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TLDR; News about OS++. Updated weekly, so you know the hamster wheel is still turning.
I found that as my code became more complex the length of time between each release of the code became longer. Soon I was missing monthly deadlines and the project appeared stalled to the casual observer. The idea of this page is to update the news section each week and so, show that the project is live. The news itself is probably drivel. Effectively me talking to myself, about how an idea should develop. With this project, there have been many dead ends that I needed to go down, before I found “The true way” and release it. Please forgive my stupidity.
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News to go here on Friday 2023_05_19.
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Well it’s another week. I have completed my 4D graphic. I call it the “Circle and Square diagram”. I did want to call it the “Sircle and square diagaram” but the GF said no, it looks silly.
Really happy with it. I think it will be a big hit. Produced a youtube video to explain it. GF said she was very impressed. Apparently I made it clear and understandable. Well, we are about to see what other people think of my 7 minute wonder video. Because I just emailed my family and told them to go watch. The link is below.
Circle and Square diagram youtube video
I used Qcad (which I previously bought) / librecad (because I can’t find the qcad executable file and it is very similar to qcad) to create the outlines. I used inkscape to colour it in. I used Libreoffice to create a 600dpi pdf file from the monster it had become (6GB). I used gimp to produce the png file from the pdf file. I used Kazam to record the video of the png file into a Mpg video file and finally I used openshot to stitch it all together for final upload to youtube. Of course this was all done on Linux OS, itself an opensource product.
My point, There is no way I could have done this without a lot of opensource and free software. There isn’t a capitalistic product that has all that functionality. And even if there were, I would not be able to afford it. So while the capitalist economic model might be good as an incentive structure for the winners of society, I have come to the conclusion that we need a basic income to keep the future struggling opensource projects alive while they make that moonshot. Let “get a proper job” or more accurately “Work for the man, or starve”. be consigned to history. And be replaced with “Van Gogh basic income”to keep the body alive. Perhaps those dead NPCs (People stuck in the same daily routine, decade after decade) that we meet in life, might have something worthwhile to contribute. If given the chance to try without risking destruction and starvation for their families.
Then again, with money running low, I will probably need to get off my butt and do some real work. It was fun while it lasted.
Well signing off.
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Well it’s another week. The hamster wheel has stopped dead. Just 2 seconds ago. I have finished. I am pretty happy with the result. Behold the Sircle and Square Diagram. The final Vertex diagram. It only took me a whole month to get right.
The corner maths for the cube now work. The magic number was 6. 2^3
1 * 3/6 = 3/6 (Corner furthest away)
3 * 2/6 = 6/6 (Layer closest)
3 * 1/6 = 3/6 (2nd layer)
1 * 0/6 = 0/6 (corner which comes form another cube)
Total = 12/6 = 2The corner maths of the squares also works. The magic number was 4. 2^2
1 * 4/4 = 1 (Corner furthest away)
2 * 2/4 = 1 (2nd layer)
1 * 0/4 = 0 (corner pointing in)
Total = 8/4 = 2The corner maths of the line also works. The magic number was 2. 2^1
2 * 2/2 = 2
Total = 4/2 = 2Looking to produces a youtube video to explain
Well signing off.
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Well it’s another week. The hamster wheel is turning even faster.
GF asked me to redo the vertex diagram. Added green so you could see the progression better. Also I added a close up. I am happy with the 2D/3D vertex square in the middle. I think the 3D/4D vertex has problems. Happy that the corner facing away is fully owned by the cube (1) and it’s opposite is fully not owned (0). Not so Happy with the 3* 1/4 and 3* 1/12.
Obviously 1 + 3* 1/4 + 3* (1/12) = 2. But is it the right distribution? I will have to think this through and see if I can get the answer. Below is my updated picture.
Well signing off.
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Well it’s another week. The hamster wheel is turning even faster.
