Some New Data Carts to chew on

[Badboy]

Originally posted by Dwarf:


But as Zigrat points out, they all make simplifying assumptions, and provide only estimates.  Sometimes questionable estimates.

Dwarf

In the case of the Ps equation, the difficulty only arises when you attempt to evaluate it. You might choose to introduce simplifying assumptions at that stage, as Zigrat has done in his spreadsheet by assuming lift and drag coefficients that are constant with mach number, as just one example. But that was his choice, the Ps equation itself is exact and conclusions drawn from it are sound.

Badboy

[Badboy]

Originally posted by Dwarf:


Prove is too strong a word in my book. Your equations certainly say they are "directly and linearly linked". However, like all currently existing equations, they contain simplifying assumptions that gloss over parts of the problem that have, so far, been opaque to analysis.

The Ps equation can be found in almost every aerodynamics text book, with all the associated mathematics. Nothing is glossed over or opaque. The equations are simple and crystal clear.

 

Probably the truest statement we could make is, "From the parts of the picture that have so far been revealed, it appears that the two problems are linked."

The simple truth is that the whole picture has been revealed and they are linked, absolutely and fundamentally.

 

As you point out, climb is a problem that is only amenable to integration.  Acceleration is soluble with the Ps formula you have supplied.  That seems a significant difference to me.

Both climb and acceleration can be solved for using the Ps equation. For a fighter pilot there is no real difference, because positive specific excess power means that he has the ability to climb or accelerate or both.

Badboy

[Badboy]

Originally posted by HoHun:
So this is the point where you'd either have to accept the connection I proved, disprove it formally, or state that you believe I'm wrong whereupon we could both move on to further aspects or the performance comparison :-)

Henning (HoHun)

Or, as more usual with Dwarf, he will continue to offer specious argument, ad nauseam :-)

Badboy

[Dwarf]

Originally posted by HoHun:
Hi Dwarf,

So this is the point where you'd either have to accept the connection I proved, disprove it formally, or state that you believe I'm wrong whereupon we could both move on to further aspects or the performance comparison :-)

Regards,

Henning (HoHun)

It seems my meaning in an earlier post got misunderstood.  

What I meant by: "Since the equations to prove it don't currently exist, it's rather hard to conclusively demonstrate that they are not.", is that the formal proofs are all on your side.

Could we agree that climb is a special case of the broader acceleration problem?

Dwarf

[F4UDOA]

Heya Wells,

I wasn't speaking of your protracter. However there seems to be a minor conflict going on hear that is somewhat off topic.

Anyway my chart and Zigrats Spreadsheet numbers do give a climb rate of 3160 Per minute. However if you look at the AH charts they show a climb rate of nearly 3500FPM.

My question is based on the numbers used in AH more than IRL. I already have data showing otherwise such as the F4U-1D/F6F-5 vrs the A6M5 and the FW190A5 vrs F4U-1D and F6F-3.

Since you are involved in another simulation what do you feel is the best means of FM modeling.

1. FM based on calculation such as Zigrat's and your spreadsheet.

2. Based on test data such as NAVAIR docs and test trials documents.

3. Based on anecdotal evidence. Such as testamony from pilots.

Just curious.

[Zigrat]

dont use my stuff and expect any type of accuracy greater than 90%. The fidelity just isn't there and I don't want people running around using those numbers as bible. Now if something is like 50% off of what it calculates or something, then mabye something is up. If we could predict airplane performance with an excel spreadsheet i wouldnt have to school for four years to concentrate in one discipline of aerospace engineering.

as for dwarf, the other two are correct. Ps translates directly to either climb or acceleration, they are interchangable. thats why Ps vs mach number and altitude is such a valuable plot, since it shows both climb rates at any gives speed, maximum speed versus altitude, and aircraft ceiling. Its basically everything you would want to know about flight at a single level of g.

As for why the f6f climbs better, I don't know. Mabye something about the airplane? The propeller or the cowl flaps? Dunno.

