Author Topic: Beating A very dead horse...AGAIN:)  (Read 5632 times)

Offline Urchin

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« Reply #45 on: November 14, 2003, 01:25:52 PM »
Quote
Originally posted by GODO
So N1K2 is slow, right? Well, straight and level speed is the most common way to measure how fast is a plane, but not the only way. Speed after a 5g 180d turn also counts, and mostly when you are able to repeat that turn several times. Try to set up a loop circuit 1km long and put N1K2 and P51 on a race of 10 laps, I bet N1K2 will be the winner by far.


That'd measure acceleration, and yes, the N1K has better acceleration than the P-51 does.

Offline kj714

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« Reply #46 on: November 14, 2003, 02:15:23 PM »
So does lag always work to the advantage of the pilot who'
s laggy? Why wouldn't lag coming and going just end up making the game unplayable for the lagger?

Offline mold

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« Reply #47 on: November 14, 2003, 03:38:55 PM »
Quote
Originally posted by kj714
So does lag always work to the advantage of the pilot who'
s laggy? Why wouldn't lag coming and going just end up making the game unplayable for the lagger?


I think you are correct that the lagger doesn't get too much of an advantage.  However, I DO believe that lag gives an advantage to the guy shooting the guns.  Any response or extension by the guy in front will be delayed to the FE of the guy in back, and the shooter's FE gets to determine whether any hits were made.  The laggier the connection, the worse this effect is.  I've been on both sides of this equation, BTW, and to me neither case is satisfactory (although I will admit I prefer winning to losing). :)

One small advantage for the lagger is that the lagger sees all planes lagging and therefore expects this behaviour all the time as he is fighting; while to the non-lagger such behavior comes as a surprise.  I hesitate to call that an advantage, though, since it must suck to play the game where eveyone else is laggy all the time.  However, someone who expects lag can use that information to their advantage--e.g. they might opt to take an immediate shot rather than manuvering for a better shot (and temporarily presenting their back to a con), because they know that having the enemy in front is an advantage.  While the non-lagger doesn't know what the lag is with the con he is fighting, so he might unwittingly make a decision that puts him at a lag-induced disadvantage.

Edit:  One corollary of all this is that lag gives some advantage to TnBers over BnZers--whether they are the laggy con or the good con.
« Last Edit: November 14, 2003, 03:43:29 PM by mold »

Offline GODO

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Beating A very dead horse...AGAIN:)
« Reply #48 on: November 14, 2003, 04:08:42 PM »
Quote
Originally posted by Urchin
That'd measure acceleration, and yes, the N1K has better acceleration than the P-51 does.


All in all it may be considered "acceleration", but the previous case is more related to E-retention. My point is that raw top speed does not convert a plane into a slow or a fast one during combat. Not considering the "dive away" tactical move (where the diving plane is not more a danger), we may take 300 mph as a "top" speed for a close fight, so, the plane that is able to keep more time closer to that speed while maneouvering (and not diving away) would be considered the fastest one. Key factors for that are acceleration and E-retention, and N1K2 shines at both.

Offline HoHun

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« Reply #49 on: November 14, 2003, 05:34:51 PM »
Hi Widewing,

>Maybe we can put this stuff to rest with a simple test. As I have shown elsewhere on the BBS, mass is the primary factor in a zoom climb.

While it's a great idea to do tests, and I prefer tests over "common knowledge" any time, I'd like to point out that your observation is not strictly correct. (That doesn't mean it can't be an accurate summary of your test results anyway.)

How does mass influence zoom climbs without air resistance?

Assuming an engine-less zoom climb to a standstill:

E kin = E pot

m/2 * v^2 = mgh

h = 1/(2g) * v^2

Zoom height is independend of mass.

What does the introduction of air resistance change?

1) We've got to subtract the energy lost in the pull-up from the total energy. The more mass an aircraft has, the more energy it's going to lose. This means increased mass leads to decreased zoom height.

2) In the straight part of the zoom climb, we've to subtract the energy lost to drag from the total energy. One part of the drag consists of the induced drag, that increases with increased mass. Again, increased mass leads to decreased zoom height.

What does the introduction of engine power change?