Worm_world is 1D, Flat_land is 2D, Sea_space is 3D and Hyper_space is 4D. The 4 dimensions are Length(Ln), Width(Wd), Height(Ht) and Extent(Ex) aka the EXtra dimension. In the 20230407_weekly_update I used a 3d cube to show a 4D object. It was a visualisation exercise but the maths strangely worked for lower dimension objects like corners. The question I had to ask myself was > “Am I suffering from self delusion by shoe_horning the visualisation into known maths or is my visualisation valid?” I thought the best idea was to unit_test the idea by running it’s method through the lower known dimensions. If there are any changes in the method, then incorporate this into our hyperspace visualisation.
This is what I came up with. Perhaps I can put this on a T-Shirt
Well signing off.
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Well it’s another week. The hamster wheel is turning faster but I do not seem to be getting anywhere. I did however get my next instructional video compiled for os++. All I need to do is to upload it. When I have access to cheap internet.
My work on Pythag2d is going round in circles. I think I have done it and then the whole thing just falls apart. I did, However, make some progress on visualising a cell-8 (hyper cube). It goes as follows.
Take a 3d cube in space. Now replace each corner with a small cube. Since a cell-8 has 8 cubes in it and a cube has 8 corners, which you just replaced with a cube, this makes mathematical sense. No idea what I am wittering on about? Maybe this picture will help. It’s 3 pictures in 1. Explanations for each picture is below.
Think of a 3d Cube with 8 corners. Replace each corner with a cube. You now have 8 Blue cubes on the end. Just like a Tesseract. (A Tesseract is made up of 16 corners, 32 edges, 24 faces and 8 cuboids). We now have 8 cuboids in our visualisation construct to build upon.
Clear your mind of picture 1. And rebuild using the following info.
In 3d space we have a cube. A cube has 6 faces. Each “side” is square. This square shares it’s edges with other faces. Are you with me so far? The 3d cube has 12 edges and 6 faces. Therefore a face “owns” 2 edges and 2 edges are contributed by other faces. ie 12 edges / 6 faces = 2 edges per face.
Now let’s take that sentence above and replace it with 4d words
In 4d space we have a Hyper cube. A Hyper cube has 8 cubes. Each “side” is cuboid. This cuboid shares it’s faces with other cuboids. Are you with me so far? The 4d cube has 24 faces and 8 cuboids. Therefore a cuboid “owns” 3 faces and 3 faces are contributed by other cuboids. ie 24 faces / 8 cuboids = 3 faces per cuboid.
Take the red cuboid. it is showing you the faces it owns. The Yellow cube is exactly the same. I have only put the faces in that it “owns”. But as it’s not face on, you can see the missing yellow sides that are contributed by other cuboids around it.
So a Tesseract is made up of 8 cubes. We have visualised that. A Tesseract has 24 faces. Each of which 3 are owned by a single cuboid. ( ie 8 * 3 =24). We have visualised that with the yellow cuboid. So what about the corners on our visualisation? Where are they? If you take the red cube. You can see a corner where the 3 red faces intersect. Since there are 8 cuboids in our Visual construct. We now have 8 corners. But where are the other 8?
The red corner has 3 lines coming out of it. We know that an edge is made up of two points, one at each end. So if we start with our red point (in the middle) and follow an edge to the end of the red cube, we can see there should be a red point there. But this red point is only supported by one red face. We need 2 other faces. One is supplied by the yellow cuboid the other by a neighbouring blue cuboid. This point can only be 1/3 red as 1/3 has been supplied by the yellow cuboid and 1/3 has been supplied by a blue. We can see that the red cuboid “owns” 3 edges between 3 squares. Each edge fully “owns” 1 red dot in the middle and 1/3 with two other cuboids at the end. 3 edges owning 1/3 of a corner = 1. ie The red cuboid owns 2 corners. One totally surrounded by the red faces and 3 x 1/3 of a point shared with other cuboids. ie another 1. 1 plus 1 is 2. Therefore
2 Corners * 8 cuboids= 16 total corners.
LOL. I talked to my GF about this visualisation. She said her head just melted. That’s one hell of a hot flush.
