The Math Lovers Club v. 3.1416

[DT_12_Octavarium]

It's always hard to wrap my head around the fact that two separate numbers are actually the same. Because in my mind 0.999... Is still 0.000....001 less than 1.00000....

[Jamesman42]

But 0.000...001 is finite because it ENDS in 1.

[EraVulgaris]

But 0.000...001 is finite because it ENDS in 1.

It's not finite if there are an infinite number of zeros in between.

You can also think about it this way:

Let's say x = 0.999....

Multiply both sides by 10

10*x = 9.999....

Subtract x from both sides (which we defined as 0.999....)

9*x = 9

Divide by 9

x = 1

So x = 1 = 0.999....

[wizard of Thought]

Guys, I don`t want to interrupt your (in-)finite number debate (it`s almost exactly the same, everybody denying that is really nitpicking).

Anyways, I`m currently writing about the Kepler Conjecture (packings of spheres) for a school work and I`m really fascinated in that topic. I mean, the problem itself is so damn simple, but nevertheless it took 400 years to be solved and it`s interesting to see how people figured it out. It kind of demonstrates mathematician madness (really, the right answer is obvious, but not proved until 1998 which is 387 years after it was firstly mentioned), but I love it.

[EraVulgaris]

Guys, I don`t want to interrupt your (in-)finite number debate (it`s almost exactly the same, everybody denying that is really nitpicking).

Anyways, I`m currently writing about the Kepler Conjecture (packings of spheres) for a school work and I`m really fascinated in that topic. I mean, the problem itself is so damn simple, but nevertheless it took 400 years to be solved and it`s interesting to see how people figured it out. It kind of demonstrates mathematician madness (really, the right answer is obvious, but not proved until 1998 which is 387 years after it was firstly mentioned), but I love it.

Huh, I just finished working on some stuff about particle distributions in powders (exciting stuff, right?) like 20 minutes ago. Not exactly the same thing, but kind of in the same vein. What class are you taking that they want you to deal with that?

[jasc15]

After I got my engineering degree about 10 years ago, I decided I wanted to get a graduate degree in physics.  This didn't pan out, but in the process I took a few undergrad courses at Stony Brook, including complex analysis.  This was the first college class I failed.  I realized a few weeks in that I didn't have the necessary math background, including real analysis and a few other subjects, since my undergrad math education was only 4 semesters of calculus and one of statistics.

Anyway, as I struggled with the material, I couldn't help but feel like the subject matter was just barely out of reach, rather than totally above my head.  A few months ago I came across this youtube channel with a series of excellent lessons on introductory complex analysis.  The graphics are extremely helpful, especially in the later videos where conformal mapping is depicted real time of the lecturer drawing functions on the input plane with the output plane transformed using a pixel mapping script.  I remember the lecture in my class about this, and I just couldnt visualize what was happening.



https://www.youtube.com/watch?v=T647CGsuOVU&list=PLiaHhY2iBX9g6KIvZ_703G3KJXapKkNaF

[rumborak]

Ooooh, conformal mappings. You spend a lot of time doing that in EE because you can often simplify a problem by transforming it into a different space.

[jasc15]

The only EE application of complex numbers I have been exposed to is in AC RLC circuits with virtual power, reactance, etc.  Not surprising since I'm an ME.

Anyway, check out the whole playlist.  Should be about an hour total.

[cramx3]

I definitely forgot most if not all of my higher learning in math. The hardest math class I had was an EE class though. I don't think I was ever able to solve the problems relating to Fourier series.

[jasc15]

Speaking of Fourier:

https://toxicdump.org/stuff/FourierToy.swf

Edit:

And this https://www.youtube.com/watch?v=r18Gi8lSkfM

[Chino]

Okay DTF. I have a real brain-buster for you. What is the are of this hexagon?



(Please excuse the crudity of this representation. All 6 angles are identical)

[pg1067]

Okay DTF. I have a real brain-buster for you. What is the are of this hexagon?



