# Trurl

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1. ## Prime Products just one last time

Ok for the new year I wanted to clarify the equation I posted here. With the following equations I wanted to show the test value of x produces a PNP – the calculated PNP approaches zero that test values of x equal p, in N = p* q. So, when F approaches PNP the value of x is the p, in N = p * q. I have included 2 if-statements that test this. And as you can see, an x equal to PNP (85 in this case) is zero, but those numbers larger than the correct x of 5, increase in value as they approach 5 and those test values smaller than 5, increase until they reach zero. (That is in my IF-Statements.) Yes, I know that the test value is too small. The problem is the accuracy of the PNP calculation relies on the square root of a large value. This is causing a margin of error in values of PNP greater than 3 digits. But I show this because the estimate is significant. How do I make the square root of the polynomial in F in this code to be more accurate? I hope you agree this equation is significant. PNP = 85 x = 85 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 85 161 Sqrt[4123/2] 1/2 (85 - 7^(3/4) Sqrt[23] (589/2)^(1/4)) N[1/2 (85 - 7^(3/4) Sqrt[23] (589/2)^(1/4))] -0.249277 PNP = 85 x = 5 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x ]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 5 Sqrt[4179323/2]/17 1/2 (5 - Sqrt[-80 + Sqrt[4179323/2]/17]) N[1/2 (5 - Sqrt[-80 + Sqrt[4179323/2]/17])] 1.37825 PNP = 85 x = 7 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x ]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 7 (11 Sqrt[45773587/2])/595 1/2 (7 - Sqrt[-78 + (11 Sqrt[45773587/2])/595]) N[1/2 (7 - Sqrt[-78 + (11 Sqrt[45773587/2])/595])] 1.88414 PNP = 85 x = 3 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x ]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 3 Sqrt[847772947/2]/255 3 + Sqrt[88 - Sqrt[847772947/2]/255] N[3 + Sqrt[88 - Sqrt[847772947/2]/255]] 5.69458 PNP = 85 x = 1 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x ]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 1 (7 Sqrt[13123/2])/85 1 + Sqrt[86 - (7 Sqrt[13123/2])/85] N[1 + Sqrt[86 - (7 Sqrt[13123/2])/85]] 9.90669 Above is the input and output of my code. The test values are separated by spaces.

3. ## Introducing the Piangle

Good job. Keep up the great work. I don’t think it is a thousand years old. You have just used an angle and may have unknowingly drawn an involute which is a type of logarithmic spiral. (In case you don’t know I love logarithmic spirals.) When you unroll each outline to the right; Can someone in this forum tell me if it is a linear representation of the involute? I have seen something graphical, similar to this unroll in a math reference. I can’t remember where, but I believe it was for gears. I don’t have any pictures of a logarithmic spiral to share yet. I want to be sure it relates to your post. But even if this work is rediscovered, it doesn’t mean you can’t relate it to something new. What I think you should try is to “put space between your Pi angles.” What I mean by this is that having a “series” between where 1/3 r and 2/3 r and r would change the shape of the involute to a special logarithmic spiral. I know it sounds like I’m talking babble, but I am not. If you are confused on what I am trying to say, let me know and I will try and describe my idea better. Simple put I would shift the new larger angles a distance (determined by a series) across the x-axis from the original triangle that was at the origin. This way you can craft series and describe them in a logarithmic spiral. I will post a picture of an ellipse determined by angles. It is not a logarithmic spiral, but it will demonstrate using angles to determine geometry. I will try to work on drawing a graphic representation of the logarithmic spiral I describe here. But this will work till then.
4. ## Prime Products just one last time

I know this doesn’t solve the problem. I am only trying to solve the triangle I constructed. But is it possible to find FE by subtraction. I may not be doing it right, but is it possible to do subtraction of the triangle segments to get FE? To me it seems possible, but when you go to do it, it is confusing. But it just feels possible. Remember I am only trying to solve for FE. AC = N [Absolute value [ AC – AE – (AC-CE) – CE]] = FE
5. ## New TSP Method (p=np)