Well signing off.
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This entry is in my future. Today is 2023/03/30. Dear future stephen. I want you to consider something. A square has 4 edges and 4 corners. ie a 2Dcontent (area) is made up of 4 1Dcontents (lines) and 4 0Dcontents (points). This allows Pythag 1D to work in many proofs. Ie we chop off 4 corners (aka points) we left with 4 sides which forms a smaller square.
On 2023_03_29 I said “If a single Ln line is attached to 3 squares and 3 cubes. Then …. I don’t know what.” ie if we cust off X (aka lines) are we left with the same number of sides and cubes? Does Pythag2D work?
I asked chatgpt to give me a method to calculate Pythag1D. The idea is to take the language and replace points (corners), Lines (edges) and Squares (Faces) with the language of pythag2d. ie Lines become area and squares become hyper cubes. Here is the edited chat GPT result
Draw a right triangle on a piece of paper, with one leg horizontal and the other leg vertical.
Construct a square with sides equal to the sum of the lengths of the two legs of the triangle.
Divide the square into four congruent right triangles, with sides equal to the legs of the original triangle.
Rotate the triangles so that they form a smaller square in the center of the original square, with sides equal to the length of the hypotenuse of the original triangle.
The area of the original square is equal to the sum of the areas of the four smaller squares (each of which has an area equal to one-half the area of a right triangle).
This can be expressed mathematically as: (a + b)^2 = 4(1/2)a^2 + 4(1/2)b^2 + c^2, where a and b are the lengths of the legs of the original triangle, and c is the length of the hypotenuse.
Simplifying this equation gives the Pythagorean Theorem: c^2 = a^2 + b^2.
So, by constructing a square with sides equal to the sum of the lengths of the two legs of a right triangle, and then dividing it into four right triangles and rotating them to form a smaller square, we can deduce the Pythagorean Theorem.
So lets transpose from Pyhag1D to Pythag2D and see if we can glean anything useful. ie (Pythag1D)=(Pythag2D)
Draw a (right triangle)=(Hyper cubes corner) on a piece of paper, with (one leg horizontal and the other leg vertical)=(each leg corresponding to Length (Ln), Width (Wd), Height (Ht) and Extent (Ex)).
Construct a (square)=(Hyper Cube) with (sides)=(Faces) equal to the sum of the (lengths)=(Area) of the (two)=(Four) legs of the triangle.
Divide the (square)=(Hyper Cube) into (four)=(sixteen) congruent (right triangles)=(Hyper cubes corner), with (sides)=(area) equal to the legs of the original (triangle)=(Hyper cubes corner).
Rotate the (triangles)=((Hyper cubes corner) so that they form a smaller (square)=(hyper cube) in the center of the original (square)=(hyper cube), with (sides)=(faces) equal to the (length)=(area) of the (hypotenuse)=(?? hypotenuse area) of the original (triangle)=(Hyper cubes corner).
The (area)=(4D content) of the original (square)=(hyper cube) is equal to the sum of the (areas)=(4Dcontent) of the (four)=(sixteen) smaller (squares)=(Hyper cubes) (each of which has an (area)=(4Dcontent) equal to (one-half the area of a right triangle)=(??? to filled in later).
So, by constructing a square with sides equal to the sum of the lengths of the two legs of a right triangle, and then dividing it into four right triangles and rotating them to form a smaller square, we can deduce the Pythagorean Theorem.
This can be expressed mathematically as: (a + b)^2 = 4(1/2)a^2 + 4(1/2)b^2 + c^2, where a and b are the lengths of the legs of the original triangle, and c is the length of the hypotenuse.
Simplifying this equation gives the Pythagorean Theorem: c^2 = a^2 + b^2.