(Please excuse the crudity of this representation. All 6 angles are identical)

Since it's an equilateral hexagon, and the sum of the interior angles is 720 deg., the hexagon can be divided into 6 equilateral triangles with legs of 5.  Using the side-angle-side method of calculating area (A = ab*sinC/2, where a and b are the sides and C is the angle in degrees) gives you 10.8253 for the area of each, and multiply that by 6 gives you a total area of 64.9519.

Am I right?

[RuRoRul]

75√3 / 2 or about 64.9519 ' .

Got to this drawing the line connecting up the other two opposite vertices on the hexagon, and knowing each of them will be 10 as well since the hexagon has equal angles, equal sides etc.

This splits the hexagon up into 6 triangles, which are equilateral triangles (equal sides, equal angles), because in the centre you have 360 degrees divided into 6 equal angles, and at the outside you have the 120 degree angles of the hexagon split in two. Since each full line through the hexagon was 10, then each triangle has side 5.

So we want to work out the area of an equilateral triangle of side 5. Different formula for doing this but "half base times height" is perhaps the simplest one. If the base is 5 you need to work out the height, which you can get using Pythatogas' Theorem or trig functions but you don't need those. Divide the triangle in half and consider the right angled triangle with one side 5/2 (half of the base 5) and hypotenuse 5. The height will be the other side, and Pythagoras' theorem gives us that (5/2)^2 + height^2 = 5^2, which ends up giving height^2 = 75/4. Taking the square root of this gives 5√3/2.

Anyway, this was all to work out the area of an equilateral triangle using "half base times height", which gives 1/2 * 5 * 5√3/2 = 25√3/2. There are 6 of these triangles, so multiply by 6 and you get 75√3 / 2.

Not something I knew a forumla for so quite interesting to check how to get it using basic geometry. No idea of the method I described is needlessly long and if there's a much easier way of looking at it, but tried to do it using minimal other formula and not needing to know the values of trig functions.

[Chino]

Since it's an equilateral hexagon, and the sum of the interior angles is 720 deg., the hexagon can be divided into 6 equilateral triangles with legs of 5.  Using the side-angle-side method of calculating area gives you 10.83 for the area of each, and multiply that by 6 gives you a total area of 64.98.

Am I right?

That's what I got as well, but I wasn't sure. I think that's the first time I've ever needed the area of a hexagon. I'm putting up a new gazebo on Saturday and it looks like I'm going to have a little more than 100 extra square feet under cover now   

Thanks

[Orbert]

Nerds will inherit the earth.

I didn't check the numbers, but the geometry is sound.  And congrats on the gazebo!

[pg1067]

75√3 / 2 or about 64.9519 ' .

Got to this drawing the line connecting up the other two opposite vertices on the hexagon, and knowing each of them will be 10 as well since the hexagon has equal angles, equal sides etc.

This splits the hexagon up into 6 triangles, which are equilateral triangles (equal sides, equal angles), because in the centre you have 360 degrees divided into 6 equal angles, and at the outside you have the 120 degree angles of the hexagon split in two. Since each full line through the hexagon was 10, then each triangle has side 5.

So we want to work out the area of an equilateral triangle of side 5. Different formula for doing this but "half base times height" is perhaps the simplest one. If the base is 5 you need to work out the height, which you can get using Pythatogas' Theorem or trig functions but you don't need those. Divide the triangle in half and consider the right angled triangle with one side 5/2 (half of the base 5) and hypotenuse 5. The height will be the other side, and Pythagoras' theorem gives us that (5/2)^2 + height^2 = 5^2, which ends up giving height^2 = 75/4. Taking the square root of this gives 5√3/2.

Anyway, this was all to work out the area of an equilateral triangle using "half base times height", which gives 1/2 * 5 * 5√3/2 = 25√3/2. There are 6 of these triangles, so multiply by 6 and you get 75√3 / 2.

