the same energy height as a lightly loaded Spit VIII doing 300 mph at Sea Level.

Another trend I found was that, given the choice between increasing energy height by either altitude or velocity, velocity gives an exponential increase versus altitude.  Are there any useful comparisons to be made from this or am I merely getting confused by the terms?



 

[FTJR]

Stoney I know "nothing" about this.. but your formula doesn't have weight in it, so it cant be a factor (rightly or wrongly). But I guess you knew that.

[bozon]

The concept of energy height intrigued me, mostly because we were having a discussion a couple of weeks ago about which aircraft retained energy best.  Energy height is basically the total energy of an aircraft (i.e. energy state) divided by the weight of the aircraft and corresponds to the altitude of the aircraft at zero velocity. 
...
Energy Height Formula:

He = H + 1/2G X V^2

Where:

He = Energy height (in feet)
H = Altitude (in feet)
G = Load Factor (in Gs)
V = Velocity (in feet/sec)
...

I don't understand the definition in relation to the equation above. You say that He is defined at zero velocity and still you have V in the expression for He. Also, G has dimentions of 1/acceleration, or 1/g (sec^2/feet), so it is not just load factor which is dimensionless. Also, load factor is proportional to the induced drag, which should reduce from the potential, not contribute to it.

I am not familiar with the term energy height. The mathematical expression looks like 'specific energy', that is energy per unit mass. Perhaps the term G has something to do with power load? that makes more sense as it allows to include an energy-building factor. If I were to compare the conditions of two aircrafts, I'd compare their initial energy states, but also include energy building. I'd choose the timescale of energy build to be the time of free-fall from alt H. However, the additional factor does not go like V^2 (perhaps if using a more complicated definition?), so I am not sure why to use the expression as stated above.

edit:
ok, now I get it. You meant: He = H + 1/(2g) * V^2, where g is gravity acceleration.
This is just specific energy and just expresses the total energy in units of height (proportional to total energy). This also explain the "zero velocity" - converting all the kinetic energy to alt.

[hitech]

total energy of an aircraft (i.e. energy state) divided by the weight

two aircraft at the same altitude and velocity have an identical energy height.  Is this true, or did I do my math incorrectly?

To answer your question, yes this is true, but that is not the same as the statement 2 aircraft would have the same Energy.

Also the G part of your equation is really The acceleration of gravity not G loading.

Now if all was in a vacuum , and we were only considering Altitude , I.E. converting Potential energy to Kentic energy then the Mass of the object would have no effect on the Velocity when converting from one to the other.

But there is another equation. F = m * a. In the environment we are working in, a falling object such as a planes , 2 forces are at work.

1 is gravity, the other working in the opposite direction is drag, I.E. air resistance. And in this case the Force of Air resistance is related to Velocity & shape  and has nothing to do with mass, while the gravity has everything to do with mass/weight and nothing to do with velocity or shape.

So unlike your equation an object with more mass traveling straight up or straight down will accelerate more quickly when going down, and decelerate more slowly when traveling straight up.

If you wish to have all the equations there are many here who can post the basic equations.

But a few quick ones.

F = M * A.
Force = Mass * Acceleration

Working in Feet and slugs.
F ( Pounds) = M (Slugs) * A ( Gravity constant is 32.2)


So simply gravity creates 100 lb force downward on a 100 lb Aircraft.

And the force of drag is given by the basic equation F = Q * (1/2 Ro ) * V^2;

Q is a constant that is different for each aircraft and also changes as the AOA changes with a given aircraft, but you are only looking so far at the basic energy equation so for our sample (Straight up Straight down) Q does not change for a given aircraft.
Ro is Air density.

So you end up with the following

(Fg - Fd) / Mass = Acceleration in (Feet per second per second).

Fg = Force of gravity
Fd = Force of drag.

And hence Weight does have an effect on how high a plane can zoom, and a heaver plane with out its motor running and of the same shape will accelerate more quickly in a dive that does not change directions, then a lighter 1.

But once you start considering the 1. change in directions of each plane (Q will change) 2 . Engine performance 3 (Thrust now joins us in the equation). Starting speed & direction( now lift becomes part of the force equation), life becomes much  much more complicated.

HiTech

[Badboy]

One thing I noticed early was that regardless of the weight of an aircraft, two aircraft at the same altitude and velocity have an identical energy height.  Is this true, or did I do my math incorrectly?  Seems counter-intuitive that a fully loaded P-47 doing 300 mph TAS at Sea Level would have the same energy height as a lightly loaded Spit VIII doing 300 mph at Sea Level.

Yep, the equation you posted is correct, however, as HT pointed out the g isn't load factor it's gravity.

Let's take a look where it comes from. If you want to compare the energy of two aircraft you could simply add the aircraft's kinetic energy to its potential energy to get E = 1/2 m v^2 + mgh

The catch with that, is it isn't an accurate measure of potential maneuverability because of the inertia associated with weight. For example a B-17 weighing 40,000lbs would have a lot more energy than a P-51 at 10,000lbs but the P-51 is clearly more maneuverable. A more accurate indication of maneuvering potential is the total energy divided by the gross weight to determine the energy per pound of weight, this is called the Specific Energy and if we take the equation above and divide both sides by the weight. where W = mg we get:

E/mg = 1/2 m v^2 /mg + mgh/mg

Now E/mg is what we have defined as the specific energy (or energy height) and we write Es read as E sub s so the equation becomes:

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, and that does have interesting implications in aircombat.

