Author Topic: Some New Data Carts to chew on  (Read 3116 times)

Offline niklas

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Some New Data Carts to chew on
« Reply #135 on: January 12, 2002, 09:04:22 AM »
Quote
Ps = v/g*Vdot +Hdot


that formula does indeed mean what i tried to say. I only have a problem with the term Ps because it looks like a power. But it is a climbrate, actually itīs the climbrate in a non-dynamic, static observation.

Iīm used to the expression ((F-W)*v)/mg instead of Ps (F thrust, W drag)

W in a level flight for a given speed is not W in a climb at the same speed, so while the formula is correct you still canīt determine from a level fight acceleration your exact climbrate.

Hohun, CLmax should define your stall speed.

niklas
« Last Edit: January 12, 2002, 09:36:09 AM by niklas »

Offline Badboy

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« Reply #136 on: January 12, 2002, 10:12:29 AM »
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Originally posted by Dwarf


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

Dwarf


The only thing that surprises me is your, apparently endless, capacity to post complete nonsense.

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Offline HoHun

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« Reply #137 on: January 12, 2002, 10:18:14 AM »
Hi Niklas,

>Hohun, CLmax should define your stall speed.

Actually, the wing keeps providing lift at angles of attack somewhat higher than the one yielding Clmax ("critical" angle of attack). Though this is nice since it gives the pilot a margin of error before actually stalling, beyond critical angle of attack he'll get less lift at more drag in that region of the flight envelope.

In other words, he doesn't really want to go there :-) However, complete stall of the wing with wingdrop or even spin occur only after critical angle of attack has been exceeded by a certain margin that depends on the specific wing design.

Regards,

Henning (HoHun)

Offline Badboy

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« Reply #138 on: January 12, 2002, 10:57:03 AM »
Quote
Originally posted by niklas
that formula does indeed mean what i tried to say. I only have a problem with the term Ps because it looks like a power.


That's because it is a power, P sub s  (Ps) is the term used for specific excess power, that's the quantity we are discussing.

Quote

W in a level flight for a given speed is not W in a climb at the same speed, so while the formula is correct you still canīt determine from a level fight acceleration your exact climbrate.


Of course you can, if you are in level flight, regardless of acceleration, your climbrate will be exactly zero! I do see your point though, if you enter a climb, things will change, and of course so will the thrust and drag values in that equation. So the fact remains that you can always determine your instantaneous climbrate and acceleration from that equation. Naturally, it will be different for every flight condition, and even when you are established in a climb, the climbrate given by the Ps equation will only be correct for the instant at which the values were taken, that's why it isn't very helpful to consider isolated values. To be useful you really need a map of the Ps contours over the entire envelope. That way you can get a true impression of the behavior of the aircraft. In the same way that with a little experience, it is possible to obtain a 3 dimensional image of a hill by looking at contours on a map, it is also possible to obtain a similar perspective from an aircraft's Ps contours, or curves. Some of the work I'm doing involves a three dimensional approach to aircraft performance comparison, so instead of looking at the Ps curves or contours, I compare three dimensional surfaces, a bit like looking at a hilly terrain. It's exciting stuff :)

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Offline Badboy

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« Reply #139 on: January 12, 2002, 11:03:48 AM »
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Originally posted by F4UDOA
While everybody is checking to see who has the biggest protractor. Riddle me this...

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



I believe I can offer you some insight into this, if you would care to take it email. Drop me a note and we can talk about it.

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Offline Badboy

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« Reply #140 on: January 12, 2002, 11:46:19 AM »
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Originally posted by F4UDOA
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.



The best method would probably involve all three, where the first two are used together combining as many sources as possible and some statistical analysis, with the third used for validation and bias. Also, there are other methods that can be used, but unfortunately, regardless of the methods chosen, any expectation of results better than 10% either way would probably be naive. That's why there will always some controversy involved.

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Offline Dwarf

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« Reply #141 on: January 12, 2002, 08:59:36 PM »
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Originally posted by HoHun
Hi Dwarf,

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)


And whichever you're doing, this formula will work to determine the power required.  Just eliminate the surplus term.  If you're climbing, the formula becomes Ps = dh/dt + V/g and if you're accelerating the formula becomes Ps = VdV/gdt.  

If we set dh/dt = Hdot, and dV/dt = V dot, it's the same equation Badboy posted.

Dwarf

Offline Dwarf

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« Reply #142 on: January 12, 2002, 09:44:18 PM »
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Originally posted by 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


OK, assuming you want to make the 227 fps velocity (in the case of the F4U) IAS while flying the hypoteneuse of the triangle, then you can still use Trig to give you your new ground speed during the climb.  (221 fps).

Either way, the climb in question requires a Ps of 59 fps to sustain.  You don't get a 52 fps ROC by only spending 52 fps worth of power.

Dwarf

Offline gripen

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« Reply #143 on: January 12, 2002, 11:55:48 PM »
Dwarf,
Nonsense, the starting values for these calculations are known ROCs at known airspeeds ie steady climb rate at steady airspeed. Of course you can calculate ground speed but there is no sense att all to mix grounspeed in. I quess you are just trying to continue until others give up.

gripen

Offline Badboy

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« Reply #144 on: January 13, 2002, 06:10:27 AM »
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Originally posted by Dwarf

If you're climbing, the formula becomes Ps = dh/dt + V/g and if you're accelerating the formula becomes Ps = VdV/gdt.  



Nope, the first expression you posted assumes a climb with acceleration equal to unity. A climb with no acceleration would just give dh/dt.   Also dv/dt is a single term, so the second expression you posted is just meaningless nonsense.  


Quote

If we set dh/dt = Hdot, and dV/dt = V dot, it's the same equation Badboy posted.


You don't have to set dh/dt = Hdot because they are both well known notation (courtesy of Leibniz and Newton respectively) for the same thing.

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