Help with Trig (Updated drawing)

[rocketjet787]

I am pretty sure you need to know more about the angles, just fixing side lengths does not give you enough information. I have attached 3 examples of this shape all with the same side lengths, just pulled to different angles.

 

[GrouchoDuke]

I am pretty sure you need to know more about the angles, just fixing side lengths does not give you enough information. I have attached 3 examples of this shape all with the same side lengths, just pulled to different angles.

The 11.5 & 3 dimensions on the 14.5 length side are the key to fixing those angles.

EDIT: "finding those angles," not "fixing those angles."

 

[Jay Rairigh]

(...adjacent angles to a and b...as it were Groucho.)

Well you can't solve an unknown WITH an unknown. Even pythagarus needs 3 known entities to solve a triangle. You can write the equations with abc, XYZ, but you need known data (3 pieces in this case) to actually solve it.

Somebody said at the start of this thread, that if we assume those 2 things above it is easy, and that was correct.

I would send this drawing right back to engineering for repair/revision! And tell them how to fix it to boot.

 

[Jay Rairigh]

Exactly what rocketjet shows! The drawing needs clarified.

 

[rocketjet787]

The 11.5 & 3 dimensions on the 14.5 length side are the key to fixing those angles.

You are correct, good catch. If following with the 3" to 11.5" matching with the end of the 45" you get an x of ~34.4

 

[GrouchoDuke]

(...adjacent angles to a and b...as it were Groucho.)

Well you can't solve an unknown WITH an unknown.

Yeah, you can. In this case, you can solve for all 4 of the unknowns that Kelly pointed out.

 

[GrouchoDuke]

You are correct, good catch. If following with the 3" to 11.5" matching with the end of the 45" you get an x of ~34.4

Make that 3+11.5 line straight & you'll get 33.2. (@Banzai88 said it was straight)

 

[rocketjet787]

Make that 3+11.5 line straight & you'll get 33.2. (@Banzai88 said it was straight)

WIP

 

[GrouchoDuke]

...breaks...

Fix the tip of the "fin," make the joint between the 7.5 & (3+11.5) lines flexible, then move the "fin root" up & down the mid-line until the 3 cuts the wingspan line.

That is the best I can straighten it, ~31.4".

Yep, that's pretty close.

 

[Jay Rairigh]

The 11.5 & 3 dimensions on the 14.5 length side are the key to fixing those angles.
[/QUOTE]

rocketjet has those dimensions on ALL of his examples. Apparently they do not 'fix' any angles. They can't. It's simple math. You need 3 knowns, we have 2.
3-2 = 1. We are missing exactly 1 piece of information.

 

[GrouchoDuke]

We are missing exactly 1 piece of information.

If that line is straight like @Banzai88 said, I don't think there's anything missing.

 

[Jay Rairigh]

Fellers...you now have a right angle between sides 45 and 7.5, is that not right rocketjet?

 

[Jay Rairigh]

Actually ALL of rocketjets sample are correct!

Which configuration are we going to make from the drawing though?

 

[GrouchoDuke]

Actually ALL of rocketjets sample are correct!

Which configuration are we going to make from the drawing though?

If you're looking at his post 31, I think you're missing that the 14.5 length straight line has 3 units on one side of the 34 unit wingspan line and 11.5 on the other. If you apply that constraint, then only one X can be correct.

 

[Jay Rairigh]

Groucho...yes that line is a straight line. The triangles between sides 45/45 is known. That angle between 45/45 is known. That's not the problem.

Rocketjets examples ALL had this same big triangle! The little ones are the unknowns.

 

[GrouchoDuke]

Groucho...yes that line is a straight line. The triangles between sides 45/45 is known. That angle between 45/45 is known. That's not the problem.

Rocketjets examples ALL had this same big triangle! The little ones are the unknowns.

Yep, the big 45/45/34 triangle is known, so the wingtip points are directly calculable. The little ones are where the fun is.

 

[Jay Rairigh]

Yes they are calculable....with 3 pieces of info....we have 2 sides....not 3....and NO angle as rocketjet rightly showed. Just because I know one triangle doesn't mean I can solve the other.

The 14.5 can be broken,bent, straight, or 11.5+3 as one straight line....none of that matters. We have no angle.

Rocketjet had all the 'constraints' in ALL his examples in post 31. Didn't do no good!

 

[Jay Rairigh]

It is not fun when there's a missing dimension groucho. I deal with this stuff at work daily.

We can only assume right angles where I proposed, and they probably are, but we really wouldn't do that in real life, we'd send the drawing back.