Not something I knew a forumla for so quite interesting to check how to get it using basic geometry. No idea of the method I described is needlessly long and if there's a much easier way of looking at it, but tried to do it using minimal other formula and not needing to know the values of trig functions.

I blew up my screen size to 500% and still couldn't read that!   

[RuRoRul]

75√3 / 2 or about 64.9519 ' .

Got to this drawing the line connecting up the other two opposite vertices on the hexagon, and knowing each of them will be 10 as well since the hexagon has equal angles, equal sides etc.

This splits the hexagon up into 6 triangles, which are equilateral triangles (equal sides, equal angles), because in the centre you have 360 degrees divided into 6 equal angles, and at the outside you have the 120 degree angles of the hexagon split in two. Since each full line through the hexagon was 10, then each triangle has side 5.

So we want to work out the area of an equilateral triangle of side 5. Different formula for doing this but "half base times height" is perhaps the simplest one. If the base is 5 you need to work out the height, which you can get using Pythatogas' Theorem or trig functions but you don't need those. Divide the triangle in half and consider the right angled triangle with one side 5/2 (half of the base 5) and hypotenuse 5. The height will be the other side, and Pythagoras' theorem gives us that (5/2)^2 + height^2 = 5^2, which ends up giving height^2 = 75/4. Taking the square root of this gives 5√3/2.

Anyway, this was all to work out the area of an equilateral triangle using "half base times height", which gives 1/2 * 5 * 5√3/2 = 25√3/2. There are 6 of these triangles, so multiply by 6 and you get 75√3 / 2.

Not something I knew a forumla for so quite interesting to check how to get it using basic geometry. No idea of the method I described is needlessly long and if there's a much easier way of looking at it, but tried to do it using minimal other formula and not needing to know the values of trig functions.

I blew up my screen size to 500% and still couldn't read that!   

Thought small size was the spoiler one here, in case mine was the first reply and others were thinking about it as well But I see yours was first so no need to be spoilered, and yeah mine pretty much what you said except more long winded.

[Lonk]

https://rechneronline.de/pi/hexagon.php

[qed]

https://rechneronline.de/pi/hexagon.php

Dodecahedron IS the answer.

[pg1067]

https://rechneronline.de/pi/hexagon.php

Dodecahedron IS the answer.

Disdyakis triacontahedron.

[Stadler]

https://rechneronline.de/pi/hexagon.php

Dodecahedron IS the answer.

Disdyakis triacontahedron.

It's getting hot in here... 

[qed]

https://rechneronline.de/pi/hexagon.php

Dodecahedron IS the answer.

Disdyakis triacontahedron.

[MrBoom_shack-a-lack]

Adam Savage builds a Rhombic Dodecahedron: https://www.youtube.com/watch?v=65r_1TzJXaQ

Really cool when you see the cube.

[Chino]

I need help calculating some cuts and I'm at a complete loss on how to figure this one out. I'll just give the high-level overview, and if it's something someone can help me with, I can get a little more granular with measurements.

 

So I'm building a 3lb combat robot, basically a miniature version of this:
 

Here is how it sits now:


I'm looking to wrap it in armor, but it turns out I absolutely suck at geometry at this level. It was evident pretty quick that me thinking all the cuts would be a uniform 45 degrees was not the case:


Basically, the angles that are flush with the top of the bot and the floor are going to be 45 degrees, but I have no idea how to calculate the angles for the corners (along multiple axis). Is this something that somebody here could school me on?

[Stadler]

Is that your kitchen/dining room table?  That's bad ass.

[Chino]

That's my dining room table. I love it. It belonged to the person who owned the house before me. At the closing, I negotiated that as part of the transaction. I really wanted it.

[Stadler]

They're dumb. 

So.... if I understand you correctly...  you're going to cut four trapezoids, with a top dimension equal to the length of the top square/rectangle piece, and a bottom dimension equal to... whatever the math says it is (I figure the top dimension is x, the height is h, and because you're working with 45^o angles, your base is x+2h). 