If the engine is running during the zoom, the total energy is increased by the sum of the energy from the engine thrust. This energy is independend of mass, but depends on engine power and duration of the manoevre.

This leads to the conclusion that zoom climb ability does not improve with increased mass, but that actually the opposite is true.

As other factors, mainly engine power and aerodynamics, influence total zoom height, it's not suprising that the test results don't yield a list of planes sorted by mass with the lightest planes as the best zoomers. It's a bit surprising it comes out in the opposite sort order though :-) Still, that must be due a combination of other factors.

(You could confirm that Aces High modelling yields realistic zoom climb abilities by performing your test with the same aircraft type twice, once at very low fuel and once with a full fuel load. The former configuration should give the better zoom height than the latter.)

Regards,

Henning (HoHun)

Offline mold

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« Reply #50 on: November 14, 2003, 07:14:36 PM »
Quote
Originally posted by HoHun
Zoom height is independend of mass.


Yes, this is absolutely true, and often forgotten.  Remember the hammer+feather drop on the Moon....

Quote
Originally posted by HoHun
1) We've got to subtract the energy lost in the pull-up from the total energy. The more mass an aircraft has, the more energy it's going to lose. This means increased mass leads to decreased zoom height.


Also true.

Quote
Originally posted by HoHun
2) In the straight part of the zoom climb, we've to subtract the energy lost to drag from the total energy. One part of the drag consists of the induced drag, that increases with increased mass. Again, increased mass leads to decreased zoom height.


This is not quite true, and not quite the right way of looking at it.  It is better to start from the kinetics and then arrive at energy, for this problem.  Let us simplify this case a little and pretend we are zooming straight up (AOA=0, CL=0, induced drag = 0).  Now also pretend the total drag is the same for both airframes.  The drag force will be the same, but the force is being applied to a smaller mass.  The kinetic equation is:

x(t) = V(0)t - 0.5*a(t)*t^2
x(t) = V(0)t - 0.5*(F(t)/m)*t^2

Now F(t) is not a constant with time, since drag decreases with decreasing speed, but since we are comparing 2 planes with the same drag, we can take drag as constant wrt speed for comparison purposes.  From this equation, we see that to maximize zoom distance, we have to minimize F/m, which means higher mass gives us a higher zoom climb.

Now we can look at induced drag separately, which is of course worse for higher mass.  Perhaps we can even come up with a formula, which determines what AOA/CL (if any) is enough to cancel the mass advantage in a zoom climb.

Quote
Originally posted by HoHun
If the engine is running during the zoom, the total energy is increased by the sum of the energy from the engine thrust. This energy is independend of mass, but depends on engine power and duration of the manoevre.


This has a similar effect as drag, but in the opposite direction.  So, less mass has higher zoom given the same engine power.

Both of these things (drag and thrust) need to be put together to determine whether mass aids or hinders the zoom.  In short, when drag is greater than thrust, higher mass is better.  When thrust is greater, lower mass is better.  Drag vs thrust changes with time as you are zooming.

Offline mold

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« Reply #51 on: November 14, 2003, 07:31:10 PM »
And now that I think about it, what that means is that lighter planes are better zoomers at 100% power (all else equal), since 100% thrust is greater than drag at all speeds below Vmax.

A good test to see if this jives with reality is to take up two planes, one with 10% fuel and the other with 100% fuel (no drop tanks of course).  See which one zooms better at 100% power.  That way all else is equal, except the mass.

Offline Widewing

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« Reply #52 on: November 14, 2003, 10:33:29 PM »
Quote
Originally posted by mold
And now that I think about it, what that means is that lighter planes are better zoomers at 100% power (all else equal), since 100% thrust is greater than drag at all speeds below Vmax.

A good test to see if this jives with reality is to take up two planes, one with 10% fuel and the other with 100% fuel (no drop tanks of course).  See which one zooms better at 100% power.  That way all else is equal, except the mass.


I've done this offline with and without power, at 25% and 100% for most of the fighters in the plane set.

Without power, the A-20G beats every fighter. Here the 109G-10 and Tempest did poorly. Add fuel to weight to increase zoom height.

With power (I have tested just 14 types), the 109G-10, P-38L and F4U-4 are at the top of the heap. Subtract from fuel weight to increase zoom height.