To be completed
To be completed
If you were a 1D worm, who only experienced length, how could you prove that width and height were real in a 3D world? The answer I got from my friend was 1D worms don’t exist. And he is right. But that is not very helpful. If I armed my 1D worm with high explosives and a timer he should be able to work it out. It’s called the inverse square law. The thought experiment is as follows: The worm sets the explosive timer and runs away. Assuming the worm is not blown up, the longer the timer runs the further away he is. The further away he is the less the explosion effects him. The distance and effect are not linear. The effect drops off at the inverse square. Double the run distance and the effect is divided by 4. QED We live in a 3D world. If we lived in a 4D world we would have the inverse cubed law.
But we do live in a 4D world. Special relativity tells us so. Let’s pretend we humans live in a 2D world sensory world. That is, your ears are not points. They are ear drums. Drums have a 2D surface. Your hearing is measured in 2D. Your eyes have retinas. Retinas are 2D. If I divide a 4D world by area. Wouldn’t the output be 2D, as in the case of your eyes? And if the output was not 2D but 1D, as in the case of an ear drum, wouldn’t that mean to double the input to your brain you would have to square the loudness of the sound? It would not be linear it would be, well, measured in decibels. And it is. But what about the 1D worm Experiment? Well as my friend says, they don’t exist. and he is right. He must be, he used to work for the government.
I think this idea is about as sound as my idea that a 4D hyper cube only has 3 node vertices and therefore we can only ever experience it in 3D. Interesting idea but quite wrong. Perhaps I should do some actual constructive work today? Rather than faffing around with 4D space and Pythag 2D.
The drawing a Tesseract hypercube is not going well. I am looking to see If I can glean some insight by calculating the number of lines and areas and volumes by dimension.
|
Dimensions (D) |
Lines by D | Area by D | Volume | Hyper Volume | |
| Line | 1 |
1xLn ======= Total=1 |
|||
| Square | 2 |
2xLn 2xWd ======= Total=4 |
1xLnxWd ======= Total=1 |
||
| Cube | 3 |
4xLn 4xWd 4xHt ======= Total=12 |
2xLnxWd 2xWdxHt 2xHtxLn ======= Total=6 |
1xLnxWdxHt ======= Total=1 |
|
| Hyper Cube | 4 |
8xLn 8xWd 8xHt 8xEx ======= Total=32 |
4xLnxWd 4xWdxHt 4xHtxLn 4xLnxEx 4xWdxEx 4xHtxEx ======= Total=24 |
2xLnxWdxHt 2xWdxHtxEx 2xLnxWdxEx 2xLnxHtxEx ======= Total=8 |
1xLnxWdxHt ======= Total=1 |
So, in 4d space, I have 8xLn. In the lines space. This must be spread over 4xLnxWd, 4xHtxLn, 4xLnxEx in square space. And also spread over 2xLnxWdxHt, 2xLnxWdxEx, 2xLnxHtxEx in volume space. That is each Ln must be used more than once.
A square has 2 Ln sides. Meaning each square (eg LnxWD) has 2xLn. So 4xLnxWd, 4xHtxLn, 4xLnxEx. is 24 Ln’s. But I only have 8 Ln’s. Meaning Each Line Must have 3 squares attached to it. One in the Ht,One in the Ht, and One in the Ex? Well there goes my idea of a 3 noded vertex in 4d space.
A cube has 4 Ln sides. Meaning each cube (eg LnxWdxHt) has 4xLn so 2xLnxWdxHt, 2xLnxWdxEx, 2xLnxHtxEx means we have 24 Lns. But I only have 8 Ln’s. Meaning each Line must support 3 cubes
If a single Ln line is attached to 3 squares and 3 cubes. Then …. I don’t know what.
I went to Perth state library in search of a Tesseract book that did not include the words “doctor who” or “Jesus”. I was unsuccessful, despite the lovely librarians helping me out on the top floor of the library. Section 520 was good though.
I have had a revelation. My revelation is that 4 can be divided by 2 exactly. This means a 4D image should be easily viewed on a 2D plane called a piece of paper. But where to start? I decided to start with a corner of a Tesseract and work outwards. I decided that green would be a good colour.
