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XLFR5 Reverse Engeneering Verhees Delta

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berridos

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Oct 10, 2009
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995
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madrid
Hi everybody

I am having a hard time understanding the verhees delta in xflr5.
Basic parameters of the original: CoG between 1270 and 1390 cm
Cruise speed 220kmh = 62 ms
Nowhere is stated the reflex used in cruise, but on the main plans, i measured on an uncommented picture, an up reflex of 1,5º
After doing hundreds of trial and errors i tried to put some system into the exercise.
I launched three packs:
The first pack shows the plane on type 2 analysis with the elvon at 3º reflex and varying the cog from 1050 up to 1550cm, beeing the white one the lowest with the most forward Cog.
The second repeat the same exercise but with 2º reflex.
The last picture repeats the exercise with 0 reflex. (this pic is the only useful i guess and its the last one)
The airfoil at the root is a 63020A, thinning to 63011A at the tip.
Anheadral is 7 or 7,5º, washout 1,5º.

Appearently this plane doesnt need reflex, as the fin is pitching up more than enough so my previous research was superflous. Is my interpretation right?
The model is incomplete by a short stubb fin that protrudes below starting at the thickest airfoil section and that hosts the rear landing gear.
We do i only get all the curves on the CM vs AOA graph?

Is the nose attitude really 4,5º uop in cruise flight in the 0 reflex version?

In the 0 reflex version i always have pitching moment. Shouldnt i have steady flight with neutral pitch?
For a plane to be stable, the cm curve has to be downward sloping?
When the cm curve is horizontal, is that the neutral point?
I have a spanwise lift distribution. Does this graph have any value without the local cl max as a comparison in order to evalue the quality of the bell shaped distribution?
How does the lift distribution look like according your experience.?

I have tried reflexed elvons from 5º up to -1º degree down, but xlfr5 will always stop calculating at an AoA of 3º because the lift turns 0. Shouldnt the up reflex allow lift below 3º in steady flight?
I have a thousand more questions but hope some of these will be commented on. Thx
reflex 3.png


reflex 2.png
iso.jpgside view.pnglift distrib.jpgtopview.jpg

0 reflex.jpg
 
Last edited:

berridos

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Joined
Oct 10, 2009
Messages
995
Location
madrid
I already bought the real sized plans, The airfoil down on the left shows a 1,5º reflex. According to xflr5 thats too much in cruise.
 
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Scheny

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Feb 26, 2019
Messages
222
Location
Vienna, Austria
XFLR5 uses VLM to calculate and therefore my guess is, that the influence of the huge rudder is not fully accounted. Both rudder and wing create low pressure areas and in the lift diagram out is clearly visible that this influence is not taken into account.

I had a similar problem when calculating neural point for my plane. XFLR5 is nice for a good educated guess, but has limitations.
 

berridos

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Oct 10, 2009
Messages
995
Location
madrid
So the hypothesis that the rudder is producing pitch up is wrong? How otherwise would the plane be capable of level flight with a simetrical airfoil?
 

Speedboat100

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Nov 8, 2018
Messages
1,464
Location
Europe
XFLR5 uses VLM to calculate and therefore my guess is, that the influence of the huge rudder is not fully accounted. Both rudder and wing create low pressure areas and in the lift diagram out is clearly visible that this influence is not taken into account.

I had a similar problem when calculating neural point for my plane. XFLR5 is nice for a good educated guess, but has limitations.

I hope so..at least at greater reynolds numbers the drag is lesser...like it should be.
 

berridos

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Oct 10, 2009
Messages
995
Location
madrid
Thanks Brandon, i started at the center section. Will correct the exercise and add also the ventral fin . However got the feeling that the later jump to open foam will be unavoidable in order to tailor a better spanwise lift distributions.
Verhees told me he used a modified 63020. Did you do some work on the degree of modification of the airfoil? Is it still a 63020 to 63011?
The intersection of the wing/fin is so large that some area rule optimisation on this kind of configuration could be worth it.
Do you agree on the 7,5º degrees anhedral?
 

Deuelly

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Nov 6, 2012
Messages
233
Location
Marshall, MN
I'll have to check my notes on the airfoil when I get home. If i remember right the only difference was a smaller leading edge radius on the root airfoil.

The anhedral should be 7.2°.

Brandon
 
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