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washout question

Old 06-26-2016, 08:41 AM
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alex5
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Default washout question

If the wingtip washout is 2 degrees,and the wing dihedral is 2 degrees, does that mean that the rear edge of the wing has a 4 degree dihedral,as opposed to the 2 degree wing chord dihedral?? Alex
Old 06-26-2016, 10:23 AM
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No I means the tip rib would be 2 degree higher The want ever the angle of incident the root rib is. So lets say the root rib angle is 2 degrees and the tip rib angle would be 0 degrees. The change in the dihedral will vary at the trailing edge, the important thing is the relationship of the root and tip ribs!!
Old 06-26-2016, 02:13 PM
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that is what I am getting at.The sum total at the rear of the tip outer chord.is 4 degrees dihedral,whereas the chord dihedral(usually at the max Height of the chord),of the basal airfoil,is 2 degrees.The chordal dihedral ,is drawn as a line inclined at 2 degrees,and the washout is drawn as a line from the posterior tip at -2 degrees,resulting in a horizontal trailing edge,and a climbing max thickness line following the 2 degrees dihedral
Old 06-26-2016, 04:41 PM
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Chad Veich
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The trailing edge will have a greater amount of dihedral but it is not directly related to how many degrees of wash out you have. The degrees of wash out is based upon the difference in angle between the root and tip ribs and the difference in height at the trailing edge is a rotational dimension based upon the length of the tip rib and the point of rotation.
Old 06-28-2016, 12:22 PM
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The trailing edges would be higher but not by as much as you're thinking. You can't add the angles directly like you are doing.

The vertical height from an angle like this is related to a function of the angle times the horizontal distance. And for the two angles you don't add the angles. Instead you need to calculate the vertical height for that horizontal run and angle for the two cases then add the verticals together.

If we're talking trigonometry the rise related to the angle change is the sine function of the angle times the length of the lower side or, in this case, the span or chord depending on which you are looking at.

The sine of 2° can be found by using the Scientific mode of the calculator that comes with Windows. The value in this case for 2° is 0.035. So the rise at the trailing edge due to dihedral is .035 x the span of the one side. The additional rise due to your 2° of washout is .035 x the tip chord if you set the dihedral at the leading edge to 2°. Or if you set the dihedral at the main spar to 2° then it'll be .035 times the distance from the spar to the trailing edge. Add the rise from the panel's span to the additional rise from the washout and you get your total added rise at the tip of the trailing edge.

To go backwards from the added vertical heights to find the angle of the trailing edge in degrees gets a little more iffy. How we know the tip vertical and the half span lengths. To find the resulting angle we need to do a reverse or inverse "tangent". In trigonometry this would be the vertical length divided by the horizontal panel span then do an "arctan" on the resulting value to get the dihedral angle of the wing's trailing edge.

Which right about now is likely more than you wanted to know, right?
Old 06-28-2016, 12:28 PM
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Edit function is wacky. Sorry for the added reply.

The other way to do this if you don't want to bend your brain with trig is to simply draw up a small scale diagram. Like 1/8 inch per inch of the full size model. Draw a line out that is the scaled length of the half span. To that you would extend the line by the wing chord or spar to TE length at the washout angle. In this case they are the same so just make one line that is the span plus the chord or partial chord. With a line parallel to the base or "0°" line that you have coming off the center end of the "wing" line extend the end of the span plus chord end back over the tip end of the span line. That's your added height due to the washout. Draw a new line back to the center line end of the wing and measure the new angle.

Likely as not the difference will be less than 1/2 a degree.
Old 06-28-2016, 01:06 PM
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well,dont thank Sensei for input..

Last edited by alex5; 06-29-2016 at 10:12 AM.
Old 06-28-2016, 01:51 PM
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well,I did it mathematically...I drew an angle from my chord max alt,along the chord, at wing fuselage base at 2*.Call it the center of my wing spar. That made my dihedral,then rotated the whole airfoil 2 degrees to give me my angle of attack.I put a copy of that airfoil at the distal end of my spar(which I rotated about my max chord locus -2 degrees at the trailing edge).I then drew a line from the trailing edge at the fuse,to the trailing edge at the far end of my wing.I spaced my profile along those 2 lines at 3 inch intervals along the wing(48 "),using the dihedral line,and that trailing edge line,to align my profiles.Each profile is rotated -2 degrees divided by 16(the number of ribs)=.125 degrees.
Old 06-28-2016, 01:52 PM
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You guys went diving deep!

It is this simple:
Dihedral is an angle measured using the wing spar or wing saddle on the fuselage as a reference.
Washout is an angle measured using the wing Chord at the root as a reference.

Typically these two reference points are perpendicular to each other, so it is practically impossible to create a correlation without going deep into math.