So, If you look at the green blob in the middle you will see there are 4 lines coming out of the centre to form a kite like object. The lines have a ratio of 3,5,6,7 and correspond to Ln,Wd,Ht,Ex (length,Width,Height, Extend). Which I chose as they are all primes except 6. The idea is that you are looking down from a height at a corner or vertex. If you did the same with a cube you would see 3 triangles. But since we are in the 4d world you see 4. Each triangle is half of a face which forms a rectangle. The blue triangles are the other side of the face. This forms a square in the centre.
Each triangle area has a symbol in it. So we know what we are looking at. We have a star, a bowtie, a triangle and a rectangle. Currently we are at the origin point, or the centre. But we could walk down the star corner face. Through the green to the light blue corner. What would we see? That image is drawn out in the greyed out square. Why is it greyed out? Because it is represented in the original picture. ie it’s not a new face. Each square represents a map of the view we would get if we stood on a corner. And just like a cube with 8 corners, with each corner having 3 faces connected to it. Giving us 8 maps with 24 faces But only 6 would not be greyed out. I think I have made a few mistakes on this diagram. But it’s a good start. There are 24 square faces. In summary each map touches it’s neighbours map by the corner. The empty faces have been filled with arrows which depict edges and corners. These have red and yellow squares in them.
My problem is as follows.
A square has 4 corners. Each corner is where 2 lines meet. ie Each corner is the same
A cube has 8 corners. Each corner is where 3 lines meet. ie each corner is the same.
A hyper cube has 16 corners. Each corner is where 4 lines meet. ie each corner is the same. - This is wrongExcept of course my map has 2 types of corners. The red square ones and the yellow square ones. They are different. Now I believe my map is wrong. But I think it is consistently wrong. ie it might be right from a certain point of view. but I have not worked out why. So I decided to see if I could generate a equation which calculates the Points of a shape given a D=Dimension. The answer is 2^D points/corners (0D). How about lines/edges (1D)? or How about square/faces (2d) ? etc.
I gave it a go and used Wikipedia to check my answers. Unfortunately I had problems. Chapgpt lead me astray with incorrect equations. But gave me some good Key words. ie Schläfli symbol. Which lead me to https://arxiv.org/pdf/1511.02851.pdf. Looks like I have some reading to do
|
0Dcontent (Point) |
1Dcontent (Length) Ln |
2Dcontent (Area) Ln * Wd) |
3Dcontent (Volume) Ln * Wd * Ht) |
4Dcontent (Hyper volume) Ln * Wd * Ht *Ex |
5Dcontent (???) |
6Dcontent (???) |
|
|
0Dcontent (Dot) D=0 |
1 (2^D) |
0 |
0 |
0 |
0 |
0 |
0 |
|
1Dcontent (Line) D=1 |
2 Corner 2^D |
1 () |
0 |
0 |
0 |
0 |
0 |
|
2Dcontent (Square) D=2 |
4 Corner 1*2^(D-0) |
4 Edge D*2^(D-1)) |
1 Face () |
0 |
0 |
0 |
0 |
|
3Dcontent (Cube) D=3 |
8 Corner 1*2^(D-0) |
12 Edge D*2^(D-1) |
6 Face 2D*2^(D-2) |
1 Cubical 1*2^(D-3) |
0 |
0 |
0 |
|
4Dcontent (Hyper Cube) D=4 |
16 Corner 1*2^(D-0) |
32 Edge D*2^(D-1)) |
24 Face 2D*2^(D-2)) |
8 D*2^(D-3)) |
1 1*2^(D-4) |
0 |
0 |
|
5Dcontent (5-cube) D=5 |
32 Corner 1*2^(D-0) |
80 Edge D*2^(D-1)) |
80 Face 2D*2^(D-2)) |
40 2D*2^(D-3)) |
10 D*2^(D-4) |
1 |
0 |
|
6Dcontent (6-cube) D=6 |
64 Corner 1*2^(D-0) |
192 Edge D* 2^(D-1)) |
240 Face Wrong! |
160 4D*2^(D-3)) |
60 1*2^(D-4) |
12 ??? |
1 ??? |
|
7Dcontent (7-cube) D=6 |
Corner |
Edge |
Face |
??? |
??? |
??? |
??? |
So what I am doing is trying to develop checks and truths that I can use to ensure the diagram is correct. ie I should have 16 corners, 24 faces, 32 edges and 8 cubes.
