The formula for the chamfer angle (actually this is for a more general situation that includes the case in the video) is: arcsin[cos(180°/n)/cos(α/2)] (where arcsin means the same as sin⁻¹). This is the chamfer angle for the inclined sides of the triangular faces of a pyramid with a regular n-sided polygon base and n triangular sides, each of which is an isosceles triangle with apex angle α. Note that the chamfer angle is simply half the angle between the adjacent faces. In the case in this video, we have a pyramid with a square base and equilateral triangular sides, so we apply the formula with n=4 and α=60°, and we get the chamfer angle arcsin[cos(180°/4)/cos(60°/2)] =arcsin[cos(45°)/cos(30°)] =arcsin[(√2/2)/(√3/2)] =arcsin(√2/√3) =arcsin(√(2/3)) =arcsin(0.816496...) =54.7356...° The formula actually works in more general situations than these pyramids. For example, it can be used to calculate the angle between adjacent sides of any regular polyhedron. This is because to use the formula we don't need the joining faces to be triangles; it applies to n corners of angle α meeting "regularly" at a point, as we can cut off a pyramid with a regular polygonal base and isosceles triangular faces close to the point. As the angle between adjacent faces is twice the chamfer angle we have (in degrees) θ=2arcsin[cos(180°/n)/cos(α/2)] where θ is this angle. For example, for a cube, we have three faces of angle 90° meeting at each vertex, so n=3 and α=90°, and θ=2arcsin[cos(180°/3)/cos(90°/2)] =2arcsin[cos(60°)/cos(45°)] =2arcsin[(1/2)/(1/√2)] =2arcsin(√2/2) =2×45° =90°, as expected. For those interested in the mathematics behind the formula, here is an explanation of how I derived it. Unfortunately, without diagrams, it is not easy to follow. I advise readers to draw the diagrams as you go along. Let ΔAOB and ΔBOC be two adjacent isosceles triangular faces of the pyramid with apex O, where OA=OB=OC=1 (the equal sides of the isosceles triangles),
I was slaving trying to make this pyramid for the inside of a bird feeder to direct the feed towards the four sides. This was perfect and very elegant. Thanks
Great video! No distracting music, no self-important titles/logos - just the demo. However, it would be helpful if you explained how to calculate the cutting angle! Plus show how to cut the equilateral triangles in the first place.
I came up with this formula arcsin[cos(180°/n)/cos(α/2)] which leads to arcsin(√(2/3))= 54.7356...° in the particular case in this video. For an explanation, please see my comment.
Very nice execution and idea! Thank you for the excellent "how to."!!! 3 days ago, I bought the angle guage you used, specifically to build a pyramid using the table saw.
I came up with this formula arcsin[cos(180°/n)/cos(α/2)] which leads to arcsin(√(2/3))= 54.7356...° in the particular case in this video. For an explanation, please see my comment.
I came up with this formula arcsin[cos(180°/n)/cos(α/2)] which leads to arcsin(√(2/3))= 54.7356...° in the particular case in this video. For an explanation, please see my comment.
An equilateral triangle has 3 equal sides in length, and 3 equal angles of 60 degrees, so I would think that the angle would have to be 60 degrees on the miter cuts as well...
hi can i ask you a question -- will the angle changes according to high and width or remain the same --what will be the angle if there were 6 sides thanks ,,love your video
Yes the angle changes - there was a good online calculator to work out the chamfer but that has long since been taken down. I should revisit this project - making geometric 3D shapes since I've made my CNC machine.
I came up with this formula arcsin[cos(180°/n)/cos(α/2)] which leads to arcsin(√(2/3))= 54.7356...° in the particular case in this video. For an explanation, please see my comment.
I came up with this formula arcsin[cos(180°/n)/cos(α/2)] which leads to arcsin(√(2/3))= 54.7356...° in the particular case in this video. For an explanation, please see my comment.
