Polyhedral question

[Giuseppe Saroli]

Hi all I am replacing the wings on my Astro Viking because I found the original plan1942 I think of the original Cleveland Viking had a polyhedral configuration

on the plan it shows the wing centre panel not flat and with a 1.1/2" inner panels and 5" outer panel

For easy build I am right to say that if i put the left inner panel flat on board I need to double up the right inner panel to 3" and there for the tip panel would be 6.1/2 !

And if I put the right inner panel flat on the board the wing tip need to be 3.1/2"!

I hope I am making sense here math was never my best subject Thanks in advance

[Dave Hopkin]

No, if you build the inner panel flat, then the dihedral on the inner need to be subtrated from the outer to retain the same angle at the mid section, but dont forget the inner rib at the wing root need to be angled outward (top of rib towards the wing tip) to maintain the inner sections dihedral when joined.......

Edited By Dave Hopkin on 21/11/2014 10:54:41

 

To calculate the rib angle (A), you need to know the length of the inner section (H), the vertical height of the end of the inner section when at its correct dihedral (O)

Sine(A) = O/H

So Divide O/H then enter the result into this website to get the inverse sine which will be the lean angle

http://www.rapidtables.com/calc/math/Arcsin_Calculator.htm

if you dont have a calculator than does sine, then

Edited By Dave Hopkin on 21/11/2014 11:05:53 to remove a small outbreak of stupidity!

Edited By Dave Hopkin on 21/11/2014 11:14:27

[Giuseppe Saroli]

Thanks Dave

[Giuseppe Saroli]

but if i put say the inner right panel flat on the board should the outer panel be set at 3.1/2" or 2" !

[Dave Hopkin]

If the inner is flat on the board the outer tip needs to be at 3.5"

[Phil Green]

Thats not right Dave, you cant simply subtract the half-wing dihedral measurement { 1.5" }  from the full wing measurement { 5" } It would only be 3 1/2" if you lowered the wing vertically, maintaining the angle of both panels until the root of the tip panel was touching the board and the root panel was submerged in the building board!!!

 

 

Edited By Phil Green on 21/11/2014 11:35:25

[Giuseppe Saroli]

So is 2" then! if I lay centre panel flat on the board and prop up the wing tip to 2"(2 + 3 =5)

I am confused

[Dave Hopkin]

Posted by Phil Green on 21/11/2014 11:31:09:

Thats not right Dave, you cant simply subtract the half-wing dihedral measurement { 1.5" } from the full wing measurement { 5" } It would only be 3 1/2" if you lowered the wing vertically, maintaining the angle of both panels until the root of the tip panel was touching the board and the root panel was submerged in the building board!!!

Edited By Phil Green on 21/11/2014 11:35:25

There will be a tiny error because of the inner wing tips describing an arc (when lowered) but that will be almost unmeasurable in the length of arc in question)

If you want to calculate it exactly you have to use the right angle triangle rule, measuring the length of the part span (Hyponenuse) anf the height of Dihedral (Opposite Side) - divide H by Opp Side and find the inverse Sine of that number which will give you the degree of dihedral for that segment

[Giuseppe Saroli]

Ok tonight I will do it the hard way and prop it up temporally and see if I can come up with something and report on Monday thanks a lot guys Regards

[Engine Doctor]

Hello Giuseppe. If the drawing you show in the opening post is correct then of course it will work . As said the ribs at the panel roots and inner panel tip will have to be set to accommodate the angles .If it is a fraction of a degree out ,provided both sides of the wing are the same it will be fine .I have used this method when building this type of wing in the past .As you say Maths were not my strong point either but simple plans need simple methods . Give it a go .

[Biggles' Elder Brother - Moderator]

I'm sorry ED but the short answer to Guiseppe's orignal question is "no - it will not work, you cannot simply double the height and get the same angle - because sin(x) is non-linear function". To answer your question with what the height would need to be we would need the length of the inner and outer wing sections.

To see this examine the two diagrams below:

In the upper diagram we have just the inner part of the wing - positioned level - with a length of L and height at the end of 3.5 as specified on the plan. The resulting angle of dihedral is called A. The lower diagram shows what you want to achieve - ie the left wing level on the building board and the angle at 2A - the unknown height to achieve this is labelled X.

From the first diagram:

sin(A)=3.5/L

From which A=arcsin(3.5/L)

From the second diagram:

sin(2A)=X/L or,

X=Lsin(2A)

Substituting from above for A gives us:

X=Lsin(2arcsin(3.5/L)

So if for example L=10" then we would have:

X=10sin(2arcsin(3.5/10))

Which tells us that X=6.56", so not 7" as you suggested.