The squares on the diagram represent points (16). The points are at the end of edges. and the coloured faces are, well, faces (24). I am not sure where the 8 cubes are to be found in the diagram. Which is confusing as a cube has 6 square sides. So where does 6*8=48 sides fit in this shape? Clearly the cubes share faces. In the same way a square on a cube shares lines.
Well this is all very confusing and I hope to have an answer. But in my quest for Pythag2d. I am going to need a hypercube where the corners are all the same.
The rules are for my Hypercube.
- Must have 16 corners - With are all identical
- Must have 32 edges
- Must have 24 faces
- Must have 8 cubes
- While the end goal is a The hypercube (square) representation must support a hyper brick (rectangle) 3,5,6,7 edge ie (Ln=3,Wd=5,Ht=6,Ex=7)
-- Each edge of the hyper brick needs to tessellate with the edge on the other face.
-- Each face has must be a 2D rectangle
-- Shape must be logically viable and consistent.
-- Each side must be equal. ie a 3d brick has 12 sides. and 3 length (3d). Each length corresponds to 4 sides.
---Therefore if we have 32 edges and 4 lengths (4d) then each length must be used 8 times.In other news, while on holiday and not pestering the library for maths books. I had a good chat with Stevie C from Perth. He runs a free popup art exhibition. Which I enjoyed. I decided to invest and now enjoy his picture on my wall every day. I asked if I could put the image on the side of a Tesseract. He was very enthusiastic.
My definition of good art is if it makes me smile when I see it in the morning before I goto work. His picture on my wall meets that definition.
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Spent entire the week trying to get the master formulae (Pythag2d) working. - Which is a copy and paste of last week. I can’t seem to leave this alone. I am just having issues getting a 16 corner 24 or 4 face shape into my head. 24 or 4 faces depending on how you think of it. I think I am turning a little bit strange. Which is concerning because I was a little bit strange to begin with. I think I will leave this for a week and come back to it.
The basic shape is a Rectangular Tesseract with Ln=3, Wd=5, Ht=6 and Ex=7. They are all primes except 6 which is a prime of 2. I am betting that if I can get it working for this shape then it will work for everything
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Spent entire the week trying to get the master formulae (Pythag2d) working. Unfortunately it does not work when I pump in real numbers. I think I will give up for a week and do some real work as I seem to be chasing my tail. My idea of context switching from one task to another each day, does not seem to work. Perhaps context switch each week is the way forward?
Done Week - 1 End 10/3/2023 - Pythag2d
Todo Week - 2 End 17/3/2023 - Video update + Questions and answers
Todo Week - 3 End 24/3/2023 - Update code L0_cone.scad
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Finally I have the answer to the following question:
Given the area and height of an ice cream cone; What size is the ice cream ball at the top?Which is functionally the same as the more formal question:
Given the Lateral area of a cone and it's height; What is the radius of the cones base circle? Of course I want bigger better more, and over the last week I have been trying to derive Pythag2D which is the relationship between two perpendicular areas in the same way that Pythag1D is the relationship between two perpendicular lines.
This “small” project has devoured my time and there seems no end in sight. Therefore I have decided to move to the following:
Mondays is Derive Pythag2D day
Rest of week is spent on OS++;In this fashion I hope to move on with the OS++ project while still trying to resolve Pythag2D. As a result I will Run two news lists. One with OS++ and One with Pythag2D. If and when I solve Pythag2D I will link it up.