My regards,

Widewing
My regards,

Widewing

YGBSM. Retired Member of Aces High Trainer Corps, Past President of the DFC, retired from flying as Tredlite.

Offline HoHun

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« Reply #53 on: November 15, 2003, 04:57:40 AM »
Hi Mold,

>>2) In the straight part of the zoom climb, we've to subtract the energy lost to drag from the total energy. One part of the drag consists of the induced drag, that increases with increased mass. Again, increased mass leads to decreased zoom height.

>This is not quite true, and not quite the right way of looking at it.

Actually, I can't see why it shouldn't be true, or not the right way of looking at it :-)

>It is better to start from the kinetics and then arrive at energy, for this problem.  

Hm, I don't understand what your formula would tell us about energy. In my opinion, it should look like this:

Edrag = integral Fdrag ds

Even at equal drag, a lighter plane is going to zoom higher than a heavier plane. Induced drag increases the drag of the heavier plane, so it's going to be even worse. The lighter plane might spend more energy  on drag, but that's only because it goes a longer way up.

>>If the engine is running during the zoom, the total energy is increased by the sum of the energy from the engine thrust. This energy is independend of mass, but depends on engine power and duration of the manoevre.

>This has a similar effect as drag, but in the opposite direction.

Well, I'd say

Emot = integral Thrust ds

Regards,

Henning (HoHun)

Offline GODO

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« Reply #54 on: November 15, 2003, 06:20:46 AM »
Quote
Originally posted by HoHun
Induced drag increases the drag of the heavier plane, so it's going to be even worse.


Is that true when your plane is climbing vertically? How can that drag come from weight when you are in a vertical possition and ascending?

Offline HoHun

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« Reply #55 on: November 15, 2003, 06:46:46 AM »
Hi Godo,

>How can that drag come from weight when you are in a vertical possition and ascending?

Well, that's an increase by zero, which is the borderline case of an increase mathematically :-)

However, even going straight up, induced drag certainly isn't going to decrease with increased mass.

Regards,

Henning (HoHun)

Offline mold

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« Reply #56 on: November 15, 2003, 07:38:33 AM »
Widewing--

Thanks for doing those tests.  I think the tests with the fuel changes only are  pretty conclusive.   Power off, higher mass wins.  Power on , light mass wins.  However, I'd say power on is a slightly more common situation.  Therefore I think a good general conclusion is that lighter planes are better zoom climbers.


HoHun--

Quote
Originally posted by HoHun
Hm, I don't understand what your formula would tell us about energy.


My formulas say nothing about energy directly.  But realize that what matters to us is height, not energy.  The kinetic equations I mention provide us with the answer directly (the height of the zoom).  The result of those equations is clear--if the engine is off, and there is air resistance, the heavier plane zooms higher.

Edit:  But your points regarding higher induced drag and higher energy expenditure when changing directions to enter the zoom are both correct.  Both of these factors will reduce the advantage of higher mass in a no-power situation, and increase the advantage of lower mass in a power-on situation.


Quote
Originally posted by HoHun
In my opinion, it should look like this:

Edrag = integral Fdrag ds


Yes, you can also look at it this way, and you will arrive at the same result as I did with the kinetics.  Physics is consistent. :)  Let us start with your first equation, which is correct for no air resistance:

Ekin = Epot

Now, add the energy from drag to the balance:

Ekin = Epot + |Edrag|

Epot = Ekin - |Edrag|

mgh = 0.5mv^2 - |integral F ds|

h = 0.5v^2/g - |integral F ds|/m

And once again, you can see that increasing mass increases height of zoom.


Qualitatively, you can think of this experiment.  Take a solid metal ball, and a hollow plastic ball of the same size.  Drop both of them.  Which one hits the ground first?  The metal one, of course.  Why?  Because the higher mass and  higher inertia are able to "overcome" the equal force from air resistance.  But now go to the moon, and do the same thing.  Both balls hit the ground at the same time, because no air resistance.