You want to know what the angle from the face to the back is on the sides, so that when you fit all four together, the edges are flush all around on the outside and the inside.

If height h=0, then the that angle - from the face to the edge - would be 90^o.  If height h was infinite (meaning the side trapezoids weren't trapezoids, but were rectangles), that angle would be 45^o.   So if you're tipping the trapezoids on a 45^o angle themselves, wouldn't the face to edge angle be halfway between 45 and 90, or 67.5^o?   I'm sure there's math to calculate that, but it's beyond me; you're probably better off taking a couple pieces of wood and test cutting them for fit.

(Holy crap, this would be so much easier in person or with a white board; trying to put this into words is a task in itself.)

[Stadler]

Here's the other thing you could do:   create one or two jigs that are a triangle with two sides of height h.   Gently glue one or both (depending on how long the pieces are) to the backside of your trapezoids, so that they sit, propped, at that 45^o angle.  Then cut the trapezoid at a 45^o angle from either base of the trapezoid, except the trapezoid itself will not lie flat on your saw, but will be propped at an angle.  Does that make sense?   I can make a makeshift video I think, with my phone if it doesn't. 

[Chino]

They're dumb. 

So.... if I understand you correctly...  you're going to cut four trapezoids, with a top dimension equal to the length of the top square/rectangle piece, and a bottom dimension equal to... whatever the math says it is (I figure the top dimension is x, the height is h, and because you're working with 45^o angles, your base is x+2h). 

You want to know what the angle from the face to the back is on the sides, so that when you fit all four together, the edges are flush all around on the outside and the inside.

If height h=0, then the that angle - from the face to the edge - would be 90^o.  If height h was infinite (meaning the side trapezoids weren't trapezoids, but were rectangles), that angle would be 45^o.   So if you're tipping the trapezoids on a 45^o angle themselves, wouldn't the face to edge angle be halfway between 45 and 90, or 67.5^o?   I'm sure there's math to calculate that, but it's beyond me; you're probably better off taking a couple pieces of wood and test cutting them for fit.

(Holy crap, this would be so much easier in person or with a white board; trying to put this into words is a task in itself.)

That's exactly what my thought process was, but when it came time to make some test pieces out of 1/4" plywood, they weren't anywhere close.

I hear you on verbalizing this. I'm really struggling. For each armor plate, there will need to be a total of six angles cut, and even if I get the match to check out, cutting them properly with my limited tools is going to be a whole other challenge. I'm going to tinker with it tonight and see if I can get anywhere or any more pics that might be helpful.

[Chino]

Here's the other thing you could do:   create one or two jigs that are a triangle with two sides of height h.   Gently glue one or both (depending on how long the pieces are) to the backside of your trapezoids, so that they sit, propped, at that 45^o angle.  Then cut the trapezoid at a 45^o angle from either base of the trapezoid, except the trapezoid itself will not lie flat on your saw, but will be propped at an angle.  Does that make sense?   I can make a makeshift video I think, with my phone if it doesn't.

I think I'm following. I'm not sure if that's really necessary because I have the ability to angle the blade on the table saw these are getting run through. Instead of having to jig/tilt the material to get my angle, I just angle the blade.

[Stadler]

Here's the other thing you could do:   create one or two jigs that are a triangle with two sides of height h.   Gently glue one or both (depending on how long the pieces are) to the backside of your trapezoids, so that they sit, propped, at that 45^o angle.  Then cut the trapezoid at a 45^o angle from either base of the trapezoid, except the trapezoid itself will not lie flat on your saw, but will be propped at an angle.  Does that make sense?   I can make a makeshift video I think, with my phone if it doesn't.

I think I'm following. I'm not sure if that's really necessary because I have the ability to angle the blade on the table saw these are getting run through. Instead of having to jig/tilt the material to get my angle, I just angle the blade.