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I decided to do a blind test of my code over the weekend. My GF created cones with calculated Radius, Height and Lateral area. She then handed me a piece of paper with just the Height and Lateral area and I had to calculate the radius. The conclusion is that it would appear my code works. Which is good.
Code
function Di_cone(La,Ln)= sqrt(sqrt(4*Ln^4 + (La*4/PI)^2 )-2*Ln^2);
/*
echo(Di_cone(La=15*PI , Ln=4)); // expect 6
echo(Di_cone(La=20*PI , Ln=3)); // expect 8
echo(Di_cone(La=sqrt(2)*PI , Ln=1)); // expect 2
// other Pythagorean Triples
//
// (3, 4, 5) (5, 12, 13) (6, 8, 10)
// (7, 24, 25) (8, 15, 17) (9, 12, 15)
// (9, 40, 41) (10, 24, 26) (11, 60, 61)
// (12, 16, 20) (12, 35, 37) (13, 84, 85)
// (14, 48, 50) (15, 20, 25) (15, 36, 39)
//
// source https://mathsathome.com/pythagoras-theorem/
*/
// These are the numbers that GF gave me and I had to calculate without knowing the answer
// She said what if you cheat? How I asked. why would I? This is not a test to decide if I am a cheat.
// It's a test to see if I have made an honest mistake. She was not convinced.
echo ("1",Di_cone(La=921.95, Ln=3)/2);
echo ("2",Di_cone(La=683.68, Ln=26)/2);
echo ("3",Di_cone(La=51.05, Ln=6)/2);
echo ("4",Di_cone(La=121.16, Ln=12.5)/2);
echo ("5",Di_cone(La=445.29, Ln=19)/2);
echo ("6",Di_cone(La=80.59, Ln=3.5)/2);
echo ("7",Di_cone(La=20.12, Ln=2.5)/2);
echo ("8",Di_cone(La=502.27, Ln=9.5)/2);
// 
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Spent the weekend trying to derive Di (Diameter) from Ln (Length/height) and La (Lateral area - ie the trunk of the Pcone) After a weekend of many crossed out pages and an argument with the GF I came up with this: NB this is all openscad.org and ascii/127 compatible. Sorry for being verbose.
#Starting with La = PI * Ra * L
where
La = Lateral area
PI = Pi or 3.14159265 ish
Ra= Radius of base circle of Pcone (ie Di/2)
Ra=(D/2)
L = Lateral length of cone (ie diagonal line from base to peak)
L=sqrt( Ln^2 + Di^2 )
Ln = Length or height of the Pcone
^2 means to the power of 2
* means times or x
+ means add or +
= means numerically equivalent to
#Derive Ra= in terms of Ln (length/height of cone )and La (Lateral area of cone)
#Work
NB sOh cAh tOa
NB L ie sqrt( Ln^2 + Ra^2 ) but also L =(Ln/cos(theta) )
Where
Ln = Length or height of the Pcone
Cos(An) = cosine = adjacent / hypotenuse = Adj / Hyp
Where
An = 1/2 angle of the peak of Pcone.
ie An = 90-Sl
Sl= slope of the Pcone from the base
Q) Back2, the line below why do we care?
NB L ie sqrt( Ln^2 + Ra^2 ) but also L =(Ln/cos(An)) )
A) Because when I use L=sqrt( Ln^2 + Ra^2 ) I get lost in a maze of trying resolve Ra in terms of Ra. It's a bit like taking a picture of a mirror.
