L/D max

[Nich]

 Has anyone gone into the game files and figured out what L/D max is?  Also what angle is it at? If I remember correctly maximum climb occurs when you fly at L/D max.  Is it the same for all wings? Is it the same for lifting surfaces?

[NathanKell]

Wing L/D isn't the whole story, vessel L/D is. While all wings/control surfaces have the same AoA->lift and AoA->drag curves, that varies with mach anyway, and total vessel L/D at a given AoA is highly vessel-dependent.

[Nich]

8 minutes ago, NathanKell said:

Wing L/D isn't the whole story, vessel L/D is. While all wings/control surfaces have the same AoA->lift and AoA->drag curves, that varies with mach anyway, and total vessel L/D at a given AoA is highly vessel-dependent.

 But generally non lifting surfaces have a worse L/D then wings and control surfaces so tilting the wing close to this angle should increase the L/D of the vessel.  Good to know that control surfaces have the same L/D  as wings my playing testing had led me to believe otherwise but I'm sure that was just anecdotal evidence. I am very impressed that KSP has multiple curves for the different flight regimes.  Also I do believe my prior assumption is a generalization assuming jet engines have static thrust which is not the case in KSP does anyone else know any formulas for max rate of climb?

[LordKael]

The formula can be derived from the data provided in the .cfg file. For a rocket engine, it's a lot simpler, as there is only one curve to plot (atmospheric density). The .cfg file provides three points from the line at predetermined densities that highlight the curve of thrust vs atmospheric density. 

Now, looking at the .cfg for the "Wheesley", you see the following:

     // Jet params
        atmChangeFlow = True
        useVelCurve = True
        useAtmCurve = True
        flameoutBar = 0.02
        flowCapMult = 1.0
        machLimit = 0.85
        machHeatMult = 25
        velCurve
        {
            key = 0 1 0 0
            key = 0.53 0.834 0 0
            key = 1.3 0.96 0 0
            key = 1.674 0.843 -0.876726 -0.876726
            key = 2 0.1 0 0
        }
        atmCurve
        {
            key = 0 0 0 1.186726
            key = 0.072 0.092 1.339822 1.339822
            key = 0.337 0.4 0.8976688 0.8976688
            key = 1 1 0.9127604 0
        }

 

Now, this may look like a whole mess of numbers (it is), but the important bits are the parts following "atmCurve" and "velCurve". Here, the atmCurve is exactly what I described for the rockets; as the engine gets into less dense atmostphere, the thrust changes. For a jet, it lowers.

Looking at the velCurve, you can see the same format as for atmospheric density, but with a different pattern. Here, it goes up and then down, as opposed to just going down. The "key" gets looked at by another part of code, and determines the thrust output for the engine.

1 hour ago, Nich said:

 Also I do believe my prior assumption is a generalization assuming jet engines have static thrust which is not the case in KSP does anyone else know any formulas for max rate of climb?

So, to answer your question:

In order to find the formula for max rate of climb, one would have to plot the two curves on a graph and then draw a best fit line through the intersections. This would show how, as an aircraft climbs into thinner air, the airspeed will have to change to keep the aircraft in the proper speed and AoA. 

 

If a somebody who actually understands how math works, and wants to do this, I would really appreciate seeing how it looks!

Edited February 23, 2016 by LordKael

[NathanKell]

atmosphereCurve uses pressure (in atmospheres). atmCurve uses density, normalized to 1 = sea level density for standard (288K) temperature.

[KerikBalm]

Interesting... I hadn't noticed the difference between atmosphereCurve and atmCurve before.

Duna's atmosphere is a good example of why density matters. Prior to 1.05? 1.04? it had a 0.2atm pressure at 0 altitude if I recal, but the MW of the gas was set at 14, compared to Kerbin's which is what... 28 or something. Now its something in the 30's... as far as I can tell, the denisity did not change... and its the density that is important for wings and such.

Laythe went from 80% atmosphere, 80% gravity to 60% atmospheric pressure, 80% gravity... which I assumed made the air "thinner" than before, and that wings/jets would produce less than 80% (kerbin relative) lift/thrust, making it harder to lift off.

But now that I think about it.. laythe's atmosphere is significantly colder, and thus should be denser for the same atmospheric pressure.

What does this mean for lift and jet TWR on Laythe -> I don't have the exact numbers for the relative temperatures... but I can guess the values are like 250 and 300k for laythe and kerbin sea level. PV=nRT, so with that rough estimate, Laythe would have 80% the gravity, and 72% the atmospheric pressure.

Perhaps they adjusted the numbers so its still 80/80 and planes will fly the same? I had the impression that it was harder to get planes to liftoff on laythe, maybe I was hallucinating?

[Nich]

atm and gravity are not directly related.  If your atm had been stripped by solar pressure you could have an atm of .01 at sea level on a planet twice the size of earth.  Conversely if a planet forms with twice as much free gas available and has the same gravity it will have twice as much pressure at sea level. 

I believe PV=nRT is a simplification of other gas laws and only applies to the same gas.  You can not take a PV=nRT of hydrogen and compare it to a PV=nRT of oxygen or methane.

I do believe they increased the thrust on jet engines to deal with the increase drag in 1.0.5.  On laythe the drag reduction must be larger then the power loss from density change.