@@vigilante004 actually pyramids are being used as healing apparatus because they have the golden ratio,they balance energy and deflect EMF,they are very useful.
I'm not an expert at this, but I assume that you would only cut 45 degree bevels if you were making a cube. As for 54.5 degree, I guess the extra 9.5 must account for the 'pyramid' slope.
I came up with this formula arcsin[cos(180°/n)/cos(α/2)] which leads to arcsin(√(2/3))= 54.7356...° in the particular case in this video. If you set n=3 and α=90°, you get 45°, as would be expected for a cube. For an explanation, please see my comment.
Great jig! Watch your videos with fascination at every project. GemRed Digital Bevel Box and a Feeler guage! Two investments I have made just by watching you, good work.
The formula for the chamfer angle (actually this is for a more general situation that includes the case in the video) is:
arcsin[cos(180°/n)/cos(α/2)]
(where arcsin means the same as sin⁻¹).
This is the chamfer angle for the inclined sides of the triangular faces of a pyramid with a regular n-sided polygon base and n triangular sides, each of which is an isosceles triangle with apex angle α.
Note that the chamfer angle is simply half the angle between the adjacent faces.
In the case in this video, we have a pyramid with a square base and equilateral triangular sides, so we apply the formula with n=4 and α=60°, and we get the chamfer angle
arcsin[cos(180°/4)/cos(60°/2)]
=arcsin[cos(45°)/cos(30°)]
=arcsin[(√2/2)/(√3/2)]
=arcsin(√2/√3)
=arcsin(√(2/3))
=arcsin(0.816496...)
=54.7356...°
The formula actually works in more general situations than these pyramids. For example, it can be used to calculate the angle between adjacent sides of any regular polyhedron.
This is because to use the formula we don't need the joining faces to be triangles; it applies to n corners of angle α meeting "regularly" at a point, as we can cut off a pyramid with a regular polygonal base and isosceles triangular faces close to the point.
As the angle between adjacent faces is twice the chamfer angle we have (in degrees)
θ=2arcsin[cos(180°/n)/cos(α/2)]
where θ is this angle.
For example, for a cube, we have three faces of angle 90° meeting at each vertex, so n=3 and α=90°, and
θ=2arcsin[cos(180°/3)/cos(90°/2)]
=2arcsin[cos(60°)/cos(45°)]
=2arcsin[(1/2)/(1/√2)]
=2arcsin(√2/2)
=2×45°
=90°, as expected.
For those interested in the mathematics behind the formula, here is an explanation of how I derived it. Unfortunately, without diagrams, it is not easy to follow. I advise readers to draw the diagrams as you go along.
Let ΔAOB and ΔBOC be two adjacent isosceles triangular faces of the pyramid with apex O, where OA=OB=OC=1 (the equal sides of the isosceles triangles),
even without your words, I have learned quite a lot so far. Keep it up bud.
I was slaving trying to make this pyramid for the inside of a bird feeder to direct the feed towards the four sides. This was perfect and very elegant. Thanks
I was in search of the angle to cut and your video has shown me that. Thanks 54.5 :) times
Very good techniques demonstrated. Many thanks for sharing.
Great video! No distracting music, no self-important titles/logos - just the demo. However, it would be helpful if you explained how to calculate the cutting angle! Plus show how to cut the equilateral triangles in the first place.
fgriffintx Smith yes i would agree
I came up with this formula
arcsin[cos(180°/n)/cos(α/2)]
which leads to
arcsin(√(2/3))= 54.7356...°
in the particular case in this video.
For an explanation, please see my comment.
So not a great video then
Do you have plans for that jig available anywhere?
Brilliant, most Brilliant thing i have seen all year.
Very nice execution and idea! Thank you for the excellent "how to."!!! 3 days ago, I bought the angle guage you used, specifically to build a pyramid using the table saw.
Hello there! I have to do a similar project-can you tell me how to come up with the bevel angle? Won’t it change depending on the size of the pyramid?
