Real Time Pyramid Mystery

The faces of the 4 triangles must be 1000 mm base x (square root of 500² + 1000²+

= 1118.0339 mm perpendicular height.

The angles can be calculated if one has a working calculator .

All the sides must be beveled as well.

please refer to forthcoming posts.

Huh?

If 1M (1000mm) along each base edge and 1M (1000mm in height) wouldn't your "1118.0339mm" be for the length of each triangular face from base to tip.

I guess your calc is if you lay a face piece on the deck, the "perpendicular height" is from the base edge to the top point?

Sorry, I guess the wording of "perpendicular height" is confusing me a bit... :P

Thanks Bill. This has already helped a lot. If I want a slope of 51 deg. the angle is 38.2 deg. on my saw. Roughly. It shall do. With CR4= Childs play.

There is a small matter of the 4 sloped sides NOT being flat or planar. They are a bit "caved in" from the center of each base line to the apex.

This was discovered by an WWII aviator flying over the pyramids as the sun was illuminating one portion and the mating portion was in a shadow! Fortunately he or a fellow crewman was able to document this with a nice photo.

I will provide a citation for this curious quirk in the design on the morrow. SS

Hi Stan

Looking forward to the photo. I have never heard of this. I always thought they were planar. Hope all is well and the cold is not giving you a too hard time. Greetings. Ky.

IIRC the book showing the dihedral angle in the faces of the Great Pyramid is Peter Tompkins' "Secrets of the Great Pyramid."

Unfortunately it is missing or lost. SS

G'day my friend

I'll send you some other photos soon. All is good. I had to get to these dihedral angles sooner or later. Not for lift off though. Greetings. Ky.

The following calculations can easily be done by calculator or by Excel software:

a. What would be the angle if I would want to make it 1 m high?

The angle of slope Θ shall be calculated as:

Tan Θ = 1 m / 50 cm = (1x 100 cm) / 50 cm = 2

Θ = Tan^-1 2 = 63.43495^o

b.What would be the angle if I would want to make it 50 m high?

The angle of slope ø shall be calculated as:

Tan ø = 50 m / 50 cm = (50 x 100 cm) / 50 cm = 100

ø = Tan^-1 100 = 89.42706^o

Abdel

Thank you for your reply but I think you are talking about the angle of the slope and not the angle of the 8 beveled sides of the pyramid. There are really 9 if you include the one the pyramid is standing on. If I wanted a bottom plate another bevel would be needed. It was not easy to fraze the question so I am possibly to blame for not explaining what I am after. Thousands of years later with all modern tools they still give us a hard time or at least me. Thanks again. Ky.

Dear Ky,

By mathematics, it is not difficult to find any angle you need. Please point to the angle you need to find.

Dear Abdel

I'll refer you to answer #13. That link provides all the answers I need. Thank you for helping. Ky.

Hi Ky,

There is 'more than one way to skin a cat' as they say. If you're not confident with the math ( though it is relatively easy, and some good explanations have been given. If it's any consolation I have yawning gaps in my knowledge, English language in particular.) You could go for a practical approach. This all depends on it's end use.

Get a big block of clay or similar. Carve it into the desired shape. Pour plaster of Paris or similar to create a negative, and then fill the sides of the negative in with fibreglass. That's a bit simplified, and I can see you have a fairly big block, but hopefully you get the gist of it.

Working out the bevel angle for the edges was neat ( I forgot who posted that)

Hi Kris

I think you meant Bill. Great help. The real difficulty is to end up with a homogeneous structure. Same thickness and no gaps and all from the same material. Yes , sometimes I hate being a bit dyslectic when it comes to maths and its written language but usually I can compensate with a practical solution, like the one you suggested. Forming matter to any desired shape is more of a sculpture approach but who is counting? Thanks Mate. Ky.

This guide should help you scale up to whatever size you want:

Hi TVP45

Thanks a lot. This reminds me of the stuff that was drawn onto the beer coasters back then. The isometric projection shows the angle I was after. To get answers on CR4 is child's play. To be able to comprehend them takes page 7 in my book and some one must have ripped that one out. Very happy with the answer.