Just to reiterate please don't use that excel spreadsheet as a tool and expect accurate results. The prop efficiency stuff is purely fictional, as are all of the other curves I have seen here. To say that they can be aoppplied to any ww2 airplane is kind of silly.

Plus there are all sorts of factors that this type of analysis dooesnt consider. The fact that the wings are blown by the propellers, stuff like cowl flaps opening and closing based on throttle setting, variations in lift/drag due to camber, thickness, twist, airfoil selection, and planform. The list goes on and on.

Basically I just made up that sheet for fun, and to see if HTCs numbers were in the ballpark. They are.

[HoHun]

Hi Dwarf,

>Could we agree that climb is a special case of the broader acceleration problem?

Climb and acceleration are manifestations of the broader specific excess power complex.

Regards,

Henning (HoHun)

[HoHun]

Hi F4UDOA,

>Why does the F6F-5 have a higher climb rate than the F4U?

>[...]

>Using these numbers the F6F should not be able to break 3,000FPM but somehow it reaches 20,000FT in 7.0 minutes with an initial sustained climb rate of almost 3500FPM.

The Navaer charts list a combat power climb rate of 2980 fpm at 12740 lbs and 2000 HP for the F6F-5. (Time to 20000 ft is listed as 7.7 min.) However, I think this might be without water injection - at 2250 HP power as listed in the above chart, I'd expect the climb rate to be about 3500 fpm. Unfortunately, I can't confirm which power setting the 60" Hg boost listed in the above chart refers to.

Regards,

Henning (HoHun)

[niklas]

Originally posted by HoHun:

My equations are a formal proof in the true sense of the word.

The proof could be flawed, but the only way to disprove the direct connection of climb and acceleration is to do the math and find the flaw.

Actually you're right when you say power will determine acceleration and climb, but you're wrong if you think your horizontal acceleration can be easily converted into a climbspeed.
The problem is that during your climb, you gain altitude. This means your best climbspeed changes. Stay at constant TAS and your drag changes. Stay at constant IAS (usual procedure) and you have the problem that TAS becomes higher during your climb. This means your plane accelerates during your climb, and a (small) amount of your power is also used for this acceleration. So your climb performance is in a dynamic climb a bit lower with constant IAS compared to a static climb calculation (simple excess power calculation) what doesn't take this effect into account.

BTW Hohun, your drag power curve in your chart is wrong. You seem to neglect induced drag.  Drag, as a force, has a v^-2 characteristics at slow speeds (induced drag) and a v^2 characteristics at high speed (zero lift drag). This means for a power (F*V) the characteristic should be v^-1 near stall speed and v^3 at high speed. But your curve is flat (v^0) at slow speed and ~v^2 at high speed, so it's either a drag FORCE (neglecting induced drag) in a POWER chart, or its wrong.

niklas

[Badboy]

Originally posted by niklas:


Actually you're right when you say power will determine acceleration and climb, but you're wrong if you think your horizontal acceleration can be easily converted into a climbspeed.

Not climbspeed, climbrate! And that is not wrong, horizontal acceleration can easily be converted into a climbrate. Naturally the situation is dynamic, as you point out, but the relationship still holds even though the speed may be changing, and still works for any particular instantaneous airspeed.

 

The problem is that during your climb, you gain altitude. This means your best climbspeed changes. Stay at constant TAS and your drag changes. Stay at constant IAS (usual procedure) and you have the problem that TAS becomes higher during your climb. This means your plane accelerates during your climb, and a (small) amount of your power is also used for this acceleration. So your climb performance is in a dynamic climb a bit lower with constant IAS compared to a static climb calculation (simple excess power calculation) what doesn't take this effect into account.

None of that makes the slightest bit of difference to the validity of the conversion, because the Ps equation compensates for changes in speed as they occur. Here is the Ps equation in another form:

Ps = v/g * Vdot + Hdot

Where Vdot is the rate of change of velocity with time, and Hdot is the rate of change of altitude with time.