I will stay out of the conversation after this post. I just wanted to clear up what I said. It may make more sense than you think. All I am saying is that geometric constructions may simplify the calculation. Let me know if this does or does not make sense. I am saying take a compass and draw a circle that encompass 2 points. Then with 3 points draw another circle from the 2 points closest together. Keep drawing circles from all points. When there are several points, the circles should intersect somewhere along the circle. Connect those intersections with lines and you have a polygon or the shortest path. Do you remember in high school when they taught us to find the center of a line by taking a radius more than half of the line and string 2 arcs from each side? The line between those arcs passes through the center of the original line. I propose if you did the same constructors of circles on the unknown points, geometric constructions such as finding the tangent of a circle (or any of the dozens of circle constructions) you would simplify the computer process. I don’t know how to put a drawing compass into a computer program, but you could always use a CAD script. But then again, I don’t know the algorithm to such a thing, as I have only spent 2 minutes coming up with a hypothesis. I would say that a radius of the circles, and arcs of those intersecting circles, would eliminate calculations that just can’t be done. I wouldn’t scrap the polygon idea. Instead, I’d use the polygon to form a path between the intersecting arcs. Also by using circles you have the advantage of all the circular functions. Do they still teach geometric constructions in geometry? Yes, I know the problem has N points, but this is a simple approach. As more and more points are formed you would have to erase (delete) some the circles no longer needed. And no, I don’t claim I can solve this problem. But I do like how the computer was run to solve this problem. Computers have already ruined chess and that game with squares. I’m glad there is much work to be done to solve this problem. I hope this is clearer. All I am saying is use geometric circle constructions. After all, you can build almost any shape with them. Someone has probably tried it, but before modern programming, such an idea wasn’t possible. Because in CAD you could program it to draw hundreds of points.
6. ## New TSP Method (p=np)

Polygons seem so complex. I like this idea, but circles are more closely related to navigation. Here is my humble opinion (not understanding the complete traveling salesman problem): Draw circles at a start point. It doesn’t matter which one. Continue to connect points with circles. Like at technical drawing or highway design where circles intersect is the path. So you would 2 circles between 2 points. Then you would draw a circle from the 2nd point to the third and where the intersection occurs is the shortest distance. The trick is to program it on the computer, taking away circles that no longer give the shortest path. I believe this is what you did with the polygon that encircles all points. However, I believe it is overly complicating your method. Brilliant none the less. Will you post your computer program here to share? I’m not concerned it is similar to other approaches. It is your approach. I suggest you get a technical drawing book and look at the geometric constructions. Circles not polygons, because you have your polygon acting as a circle. Circles are much more suited for the task.

8. ## Intelligence test

IMHO, I think you should design your own I.Q. test. I don’t mean start from scratch, but make a hybrid test that tests the skills you are looking for. Searching for an I.Q. test on the web will lead to a bunch of garbage or some high-priced package. I don’t know what your goal is. However, usually psychologist use many characteristics to determine I.Q. Testing is just one. For instance, they can look at grades in class, or how a person interacts with others, or how they spend their time. That is why I don’t think you are going to find a credible test that allows you to test people over the web. Of course, I’m just an armchair psychologist. But I believe I am giving you good advice to make your own test. After all, you are the only one how knows what factors you are looking for. Here is some credible tests, which may or may not relate to I.Q. However, those who created them knew what they were looking for in the results of the test. · SAT · ASVAB · AHSME - http://artofproblemsolving.com/wiki/index.php?title=AHSME_Problems_and_Solutions I hope this helps. It may not be what you are looking for. But without records of I.Q. of a sample group, you must be resourceful. Note that the AHSME test is very strategic. The rules are listed on the website with previous test. It is a timed test. And you can learn a lot about the student from the problems they choose to complete, because the student losses point for wrong answers and must break 100 points. There is both knowledge and strategy involved. Also, the problems are not grouped by subject and the word problems are not just rearranging values. There is problem solving. That mean no memorizing answers or not working through the problem. In fact, I propose a challenge to the members of SFN and the Moderators. If you are the scientists you think you are: I Challenge the members of SFN to design their own I.Q. test. If you think that Chriss is not knowledgeable enough to test the I.Q. with analytics then post a question to this post that you would consider a measure of I.Q. until the community has designed a credible solution.

