Geometry

[jmagg]

So I need to reproduce this star. On this first prototype, I copied an existing star angles for the cuts. I'd like to use geometry to fine tune the cut angles for ease of glue up but the alztimer s is setting in. Formula or mathematical method pls ty.

 

[Greenman]

I got nothin. But if you crack the code on this please share it here.

 

[C1]

http://starpolygons.com/Generator.aspx

Use the drawing mode to get the geometry for the raised portions.

Good Luck

http://math.wikia.com/wiki/Star_polygon

 

[jmagg]

Thanks for the star generator but not seeing a solution there. I consider the outside angles arbitrary. There must be a relationship to the inner angles. The cut angles for the prototype are 14 outside, 41 inside and 7" x 1" legs. The inside angles are a bit off.

 

[Greenman]

Thanks for the star generator but not seeing a solution there. I consider the outside angles arbitrary. There must be a relationship to the inner angles. The cut angles for the prototype are 14 outside, 41 inside and 7" x 1" legs. The inside angles are a bit off.

I don't know the geometry, but if you're just looking for the angles a halfassed 2d cad program will give you that info.

 

[jmagg]

I'd say a single degree will fix the inside angles at the size mentioned, but I'd like to do variations on this theme. Guess I'm installing cad or someshit. Thx

 

[herm0016]

Pythagoras can help you. it will take a few minutes. draw one section, label the side you want to control your overall size as X and break it down into managable triangles. once you have the 2 angles you need for cutting them, you can draw the whole thing, label the center to point distance as a radius and find the overall size in realation to the X found in the previous formula. it's just algebra.

Cad will work, but only if you understand how to build the model so that you can alter one dimension to change the whole size. i would start with a circle and base all my lines off of that so that they scale properly.

 

[jmagg]

Pythagoras can help you. it will take a few minutes. draw one section, label the side you want to control your overall size as X and break it down into managable triangles. once you have the 2 angles you need for cutting them, you can draw the whole thing, label the center to point distance as a radius and find the overall size in realation to the X found in the previous formula. it's just algebra.

Cad will work, but only if you understand how to build the model so that you can alter one dimension to change the whole size. i would start with a circle and base all my lines off of that so that they scale properly.

Not sure I follow. Lets say I start with a block 8 inches long and 3 inches wide, and decide to cut this block at 14 degrees for half the outside angles. What would be the algebraic formula to find the inside angles? (Of which half would be the inside cut angles).

 

[herm0016]

R is your radius, A is the side of a pentagon included in circle of R radius. you now know the length of 3 sides of the large triangle and can use sines and cosines to figure out the angles. you will have to choose how far you want A to be from the inside point and then you can find all the rest of the angles and lengths of the triangles. write formulas based on R as the variable to make different sizes.

 

[jmagg]

TY for your work, will study
The triangles you kindly laid out are given on a finished star. I'm starting with a rectangle with an arbitrary angle cut into it. This will give the length and outside angles of two legs, but not the relationship to the next two legs intersection or the distance of A. At this point I'm looking for a common relationship between the totals of the inside and outside angles. Without mechanical drawing.

 

[jmagg]

The triangles you kindly laid out are given on a finished star. I'm starting with a rectangle with an arbitrary angle cut into it. This will give the length and outside angles of two legs, but not the relationship to the next two legs intersection or the distance of A. At this point I'm looking for a common relationship between the totals of the inside and outside angles. Without mechanical drawing.

 

[jmagg]

Outside angles = x
Inside angles = y

If y measures 100 degrees x 5 total inside angles total = 500
If x measures 28 degrees total (14 each side) x 5 outside angles total = 140.
Then 140 (given) + 360 = 500/5= 100

360 - 100 = 260/2 = 130 which shows 40 (90+40), which is the cut inside angle.

Seems right but need to test

 

[radhak]

I am interested in this: both as a woodworker and a math guy.

Trying to understand your nomenclature when you say 'outside' and 'inside' angle, what are you referring to? I'll take a guess here, but feel free to correct me.

Classical geometry tells us that in a regular pentagram, the angles in the insides of the tips is 36 degrees, the two base angles are 72 degrees each, and each of angles in the pentagon (inside figure) is 108 degrees.

Lemme use a quickly drawn pentagram to illustrate:




So your 'outside' angle X must actually be 36 degrees, which means 18 degrees on each piece.