You should be able to see from that equation that variations in one term, automatically effect the others so that if you evaluate it at any particular moment, you will always get a useful and meaningful answer, and you can always determine what either the climb rate, or acceleration might be from any particular Ps value, or specific excess power. If you set Hdot = 0 you can find the acceleration, and if you set Vdot = 0 you can find the climbrate. Alternatively, you can determine what the climbrate would be for any given current acceleration, or visa versa.

Hope that helps.
 

Badboy

[HoHun]

Hi Niklas,

>Actually you're right when you say power will determine acceleration and climb, but you're wrong if you think your horizontal acceleration can be easily converted into a climbspeed.

I'm sure you noticed the bullet-proof proof up there :-)

>The problem is that during your climb, you gain altitude. This means your best climbspeed changes.

If altitude changes, so does Ps. Ps is specific to the precise flight condition, and the connection of climb and acceleration is a fact as long as you compare them for the same precise flight condition. If you don't observe this, the results will seem like a contradiction at first, but they really aren't.

>BTW Hohun, your drag power curve in your chart is wrong.

It's correct, but I still think you're a good observer :-) The left border actually is not the stall limit, but the Clmax limit so you would see the behaviour you described if I hadn't simply omitted it from the graph as uninteresting.

Regards,

Henning (HoHun)

[Dwarf]

Originally posted by HoHun:
Hi Dwarf,

Climb and acceleration are manifestations of the broader specific excess power complex.

Regards,

Henning (HoHun)

     

I LIKE it.

Dwarf

[Dwarf]

Originally posted by Badboy:


Here is the Ps equation in another form:

Ps = v/g * Vdot + Hdot

Where Vdot is the rate of change of velocity with time, and Hdot is the rate of change of altitude with time.

You should be able to see from that equation that variations in one term, automatically effect the others so that if you evaluate it at any particular moment, you will always get a useful and meaningful answer, and you can always determine what either the climb rate, or acceleration might be from any particular Ps value, or specific excess power. If you set Hdot = 0 you can find the acceleration, and if you set Vdot = 0 you can find the climbrate. Alternatively, you can determine what the climbrate would be for any given current acceleration, or visa versa.

Hope that helps.
 

Badboy

Some folks may be more used to seeing the equation in this form:

Ps = dh/dt + (V/g) * (dV/dt)

Much as it will surprise him to read it, I agree with Badz here.  This equation (in either form), is somewhat more accurate.

Applying this formula to the P-38 and F4U climb/accel example WAY up-topic gives results of:
Climb -
F4U = 59 ft/s (meaning it has to fly 59 feet in order to climb 52 feet)  making its actual airspeed 233 ft/s.
P-38 = 70 ft/s and airspeed 271 ft/s.

Accel -
F4U = 8.43 ft/s^2
P-38 = 8.75 ft/s^2

If you prefer something more familiar, simple Trigonometry gives an equally valid answer to the climb problem.  

When you climb, you actually fly the hypoteneuse of the triangle defined by both the horizontal distance you cover and the vertical distance you cover.  Climbrate just tells part of the story.

Dwarf

[gripen]

Dwarf,
Hm... AFAIK the speed discused here have been all the time airspeed (IAS or TAS, not ground speed) so there is no reason for angle correction. There might be a little difference in the planes angle of attack (Cl and therefore also in the drag) during full power climb or acceleration depending on case and therefore also in the IAS/CAS correction, but those difference are small and errors caused by these are also small at the best climb rate speed.

gripen

[HoHun]

Hi Dwarf,

>If you prefer something more familiar, simple Trigonometry gives an equally valid answer to the climb problem.

I'm afraid you're making no sense at all.

>Ps = dh/dt + (V/g) * (dV/dt)

If you're accelerating with no change of alitude, dH/dt = 0. If you're climbing at a constant speed, dv/dt = 0.

Regards,

Henning (HoHun)