10. ## The Physics of Star Trek

Here is my attempt at science fiction. It is a short story I wrote for a creative writing course over 7 years ago. I reread it. I didn't remember the story. It was like reading it for the first time. But be honest and let me know if it makes you think. You can be honest because I won't be offended. I am not a writer by trade. story013_final_webcopy.docx
11. ## The Physics of Star Trek

Excellent reply. I just was thinking that instead of anti-matter, anti-gravity would more efficiently propel ships. Instead of breaking up energy and using that massive amount of power released to power a warp engine, using the gravity that already exists in matter, would make more sense. We already sling-shot satellites off planets orbits to speed them up. If gravity was controlled there would be more versatile and realistic solutions. Also, if artificial gravity did exist, wouldn't it just cause forces that would terminate the inhabitants of the ship. When a rocket leaves for space the G-forces exerted on the astronauts are significant. So artificial gravity on the ship would mean that a warp drive would increase its forces. But again, if there were artificial gravity, it would be a better alternative to navigate the ship through space by using gravity. Of course, as you guys mentioned fictitious physics can't be proven. I'm just making this stuff up. But has anyone here ever thought to use their science knowledge to write fiction? I know it's been done before. But what about in college when you had to take English classes? I wrote a short story for class years ago. It didn't get its message across. I had a lot of symbolism in it, but unless you knew why I wrote it the way I did, the point was lost. I'm going to read it again and see if it is worth anything.
12. ## Prime Products just one last time

Attached are 2 important PDF’s that have been on my site for years, but are probably only seen by a few because of the difficulty of going through all the information. Yes, some patterns may repeat. But here in the first PDF, 3 equations are presented. Pay close attention to: Sqrt[(N*y – x^2)/x] = y This is one pattern for sure that I know isn’t just repetition. It is used in the equation at the beginning of this discussion here at SFN. The other 2 equations are just hypothesis. You will see this in the second PDF: I believe with most symmetric key ciphers, the fact that Prime numbers equal a one way function does not mean that there isn’t a pattern in the one way computation. So if N equals the product of 2 Prime numbers, Prime numbers have no pattern, but multiple 2 Prime numbers together there is a pattern. Where x = 571 and y = 1381 1381 / 2p = 219.7929764 2p * (remainder(1381/2p) * 2p) * 571 = 13208.65186 rewritten: 2p * ((1381/2p-whole number part) * 2p) * 571 = 13208.65186 13208.65186 / (1381/2p) = 60.09 13208.65186 / 60.09 = 219.7929764 219 = 219 And another relationship: New equations: 2p * (remainder (17 / 2p)) = 4.42 4.42 * 5 = 22.1 22.1 * (5 / (17* 2p)) = 0.884 0.884 * 5 = 4.42 4.42 = 4.42 See what you can do with it and share it. I believe p stands for Pi. Some of these write-ups are several years old and after making so many I don’t always know what I intended to communicate, without studying them. But I feel these PDF’s are important. That is important enough these write-ups may give meaning to the work. Either way, let me know if you think they are trash, or if it gives you any ideas. 20140519TrigPrimes005secCopyCC.pdf PrimeProductSolutionFlyerCopyCC.pdf
13. ## The Physics of Star Trek

I think “Space” should have its own thread here at SFN. I want to do a lot of reading on space. Such topics as: · Space Simulators · Space Dynamics · Terraforming · Mining Space · Space Station Design · Moon Bases · The Apollo Computer · Space Navigation · A Video Game of Space similar to Mine Craft But I have a question that relates to space. And is more serious than it sounds: When the Enterprise goes warp speed how does it avoid objects in space, such as comets, dust, and planets? Also, the Enterprise shots photon torpedoes without a reaction force sending the ship with an opposite force. I don’t understand photon torpedoes, but wouldn’t it be simpler to send a torpedo at warp speed, that instead of causing a small part of the ship to be damaged, would totally obliterate the enemy ship. This torpedo would act as F = ma and the resulting energy release would destroy everything. Also, the shields can’t stop such a torpedo. There are “shield harmonics” (like in sound) that block the energy of phasors. Stopping a solid object would be more difficult. So, lasers and phasors are blocked, but traditional projectiles such as missiles are not. Finally, when a space ship is hit by a projectile, I would like to see it spin end over end out of control in space, before the propulsion engines could make the necessary adjustments. I know there is a book called “The Physics of Star Trek.” It may have these answers. All that I have read so far is the inertia dampers prevent the force of coming out of warp speed from acting on the passengers. Has anyone here read the entire book? I don’t know if my questions are answered there. I shouldn’t question the physics of the ship, since Scotty and Spock make it work. It is shows like these why we are interested in science in the first place.
14. ## Prime Products just one last time