Rafael
Old 06-28-2016, 02:08 PM
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this is so easy ,since I use cad.I place my base airfoil at 0,0,0.Including my chord line.I have max chordal perpendicular in my airfoil,which intercepts my chord.I draw my dihedral line,2 degrees from horizontal,starting at 0,0,0.I rotate the source profile 2 degrees about the chordal intercept for angle of attack. I create a non associative path array,using my dihedral line as the path and my base point as the chordal intercept point.this creates the distal profile at 48 inches. I rotate the entire distal profile -2 degrees about the trailing edge point,and move the profile vertically, to align with the chordal intercept.I draw a line from the trailing end point of the profile at both locations.Then I create an array(16) along the path,of the trailing edge lie,and manually rotate each profile to align with the chordal intercept,and the trailing edge line.the end result is profiles that are centered about the chordal intercept(dihedral) and rotated to suit washout at the trailing edge.
Old 06-28-2016, 04:26 PM
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Default wing rib washout

here is an image of the layout.Note the base rib has a 2 degree attack angle,and the outer rib has 2 degree washout Total dihedral 2 degrees....but centered on the max chord height point on the chord line,after rotating on the trailing edge.
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Old 06-29-2016, 04:26 AM
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Originally Posted by alex5
If the wingtip washout is 2 degrees,and the wing dihedral is 2 degrees, does that mean that the rear edge of the wing has a 4 degree dihedral,as opposed to the 2 degree wing chord dihedral?? Alex
Originally Posted by alex5
this is so easy ,since I use cad.I place my base airfoil at 0,0,0.Including my chord line.I have max chordal perpendicular in my airfoil,which intercepts my chord.I draw my dihedral line,2 degrees from horizontal,starting at 0,0,0.I rotate the source profile 2 degrees about the chordal intercept for angle of attack. I create a non associative path array,using my dihedral line as the path and my base point as the chordal intercept point.this creates the distal profile at 48 inches. I rotate the entire distal profile -2 degrees about the trailing edge point,and move the profile vertically, to align with the chordal intercept.I draw a line from the trailing end point of the profile at both locations.Then I create an array(16) along the path,of the trailing edge lie,and manually rotate each profile to align with the chordal intercept,and the trailing edge line.the end result is profiles that are centered about the chordal intercept(dihedral) and rotated to suit washout at the trailing edge.
From a simpleton question to this bla,bla,bla. This is important why?

Bob
Old 06-29-2016, 05:08 AM
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Really Bob.I would have rotated the entire wing about the trailing edge,had many people not stepped up to give valuable input.
Old 06-29-2016, 05:15 AM
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Lol...
Old 06-29-2016, 09:19 AM
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Just use flaps, it accomplishes the same thing.
Old 06-29-2016, 10:13 AM
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alex5
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Washout accomplishes a dynamic task as the airplane reduces speed...it lowers the stall speed,without using servo input implied by the use of flaps,and enhances the stability(again at lower speed)
Old 06-29-2016, 12:11 PM
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Chad Veich
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Originally Posted by rgburrill
Just use flaps, it accomplishes the same thing.
Either I don't understand how wash out functions or you don't.
Old 06-29-2016, 12:26 PM
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alex5
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flaps perform the function of altering the airflow...creating lift. Washout ,on the other hand prevents the tip stall at a given speed,in relation to the root.It is a different effect.Washout allows the wing to regain controlled airflow,by preventing an uncontrolled spin,often generated by a tip stall. Try this link,as an explanation...since you dont seem to get it.
http://www.rcscalebuilder.com/tutori...ut/washout.pdf

Last edited by alex5; 06-29-2016 at 12:35 PM.
Old 06-29-2016, 05:55 PM
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To be very honest with you flaps or washout isn't needed if you build very light, I have built many models over the years, and I don't think I have built even one with washout in over 30 years. I keep everything I build in the low to mid 20 oz. wing loading and all those nasty bad habits go away. Oh sure you can use flaps to slow down but you can also cross up the airplane to slow down as well...

Bob
Old 06-29-2016, 06:36 PM
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by incorporating washout (altho not totally necessary) in the design,I can remove some unwanted low speed flight characteristics,and gain a more stable aircraft ,while doing so.Like if I happen to have a higher wing loading.Just because it wasn't in your arsenal,doesn't mean you shouldn't be using it.
Old 06-30-2016, 04:32 AM
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By learning to build lighter (although not totally necessary) and getting the wing loading down you also get rid of those unwanted low speed characteristics and gain stability, only you gain in inverted flight, right side up flight, and well... The entire flight envelope really, unlike a more porky counterpart with built in washout as a band aid. Just because you most likely have not pushed the envelope into lightweight construction methodology doesn't mean you shouldn't be at least exploring it. Sport aircraft, aerobatic aircraft, and warbird aircraft gain all around stability and overall performance with a lighter loading.

Bob

Last edited by sensei; 06-30-2016 at 04:48 AM.
Old 06-30-2016, 07:08 AM
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According to all the RC build manuals,the flight characteristics of the wing are not modified by weight,although the total envelope is.A wing stalls,when it stalls,because of the design.If you build in washout,as pointing at my original question,you can create better flight characteristics.I can expand that (improve on that) by lessening wing loading,but I would be unable to change AOA generated stalls, regardless, since the wing configuration generates those,relative to wing speed and thrust. Don't get it wrong,building light is a necessary component to staying in the air,and also being able to control the flight path up or down.The plane has to be responsive to input,but the wing,as we get into larger and larger(realistic) designs,has to be optimized for airflow.That is why we talk about washout,at all.
Old 06-30-2016, 07:30 AM
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Well there you go, you follow those build manuals and I will follow my own 53 years of practical designing, building and flying experience.

Happy Flying Alex,

Bob
Old 06-30-2016, 07:52 AM
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gonna pull the "Old" card.I started modifying airplanes with my Cox PT19 in 1962.Then I designed (scratch build) a Corsair jet.and on and on...Now I just keep learning.
Old 06-30-2016, 08:23 AM
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The premier designer-builder of RC airplanes , particularly STOL and warplane types,Ivan Pettigrew, uses washout in his wing designs.There must be hundreds of his models being built or flown,and I dare say,I would not want to argue with the success of his plans and his customers' skills He uses a very light build,and still uses washout to solidify the envelope.

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