But isn't that the trick, though, to know how much to angle the blade?   I'm using brute force here to empirically cut a piece that fits, and then you can measure the angle. Because the "devil" is in that "slant" of the side trapezoids.  When you look down from the top, all the angles in 2D are 45 degrees. So you're "faking" the saw into cutting the real angle by angling the trapezoid to what it would be in real life without knowing what the resulting angle will be. 

[Chino]

Here's the other thing you could do:   create one or two jigs that are a triangle with two sides of height h.   Gently glue one or both (depending on how long the pieces are) to the backside of your trapezoids, so that they sit, propped, at that 45^o angle.  Then cut the trapezoid at a 45^o angle from either base of the trapezoid, except the trapezoid itself will not lie flat on your saw, but will be propped at an angle.  Does that make sense?   I can make a makeshift video I think, with my phone if it doesn't.

I think I'm following. I'm not sure if that's really necessary because I have the ability to angle the blade on the table saw these are getting run through. Instead of having to jig/tilt the material to get my angle, I just angle the blade.

But isn't that the trick, though, to know how much to angle the blade?   I'm using brute force here to empirically cut a piece that fits, and then you can measure the angle. Because the "devil" is in that "slant" of the side trapezoids.  When you look down from the top, all the angles in 2D are 45 degrees. So you're "faking" the saw into cutting the real angle by angling the trapezoid to what it would be in real life without knowing what the resulting angle will be.

Oh I see. You're talking about a reverse engineering approach - get it to work and measure after. That's probably what I'm going to have to resort to.

[Stadler]

Yes, but get it to work without a ton of trial and error.

How thick are your final pieces going to be, do you know?  Let's say they are 1/8".   Get a 1/8" (actual, not nominal) thick piece of balsa or plywood, and cut the trapezoid, worrying ONLY about the top and bottom lengths, not the angles (yet).  So you'll end up with a trapezoid of top base "b" - which is the length of the sides of that top square plate - and bottom base "b+2h", where "h" is the distance from the bottom of your trapezoidal prism to the very top of that top square plate.   You can, if you want, cut the 45 degree angles of the top and bottom bases if you want, because you know those.    That leaves you with a trapezoid that is too "thick" on the interior side of each side edge, and that is what has to be angled.  It's that angle - from the front face to the back - that you don't know.   So you trick the saw by propping it up at a 45 degree angle from the plane of the saw deck, like it would be in the final assembly, and cut it at a 45 degree angle from either of the top or bottom base lines.

Let me know how thick your plates are; I might have some wood in the garage I can play with to get the angle.

(You've piqued the engineer in me; I want to know this now, not just for you, but to satisfy my own curiosity!).

[Chino]

Yes, but get it to work without a ton of trial and error.

How thick are your final pieces going to be, do you know?  Let's say they are 1/8".   Get a 1/8" (actual, not nominal) thick piece of balsa or plywood, and cut the trapezoid, worrying ONLY about the top and bottom lengths, not the angles (yet).  So you'll end up with a trapezoid of top base "b" - which is the length of the sides of that top square plate - and bottom base "b+2h", where "h" is the distance from the bottom of your trapezoidal prism to the very top of that top square plate.   You can, if you want, cut the 45 degree angles of the top and bottom bases if you want, because you know those.    That leaves you with a trapezoid that is too "thick" on the interior side of each side edge, and that is what has to be angled.  It's that angle - from the front face to the back - that you don't know.   So you trick the saw by propping it up at a 45 degree angle from the plane of the saw deck, like it would be in the final assembly, and cut it at a 45 degree angle from either of the top or bottom base lines.

Let me know how thick your plates are; I might have some wood in the garage I can play with to get the angle.

(You've piqued the engineer in me; I want to know this now, not just for you, but to satisfy my own curiosity!).

Armor is going to be 1/4" thick (final material will be HDPE). The offer is much appreciated. Hold off on firing up your tools though. I've got a few ideas I'm going to test out after work that might eliminate the need for two of the angles on each side. I can report back on how that works out.

[Stadler]

Roger that; here to help if you need it.