PNP = 85 (((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/ x^2))) == PNP^2 85 (7225 (-1 + x^2/170 + (52200625 x^2 + 14450 x^5 + x^8)/52200625))/x^2 == 7225 Solve[(7225 (-1 + x^2/170 + (52200625 x^2 + 14450 x^5 + x^8)/52200625))/x^2 == 7225, {x}] {{x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 1]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 2]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 3]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 4]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 5]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 6]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 7]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 8]}} N[%3] {{x -> -24.3484}, {x -> 4.97889}, {x -> -4.2072 - 2.87925 I}, {x -> -4.2072 + 2.87925 I}, {x -> 1.71775 - 5.00069 I}, {x -> 1.71775 + 5.00069 I}, {x -> 12.1742 - 21.0804 I}, {x -> 12.1742 + 21.0804 I}} f[x_] := (((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2))) == PNP^2 b = [Integrate[f, {x, PNP - 1, PNP}]] b = f[85] The above equation is simplifying the PNP polynomial equation from SFN post 1. This equation proves to be true but there is no way to solve it. The equation shows where x is as x approaches N (or PNP). Taken this feature and using the integral to simplify the original equation where PNP is 85 (in this example), we know that the integral from N - 1 (that is 85 - 1) to N (which is 85) is equal to f(85). Of course some figuring is wrong here. I haven't taken a calculus class in almost 20 years. I don't know if it is possible to have the integral equal the original function. But intuitively it seems to work. I know fault will be found here. I'm just bouncing off ideas. I would have formatted this post better, but Big Bang Theory is coming on.
15. ## Prime Products just one last time

PNP = 85 (((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/ x^2))) == PNP^2 85 (7225 (-1 + x^2/170 + (52200625 x^2 + 14450 x^5 + x^8)/52200625))/x^2 == 7225 Solve[(7225 (-1 + x^2/170 + (52200625 x^2 + 14450 x^5 + x^8)/52200625))/x^2 == 7225, {x}] {{x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 1]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 2]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 3]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 4]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 5]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 6]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 7]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 8]}} N[%3] {{x -> -24.3484}, {x -> 4.97889}, {x -> -4.2072 - 2.87925 I}, {x -> -4.2072 + 2.87925 I}, {x -> 1.71775 - 5.00069 I}, {x -> 1.71775 + 5.00069 I}, {x -> 12.1742 - 21.0804 I}, {x -> 12.1742 + 21.0804 I}} f[x_] := (((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2))) == PNP^2 f'[x] (7225 (x/85 + (104401250 x + 72250 x^4 + 8 x^7)/52200625))/x^2 - ( 14450 (-1 + x^2/170 + (52200625 x^2 + 14450 x^5 + x^8)/52200625))/x^3 == 0 g[x_] := f'[x] 1 + (85 - f[1])/f'[1] False 313875685/208889212 N[313875685/208889212] 1.50259 Again the above equation has 5 as a solution knowing only N. I know this is not enough until tried with all values, but it seems that Mathematica has a solution for this instance. I want to show what I was trying to do at the bottom (end) of the equations. I was trying to use Newton's Method. I studied the equations from Wikipedia and felt that maybe applying the method would simplify my equations. I used an y of 85 (a y of N, instead of zero) and placed the slope (the derivative of my PNP equation) in order to solve a test value of x. This test value x is intended to be an estimate, however I tried to find a modified the equations to find x based on a given number (start point; 1) and the slope (derivative) of the original equation (from the first post of this thread). I know this didn't work, but I share just to walk through the idea. If there was a way to find a given value of an equation using the equation and its slope, it would solve my original equation. Below is the equation. 0 = f ' (xn) *(x - xn) + f(xn) 85 = f ' (xn) *(x - xn) + f(xn) xn = 1 ---- This is the start value. 85 = f ' (xn) (x - xn) + f (xn) 1 + [85 - f(xn) / f '(xn)] = x I am aware this doesn't work. However I thought the idea was so simplistic it could work. This is what I am proposing: The slope is known and any start number can be used, so finding the value of x where the y-coordinate equals N can be found without the complexity of the original equation. Obviously, it has been done before. But I am asking for help in trying to apply it to my equation. 20170916SFNnewtonMethod008b.nb