Note that if you did just use 7" - simply doubling the value - then for this length - ie 10" of span - your error in the dihedral would be about 2-3 degrees. Which you might not think too bad anyway!

But the bottom line is that without the lengths of the sections - inner and outer - we can't work out the exact value you need to pack the wing tip up to. But with that data it is possible to work out the the exact values.

BEB

[Phil Green]

Exactly. Agree 100%.

[Chris Barlow]

From a workbench point of view you could just prop it up as in the original post (the easy maths is correct). The angles will change a bit due to rotating the wing to get one panel flat on the bench but really the important bit here is that both angles on both sides are the same!

I know, I know. People will be protesting the angles are wrong and it's not that simple but it's not a 200mph jet or contest level scale job so really would it matter?

How many designers and builders have put a wing together on an own design and thought "that looks about right" ? Sometimes sticking wood together isn't that technical!

[Phil Green]

Agreed small errors wont make any difference, but my point was that Daves 3 1/2 inches was a mile out!

If the polyhedral break is halfway along the wing, ie root and tip sections are the same length, then extrapolating the 1 1/2 inches becomes 3 at the tip, the tip dihedral measured with the root panel flat on the board is therefore 2 inches.

It looks like the break is at 50% but we dont know that for sure, even so 2" is a better estimate than 3.5". !!!

 

Edited By Phil Green on 21/11/2014 22:00:47

[Chuck Plains]

tandysmodelplanes.com Show lots of info apart from, 'how high to prop up the wing tip'.

This online trig calculator says prop up the polyhedral wing tip with a 1.56745902104171 inch high block. That is if you are making it to match the info in the Tandy's article which has a 44.5 inch wing span. (if I got my angles right that is) (feel free to correct me if I'm mistaken)

(screen shot from the onling trig calc)

Edited By Chuck Plains on 21/11/2014 18:13:45

[Phil Green]

Posted by Giuseppe Saroli on 21/11/2014 12:11:30:

So is 2" then! if I lay centre panel flat on the board and prop up the wing tip to 2"(2 + 3 =5)

I am confused

No you're not confused, you're quite right (assuming the panels are equal length)

[PatMc]

According to the plan on this site they're not equal. The inner panel is 13.25" the outer 11.25". (It seems to suggest that the 2.5" centre section may be flat but Guiseppe's drawing shows it not to be, so I've included it with the inner panels.)
I didn't bother to try and follow the calculations shown but did my own to work out the required measurement.

Inner DH angle = 6.5º
Projected outer DH angle = 18º
Angular difference = 11.5º
Sin 11.5 = 0.199 (Say 0.2)

Therefore with inner pannel flat outer panel needs to be raised 11.25" x 0.2 = 2.25" but 2" wouldn't make any significant difference.

[Engine Doctor]

I will eat my words I tried it with a mock spar with inner spar at 13 "long and outer at 11" and it works out at approx 2.25 " under the tip with the inner panel flat.

[Giuseppe Saroli]

Good morning all and thanks for all you feed back

The discovery of the original Cleveland Viking polyhedral wind was when I come across Tandy model site he has build the half size of it that's different on mine because I am doing the full size plus extra rib on inner wing that is 18" inner 16" outer I am not looking for perfection just a easy build

Anyway my wings are going to be in 2 pieces so last night I joined the inner panel one flat on board and prop up the other at 3" which we new, when dry I prop up 1.1/2" under each inner wings tips and temporally attached the outer wings panels to make 5" as per original picture that I posted, and made a note on a reference stick , then I put the right inner panel flat on board with right wing outer panel still attached with pegs and the measurement show me to be approx. 2.1/2"

Tonight I will double check before I do the ply brace and report back with more accurate measurement but it seem very close to 2.1/2" how it came up to this is too much for me to comprehend , 2.1/2" +1.1/2"=4"not 5" maybe is to do with the wing span but on the building board is showing me 5" when assembled as per plan, locks like ED is almost spot but he dint know my wing span and the measurement of inner and outer panels

Thanks one again Regards Giuseppe

[Chuck Plains]

Posted by Giuseppe Saroli on 22/11/2014 09:01:44:

Good morning all and thanks for all you feed back

I am not looking for perfection just a easy build

Thanks one again Regards Giuseppe

Giuseppi my friend, if the wings are nearly the same on both sides, she will fly!

I have to confess that, although I do precision engineering at work, I prefer to make things quick and easy at home.

All my built up planes so far have used flat foam and tape covering. I have some ideas on how to improve the folded foam genre and will post anything that works on the forums.