 

[GrouchoDuke]

Rocketjet had all the 'constraints' in ALL his examples in post 31. Didn't do no good!

Almost...His post 31 doesn't have all the constraints. It's missing the 3 on one side & 11.5 on the other of the wingspan line constraint. Post 35 has that ratio correct, but the 14.5 unit line isn't straight. His post 38 was really close when he had the image on that post (currently says "WIP").

 

[GrouchoDuke]

It is not fun when there's a missing dimension groucho. I deal with this stuff at work daily.

We can only assume right angles where I proposed, and they probably are, but we really wouldn't do that in real life, we'd send the drawing back.

Oh, I totally agree that we'd all send the drawing back!

 

[Jay Rairigh]

Understood with your 3 and 11.5 intersecting at the 34 line.

I'll sleep on that one and figure out why I can't be wrong tomorrow...lol. Thanks.

 

[GrouchoDuke]

I'll sleep on that one and figure out why I can't be wrong tomorrow...lol. Thanks.

I redrew the problem statement. It's not perfectly to scale, but it's hopefully close enough to make this more fun.

 

[user 30299]

I redrew the problem statement. It's not perfectly to scale, but it's hopefully close enough to make this more fun.
View attachment 432960

Does the solution rely on Chords and Bisecting circles? This should allow you to side step the need for an angle, which Jay Rairigh correctly points out. The 7.5, 3, 11.5 and 34 can be considered radiuses, diameters or chords of circles - so can 14.5, the sum of 3 & 11.5. I haven't worked with chords and bisecting circles in awhile.

 

[Banzai88]

I redrew the problem statement. It's not perfectly to scale, but it's hopefully close enough to make this more fun.
View attachment 432960

THIS. Upon consult with my son, this is it.

* Extending x would bisect the 34 side and form a right angle.
* The triangle 7.5, 3 MUST be above the 34 side.
* The line 3, 11.5 is a straight line.
Any other angles are effectively unknown (I guess we would have to do it in CAD?).

Neither of us realized that this would be quite as involved! Thanks for working this! Sorry if the drawing was unclear, I really think that he's genuinely trying to figure out something for one of his 100000000 thought projects (that often turn out some sort of cool thing from his workbench).

I think that I know what he's working on, but not entirely sure yet........

 

[bobby_hamill]

I used solidworks 2011 and drew a horizontal line 34 "
then I went to tools then sketch tools and finally sketch picture
after you insert the sketch picture you take the known 34 " horizontal line
and line up the 2 ends of the 34"line in the photo with the drawn 34 ' line

Then I drew a vertical line from the top to the bottom of the line X
and measured the resulting line and came up with 35.71 " it is not exact but its close

 

[Fattbank64]

The kid's diagram has all the info to solve the problem.

1.

* Extending x would bisect the 34 side and form a right angle.

2. Definition of altitude (triangle)
https://en.wikipedia.org/wiki/Altitude_(triangle)

It's easier to 'see' the wings/fins after rotating the diagram 90 degrees clockwise.

It is similar to F-117 Nighthawk wings or rocket fins (map view).

 

[mbeels]

The 11.5 & 3 dimensions on the 14.5 length side are the key to fixing those angles.

Yeah, this is an essential piece of information. None of the examples in post #31 satisfy that constraint.

 

[user 30299]

THIS. Upon consult with my son, this is it.

* Extending x would bisect the 34 side and form a right angle.
* The triangle 7.5, 3 MUST be above the 34 side.
* The line 3, 11.5 is a straight line.
Any other angles are effectively unknown (I guess we would have to do it in CAD?).

Neither of us realized that this would be quite as involved! Thanks for working this! Sorry if the drawing was unclear, I really think that he's genuinely trying to figure out something for one of his 100000000 thought projects (that often turn out some sort of cool thing from his workbench).

I think that I know what he's working on, but not entirely sure yet........

Can you post a screen shot or a picture of the actual math problem?

 

[Jay Rairigh]

Fattbank...33.2125 is the correct answer. Tell me, did you get this answer using the altitude?

I got the answer using cad. Our answer is correct (if calculated down to 4 decimal places), but I had to move circles and lines around until I came up with the 3 dimension.

 

[Alan15578]

Yep. It's one of the unknowns. Like Kelly said, it's asin(d/11.5). (It's the same angle on both sides.)

It makes a good lookin' rocket.
View attachment 432955

But now the questions becomes where is the fin CP, and what are the equivalent values of of the tip chord, semi span, etc. of the classic Barrowman Equations?