Cool!! Savvas, did you use the great Keop's proportions to build this one? Thx a lot for your time.
Cool jig. You could add different types of angles to make different types of triangles. Scroll saw to make a cool lamp too. Yet another subscription!
I came up with this formula
arcsin[cos(180°/n)/cos(α/2)]
which leads to
arcsin(√(2/3))= 54.7356...°
in the particular case in this video.
For an explanation, please see my comment.
Cool jig! Subscribed.
Thank you for the demo. What is that digital gizmo that you used to show the angle of the saw blade on your table saw?
GREAT VIDEO
Do you sell the precut wood? And in different sizes?plmk thx Glenn
Excellent.
So nice.
so what are the demission and how tall was that?
Thanks for the great video! How did you come up with the angle for the bevels? I have a similar project I have to do.
I came up with this formula
arcsin[cos(180°/n)/cos(α/2)]
which leads to
arcsin(√(2/3))= 54.7356...°
in the particular case in this video.
For an explanation, please see my comment.
You didnt quite show what the angle was that you set your saw to.....it seemed like you were still adjusting as the camera changed.
Nice. Very nice!
BRAVOOOooooo,congratulations!!!!!👏👏🎯💯✌️🎩👍👍📸🙏⭐🥇🥇💎🥇🎩👍
Thanks so much !!!!!!
So is the angle 54.5?
An equilateral triangle has 3 equal sides in length, and 3 equal angles of 60 degrees, so I would think that the angle would have to be 60 degrees on the miter cuts as well...
thank you man
Great
Φανταστικο.Μπραβο
very good
Thanks
Impressive
hi can i ask you a question -- will the angle changes according to high and width or remain the same --what will be the angle if there were 6 sides thanks ,,love your video
Yes the angle changes - there was a good online calculator to work out the chamfer but that has long since been taken down. I should revisit this project - making geometric 3D shapes since I've made my CNC machine.
I came up with this formula
arcsin[cos(180°/n)/cos(α/2)]
which leads to
arcsin(√(2/3))= 54.7356...°
in the particular case in this video.
For an explanation, please see my comment.
How did you calculate what angle you needed for the sides?
Did you ever figure that out?
I came up with this formula
arcsin[cos(180°/n)/cos(α/2)]
which leads to
arcsin(√(2/3))= 54.7356...°
in the particular case in this video.
For an explanation, please see my comment.
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pretty cool !
was there a reason for the pyramid?
Yes, a girl!
it gave the 13 million slaves something to occupy their time with instead of ransacking the Pharoe's riches.
@@vigilante004 actually pyramids are being used as healing apparatus because they have the golden ratio,they balance energy and deflect EMF,they are very useful.
@@AndreyLit22 Oh yeah, that too. But don't forget to wrap yours with tin foil and wear it on your head when you go outside.
hello, thanks, would not ?? some copper sellotape improve the pyramid "work"...............
hello . can you explain me how ... thanks
Skill.
HOW DID YOU CUT THE EQUILATERAL TRIANGLES?????!!!!!!
Good question...
Savvas Papasavva thanks.
can this be done with 45 angles instead?
I'm not an expert at this, but I assume that you would only cut 45 degree bevels if you were making a cube.
As for 54.5 degree, I guess the extra 9.5 must account for the 'pyramid' slope.
I came up with this formula
arcsin[cos(180°/n)/cos(α/2)]
which leads to
arcsin(√(2/3))= 54.7356...°
in the particular case in this video.
If you set n=3 and α=90°, you get 45°, as would be expected for a cube.
For an explanation, please see my comment.
Great jig! Watch your videos with fascination at every project. GemRed Digital Bevel Box and a Feeler guage! Two investments I have made just by watching you, good work.
Very
The Egyptians used a giant saw.
Very dangerous and illegal in the EU. NEVER use a saw without a guard.