Also, to calculate your 'inside' angle Y, let's first add up the other angles : 72 + 108 + 72 = 252
So the angle formed at the joint = 360 - 252 = 108 degrees total

Which means each wood piece must be cut to an angle of 54 degree

Here:

 

[herm0016]

I am interested in this: both as a woodworker and a math guy.

Trying to understand your nomenclature when you say 'outside' and 'inside' angle, what are you referring to? I'll take a guess here, but feel free to correct me.

Classical geometry tells us that in a regular pentagram, the angles in the insides of the tips is 36 degrees, the two base angles are 72 degrees each, and each of angles in the pentagon (inside figure) is 108 degrees.

Lemme use a quickly drawn pentagram to illustrate:




So your 'outside' angle X must actually be 36 degrees, which means 18 degrees on each piece.

Also, to calculate your 'inside' angle Y, let's first add up the other angles : 72 + 108 + 72 = 252
So the angle formed at the joint = 360 - 252 = 108 degrees total

Which means each wood piece must be cut to an angle of 54 degree

Here:

it looks to me like its not regular, could be made that way though.

b-c is not in line with the point below it, and thus it is not regular. the inset distance between a line from each outer point to the inner point is set by the maker, and i think it makes a more interesting shape also.

jmagg, from my instructions you can work your way back to the size of each piece of wood.

 

[skyking]

I see it as an exercise of dimensioning the A-C-B triangle and the Q-S-R triangle. Even though A-C and Q-R do not correspond to cuts or pieces of wood, they are useful in calculating the measurement from the A-C to B and all the angles.

 

[radhak]

Yes, the wood in the given sample is not dimensioned accurately, nor planed. The seen irregular joints are just imprecise cuts.

I think the OP is starting anew with a regular piece of wood, so he should ensure it is cut to the same width for the entire length, then cross cut 10 pieces to exactly the same length. Then, the miter (angled) cuts should follow (exactly 18° at one end, 54° at the other).

Personally, I'd do this all on my table-saw. But if you buy a well dimensioned (Douglas Fir or Yellow Pine is good, Red Oak even better though expensive) wood from Home Depot / Lowes, the width would be tolerably same, and an accurate miter-saw might be all that's needed.

 

[skyking]

I have a very nice push miter guide for my table saw, and still prefer the laser on the cheap chop saw for this.
I also have a ton of scraps around, and would rip up some plywood for the prototype cuts.

 

[jmagg]

The cut angles on each outside leg are 14 each for a total 28.and the inside cut angles measure 41 my miter saw which is 130 actually ( 41 past 90) or 100 measured on the inside, on this (my) star project. This is rough cut pine ripped and mitered. 40 (260 remains) degrees seems like it would close the inside miters correctly.

 

[radhak]

If you do want that angle at 28°, the angles at the joint would become 76/104/76 (instead of 72/108/72), so each angle cut would be 104°/2 = 52°.

In your point of reference, this translates to 180 - 52 = 128°, or 90 + 38°
If you do 40°, you might need to use putty or something to close the 2° gap.

I wouldn't leave that gap, but that doesn't mean it has not been done.

 

[jmagg]

52degree inside cut angle is way too much, I'm confident at 40 since 41 was all but perfect. I think the difference is that it's not a regular polygon. If you look closely, the legs lift slightly, which I think is good for the asthetics.

 

[Red Squirrel]

An easier way might be to base it on distances instead. Like look at each piece individually and look at the distances between each point, then when you make your lines and cuts just match them all. Downside of working with small pieces like this is how close your hand has to be to the saw blade though... Might be one of those things that's better to do with a hand saw. If you need to make a lot of them you can make some jigs that will help you get consistent cuts.

 

[radhak]

52degree inside cut angle is way too much, I'm confident at 40 since 41 was all but perfect. I think the difference is that it's not a regular polygon. If you look closely, the legs lift slightly, which I think is good for the asthetics.

As I explained in my previous post, that 52° is when I look at it reverse - from the orientation you are measuring, it works out to 38° (90 - 52), compared to the 40° you are planning on

 

[jmagg]

Angle b in your diagram is 100-260

 

[paperfist]

If you want that in plastic I’ll whip one out on my 3D printer for you.