Now, remember why we did this. If we hadn't divided by the weight, it would have looked as though the B-17 had far more energy than the P-51, by dividing by the weight we see now that they both have the same maneuvering potential. That doesn't mean they will actually maneuver the same, the equation doesn't tell us anything about that. If you want to know how they will actually maneuver you need another term, called specific excess power, Ps read as P sub s. All the energy height equation does, is avoid the problem you get comparing aircraft of different weights, and gives you the result that two aircraft at the same altitude and same speed have the same Es. They won't both have the same total energy, the heavier aircraft will have much more. But if they are at the same altitude and airspeed, the will have the same Es and be Co-Es. Many people leave off the s and just write Co-E. When someone complains that another fighter had more energy than than they did, what they really mean is that it had more Es.

Another trend I found was that, given the choice between increasing energy height by either altitude or velocity, velocity gives an exponential increase versus altitude.  Are there any useful comparisons to be made from this or am I merely getting confused by the terms?.

No, two aircraft that are Co-E, that is they have equal Es, should both have exactly the same maneuvering potentail, that is if one is lower but faster, and the other higher but slower, the lower aircraft should in theory be able to zoom climb to the height of the higher aircraft and end up at the same speed as it. That's what being Co-E means and is the real purpose of the equation, that is the ability to compare relative energy in a more meaningfull way.

Hope that helps

Badboy

[hitech]

Another trend I found was that, given the choice between increasing energy height by either altitude or velocity, velocity gives an exponential increase versus altitude.  Are there any useful comparisons to be made from this or am I merely getting confused by the terms?

Also do not let yourself believe that it is better to increase your speed than climbing to gain Es . To gain Es the fastest you should be at your best climb speed climbing. What is simply happening at best climb speed is you have the greatest excess power condition of your plane. And Excess power/mass * TIme = gain in Es.


HiTech

[Stoney]

Also the G part of your equation is really The acceleration of gravity not G loading.

Ok.  What value would be substituted here?  32.2 or 1?

[hitech]

32.2

[Badboy]

Ok.  What value would be substituted here?  32.2 or 1?

Yep, 32.2 with the units you mentioned earlier.

Badboy

[Stoney]

That's what being Co-E means and is the real purpose of the equation, that is the ability to compare relative energy in a more meaningfull way.

If you had a La-7 at 10,000 feet and 350 TAS and a P-47 at 8,000 feet and 325 TAS, could you make a direct comparison of the specific energy (termed energy height by my author) and it be meaningful or appropriate?  I don't want to read anything into this that I shouldn't.

[hitech]

Yes it is very meaning full. But you must also know what you can and can not do with this number.

It is a great method to estimate potential zoom.
It is a great method to estimate how much extra turn you have than the other plane.
It is a great method to estimate how much running ability you have.
It is a great method for understanding conversion of alt to speed and speed to alt.

But note I said estimate in all these, if you wish to do precise comparisons you must in one way or another factor in drag and thrust. And when doing raw power and force calculations you will have to change the power and force numbers to accelerations and this will mean you will have to put mass back into your calculations.

HiTech

[Badboy]

If you had a La-7 at 10,000 feet and 350 TAS and a P-47 at 8,000 feet and 325 TAS, could you make a direct comparison of the specific energy (termed energy height by my author) and it be meaningful or appropriate?  I don't want to read anything into this that I shouldn't.

Ok, take a look at the diagram below. You will notice that the two aircraft at A and B are Co-E. One has a speed of 250mph at a altitude of 8k, the other has a speed of 350mph at 6k, and you see that they both fall on the same line of constant Es, so they are Co-E. If A dived to B, or B zoomed to A they would both end up at about the same altitude and speed.



Now let's consider your example. The first thing to say is that the La7 is both faster and higher, so you don't even need the chart to know that this isn't a Co-E fight The la7 has about 22% more Es.

But suppose the P-47 pilot in your example wanted to zoom up to the same altitude as the La7 for a Co-Alt merge, following a line of constant Es on this chart shows that the P-47 would slow down to about 214mph during the zoom.

But that's based on their energy state at the moment when the zoom begins, it ignores any energy loss/gain due to drag and thrust while the maneuver takes place. But it does give you a quick and easy sense of what's happening.

Hope that helps...

Badboy

[Stoney]

Absolutely that helps, and yes, in my example I transposed the altitudes; it wasn't a very good example.  You lost me though when I got to part about how the P-47 would slow to 214 after a zoom to 10,000 feet.  I don't see how the chart tells me that.

And yes, Hitech, I'm starting with mere static comparisons just to understand the concept.  Once I feel like I've mastered the concept, then I'll see if I can add in some more dynamic comparisons if its any value added. 

[Badboy]

You lost me though when I got to part about how the P-47 would slow to 214 after a zoom to 10,000 feet.  I don't see how the chart tells me that.

Sorry, I forgot to mention that if you want to move one of the dots you need to slide it along one of the lines of constant Es. Or at least an imaginary line. I've shown that in the diagram below. 



Here you can see that I've moved the P-47 up to 10k along a line of constant Es, when it gets there you just read off the speed.

You can make a chart like that yourself using the equation you posted earlier.

Hope that helps...

Badboy

[Stoney]

Ok, that graphic clears it up.  Thanks for the explanation.