Quote
Originally posted by HoHun
Well, I'd say

Emot = integral Thrust ds


Yes, and in this case "Thrust" is acting in the same direction as "ds", so of course it adds to the energy rather than subtracting.  We have therefore:

Epot = Ekin - |Edrag| + |Emot|

mgh = 0.5mv^2 - |integral Drag ds| + |integral Thrust ds|

h = 0.5v^2/g + (integral (|Thrust| - |Drag|)/m ds)

So once again, if Thrust is greater than Drag, lower mass is better.  If Thrust is less than Drag, higher mass is better.  The kicker here is that Thrust is always more than Drag in a zoom climb if you still have engine power, which is why the 109G10 is the best zoomer in the game. :)
« Last Edit: November 15, 2003, 07:46:18 AM by mold »

Offline HoHun

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« Reply #57 on: November 15, 2003, 01:53:58 PM »
Hi Mold,

>h = 0.5v^2/g - |integral F ds|/m

Slight correction:

>h = 0.5v^2/g - |integral F ds|/(mg)

>And once again, you can see that increasing mass increases height of zoom.

I see your point! As the second term actually isn't independend of mass, that benefit isn't "pure", though.

The conclusion probably is that a 2.5 G pull-up, 90° zoom will favour the heavier plane more than the 5 G pull-up, 45° zoom, due to the variation of the drag term :-)

>The kicker here is that Thrust is always more than Drag in a zoom climb if you still have engine power

... unless you start your zoom going faster than level top speed, like in a zoom from a dive :-)


Regards,

Henning (HoHun)

Offline WhiteHawk

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Re: Beating A very dead horse...AGAIN:)
« Reply #58 on: November 17, 2003, 04:32:01 PM »
Quote
Originally posted by humble
This is a "curiosity" whine....

Been flying the spitty abit over last couple days...1st time in a long time. One thing it brought back to me is the uber E retension of the nikki. I was tooling along at 14k or so watching inbounds coming toward base I upped from...level with wep at ~340 or so...watched slightly higher alt nikki alter to come in...turned into him at 7.0 or so and took under on merge and extended dead level as he went for bad nose down shot...any way watch him pulling max G's bending down around...and see him rocket back up my prettythang...now over the last couple days I've seen Hogs, ponies, 38's, la-7's & jugs all try that move and fail to close enough to force me to break of the rope...the nikki not only could of caught me...it would of climbed 1500 ft over me:)...course I broke of back into him and had no further problems with him...but I can't believe the nikki can pull max G turn and retain E better than any other plane in the game by that much...

This ends the latest nikki whine...

hopefully AH2 will change the FM for the better....bring back 1.3 FM:):):aok


 I lost a 262 the exact same way.  There is no possible way that niki could have  hung on its propeller for 1/10th the amount of time that it did.    I dont fly nikis, only because I am quite sure they are gioing to fix this blatant misrepresentation of the laws of physics, and I dont want the helicoptor to be my finishing move.

Offline Widewing

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« Reply #59 on: November 17, 2003, 06:49:32 PM »
Quote
Originally posted by Widewing
I've done this offline with and without power, at 25% and 100% for most of the fighters in the plane set.

Without power, the A-20G beats every fighter. Here the 109G-10 and Tempest did poorly. Add fuel to weight to increase zoom height.

With power (I have tested just 14 types), the 109G-10, P-38L and F4U-4 are at the top of the heap. Subtract from fuel weight to increase zoom height.

My regards,

Widewing


As a follow-up, I tested most of the fighters for max power zoom climb. The top 5 are pretty much what you would expect with one surprise. Overall, the best zoom climber was the Spitfire Mk.XIV, beating the 109G-10 by about 200 feet, roughly the distance that covered the rest of the group. The Spitfire Mk.XIV easily zoomed away from the Tempest, which fell well behind the F4U-4 too.

My tests showed the following order for the top five zoom monsters:

1. Spitfire Mk.XIV
2. Bf 109G-10
3. F4U-4
4. La-7
5. Dead heat between La-5FN and Bf 109G-2 and P-38L

Of these, the P-38 was in my opinion, the most dangerous due to its superior control at stall speed (no torque).

After these, there is a closely bunched group including the Spitfire Mk.IX, Tempest, N1K2-J, Fw 190D-9 and Bf 109F.

My regards,

Widewing
My regards,

Widewing

YGBSM. Retired Member of Aces High Trainer Corps, Past President of the DFC, retired from flying as Tredlite.