#Recap
La = PI * Ra * L
L = Ln / cos(An)
#starting with tan(An) = ( Ra ) / ( Ln)
* by PI/PI tan(An) = (PI * Ra ) / (PI *Ln)
* by L/L tan(An) = (PI * Ra * L) / (PI * (L ) *Ln)
Sub La = PI * Ra * L tan(An) = (La ) / (PI * (L ) *Ln)
Sub L=Ln/cos(An) tan(An) = (La ) / (PI * (Ln/cos(An)) *Ln)
* cos(An) tan(An) / cos(An) = (La ) / (PI * (Ln ) *Ln)
Simplify tan(An) / cos(An) = La / (PI * Ln ^2)Unfortunately that is where it all ended. I simply could not simplify tan(An) / cos(An). It became obvious I could hunt the value of An down by simply increasing it 1 step at a time. After all Tan(AN) starts from 0 at 0 degrees. and goes to infinity at 90 degrees. While Cos(An). Reduced but as we are / that means it steadily increased as well. This mean between 0-90 degrees there WAS a single solution.
I tried to divide the equation of La for a cone with La for a cylinder. This gave me a ratio from the LA of a cone to cylinder. The idea was to use this in the same was a trig works. With the hypotonuse being the LA for the cylinder and the Adjacent being the LA of the cone. But this leaf me in circles.
My next discover was that If you divided the Ln by 2. The La is divided by 4. This mean that I could now work with a truncated cone (CHc). I knew the Top circle had an area 4 times the bottom one and I knew the LA was 3/4 the cone. I also had the Ln or height of the cone. So what? well it appears that we need to find a way to use LN instead of Dg to resolve the cone. As Dg is calculated from Ln via Ra. Which we don’t know. So I need something for Ln. La/Ln means nothing. But if I can calculate the ratio of Ln to the average circle I could calculate the cone as a cylinder (without Dg). Lets pretend that the magic ratio is 0.75 (it’s not) of Ln. Then using LA/(Ln*0.75) would give me the average area of the circle of the cone. And all I would need to do to find tR would be to use the area of a circle formulae. Remember this is not just any truncated cone. The ratio of the Area of top circle to bottom circle is 4. I know the Height and I know the LA. Perhaps there is enough info to calculate Ra.
More next week.
If you like this, and wish to keep me out of the poor house; please would you Contribute (Buy me a coffee)
Well the “new school” and fancy way of doing cylinders is coming to an end. Da was a surprise. Da is the angle, measured in degrees, of a diagonal line from the top of the cylinder to the bottom. With this angle you can calculate a cylinder with any of the other property types. I would have thought that Dg - the length of the actual diagonal line would have been better. In other news the OSph - Outside Sphere is completed and I just need to process ISphere and I think I will be done for the time being. Next week I will update all the documentation for L0_cone_genCyl() and start work on L0_cone_genPcn. Hopefully this will go much more quickly than the previous Gen. Looking forward to this release at the end of the month. I have a $20 new Microphone which I want to test out for video. This will replace using my mobile phone as a mic. I am also thinking of Doing the questions for free but the answers video being a Premium option.
I spent the whole day on a single equation. The whole day. I had rewritten it and derived it 3 times from scratch. The maths was correct. It had to be correct. I had done it 3 different ways and had got the same answer. Obviously I had a more fundamental issue on my hands. I suddenly realised that the question I was asking the equation; was wrong. Wrong as in the expected answer of 10 was actually 2500. A whole work day gone….. to find a bug that was not there. I am never going to make this Dec 2022 deadline.
100% of my current audience, That would be Mach, asked me to get hold of a book with the title starting “Don’t suck ….”. I am sure he means well, but I realised that I was sucking, I was sucking big time. The problem was that it was Feb 16th 2023 and I still was not ready for the Dec 2022 deadline. Don’t get me wrong. Long nights and even longer equations were being written. But my code still wasn’t ready. No matter how hard I tried. No completed code = no release. And from the worlds point of view my website was dead. Nothing was happening. A reasonable person would assume I had finally come to my senses and admitted defeat and that the project was dead. That sucks. So I created this news section. The idea is to update on the weeks work. Even if there are no changes to the code itself. A heart beat if you will. Just to show that all is well and the project is progressing. I want to be clear and assure the world that, I have definitely have NOT come to my senses. Yet.
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