Depending on were this "crazy guy" lives I would love to work with him for a while so to get a more hands on experience regarding the matter. If it is not a secret, what was the material he is using? As you might have guessed it is not one but many pyramids I am after and to have them made by a "crazy guy" in his work shop could hurt my hip pocket too much. Thanks again. Ky.

Ky,

When I said this guy was crazy, I meant worst than your run of the mill engineer. He was trying to find the right pyramid shape to sharpen razor blades! At least that's how I understood it - his accent was heavy. He ended up having his pyramid built in Ukraine and I never saw it so can't say if it worked.

Yep, sharpening razor blades was one of the things in the original April fool article: see post #12

I had given you a good answer (because it was) and it came up as "nearly good". Sorry out of my responsibility. Ky.

Two separate people need to rate an answer as good for it to be given the overall good rating. I've topped him up.

Good stuff Mate. Good to keep each other happy. TV45 (thats all I remember of his name) knowledge of matters at the right time was worth a good answer point. But then, there are so many engaged participants in this forum. I'm new to computers and the 'hangaoners' routines or ways to do things with this fantastic medium are new to me. We get better every day.

I fell for the same razor blade sharpening thing back then. Tried lots of things. Glass, aluminium, cardboard, you name it. Was not only fun but good entry into the amazing facts of pyramids.

The only thing that worked was to mummify meat. We had 2 identical (if that is possible) pyramids standing next to each other under completely the same environment. (1 third from the top) . The meat in both dried out and the control sample under a glass kitchen bowl was full of maggots. This was unexplained and did get me thinking. If I would have only stuck to the given parameters would other experiments have succeeded. I doubt it. But this one did.

Now, 3 decades later, and concerning a completely different use, it is again important to do the right thing by science and be as precise as possible during proto type testing. The precision is more of a cosmetic factor now a days. Things that look imperfect can give a wrong impression although they work as well as the "real thing". Some one "in the know" could feel offended by this improvised attempt of joining 4 sides with each other. Money is mostly spent on polished jewels, if you know what I mean.

If I only had more time. Ky.

Base Angle Θ (Miter Gauge)

For pyramid with 1.0m x 1.0m base and height of 1.0m, consider one sloping edge of the pyramid. Call it vector A. In vector notation:

A = 0.5i + 0.5j + 1.0k

B = 1.0i is a vector representing the side of the base on the 'X' axis.

|A| = (0.5² + 0.5² + 1)^1/2 = 1.22475 where |A| is the length of vector A

|B| = 1.0

The dot or scalar product of A and B is 0.5*1.0 + 0.5*0 + 1.0*0 = 0.5

But A.B = |A|*|B|*cosΘ

So cosΘ = A.B/(|A|*|B|), i.e. cosΘ = 0.5/(1.22475) = 0.408248 and Θ = 65.905^o.

Angle Between Planes Φ

Vectors A and B describe a plane. The cross or vector product of two vectors produces a new vector which is normal to the plane, say N. In vector notation,

N = BxA = -j + 0.5k which is a vector normal to the plane containing A and B.

Similarly, for an adjacent side, with normal vector N'

N' = AxC where C = j (unit vector parallel to the y axis)

N' = -i + 0.5k

The angle between two adjacent planes is equal to the angle between the two normal vectors, N and N'. The dot product N.N' = 0.5*0.5 = 0.25

Also, |N| = |N'| = (1 + .5²)^1/2 = 1.118034

cosΦ = N.N'/(|N|*|N'|) = 0.25/(1.118034)² = 0.200 and Φ = 78.463^o

The saw blade tilt is Φ/2 = 39.2315^o

The above method may be used for irregular geometry just as easily. It is simple to apply and does not require memorization of formulas or the use of tables. For that reason, I much prefer it to other methods.

Thank you ba/ael. Very helpful. All will be considered. Ky.