Trisecting an angle
 

One of the oldest of the impossibles is that it is impossible to trisect an angle using only a compass and straight edge. It is very easy to bisect an angle. It is possible to trisect an exact 60 degree angle. (I've done it, its easy) So would someone care to take a stab at explaining why you can't trisect angles other than exact multiples of 60 degrees?



Fusion

Here's what it says in one of my books......

1. If we can trisect any angle, then we can trisect a 60 degree angle.
2. If we can trisect a 60 degree angle, trig shows us that x^3-3x-1 must have a constructible solution.
3. If the cubic had a constructible solution, then it must have a rational solution.
4. Assume solution c/d where c and d have no common factors.
5. Plugging in and multiplying by d^3 gives us c^3-3cd^2-d^3=0.
6. Rewriting this one way, c(c^2-3d^2)=d^3, so c is a factor of d^3, but since c and d have no common factors, c=+-1.
7. Rewriting another way, we get d(3cd+d^2)=c^3, so similarly d=+-1.
8. We conclude that c/d must be +-1, and plugging both into the cubic fail to solve the equation.
9. Since all steps up to the beginning are (supposedly) valid, we must have erred in out assumption that we can trisect any angle.

So i guess you're thing about trisecting a 60 degree angle must not work. :?

He's talking about trisecting a 180 degree angle in exact 60 degree segments, not a 60 degree angle.

First of all, that's only 1 case. Secondly, in standard geometry angles are only defined 0<theta<180 non-inclusive. A "straight angle" of 180 degrees doesn't qualify as a real angle. Neither does a 0 degree angle (which I bet I can trisect ;) ).

Its been a lot of years since I did this but I was able to trisect a 60 degree angle. Off the top of my head though, I don't remember the exact steps. I do remember that I had to convert one of the arms to a straight line in the process. I don't think it would be hard to do again. It takes advantage of dividing a circle into six equal sections when the compass is first used to draw a circle, then the point is placed on the perimeter and an arc is drawn from the perimeter on one side through the center and to the perimeter on the other side.

Surprisingly, most people approach this problem from the wrong perspective. Draw an acute angle. Then take a compass with the point on the vertex and draw an arc connecting the sides of the angle. The problem is to trisect the arc.

Fusion

I think I know the answer, but I can't prove it :)

Draw angle @ 60deg, mark vertex as A, fix compass to a length 'x', mark off B and C. Connect B & C. Now we have equalateral triangle ABC.

Bisect line BC. The mid point of BC becomes the centre with circle of radius BC/2. With compass mark point D on the external arc from B, and similarly create point E, with compass on C.

Draw AE, and AD. AE and AD trisect BAC. Angle DBA = Angle EAC regardless of the value of angle BAC, but only trisects BAC when BAC = 60deg.

-- Craig

Trisecting an angle
 

You draw an acute angle, any angle 10 degrees, 20, 35 , doesn't matter.
Now you bisect the angle, as usual. Now you extend the bisect line far
enough to draw an exact invert of the angle. Now you bisect one of the
arms of the inverted angle. If you draw a line back to the initial vertex
of the initial angle, it will be exatly one third of the angle.
I can include a graphic example, but don't know how to do it.
Maybe you can tell me how ?
Wynand

[url=http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Trisecting_an_angle.html]This[/url] might help.

[QUOTE=Wynand]You draw an acute angle, any angle 10 degrees, 20, 35 , doesn't matter.
Now you bisect the angle, as usual. Now you extend the bisect line far
enough to draw an exact invert of the angle. Now you bisect one of the
arms of the inverted angle. If you draw a line back to the initial vertex
of the initial angle, it will be exatly one third of the angle.
I can include a graphic example, but don't know how to do it.
Maybe you can tell me how ?
Wynand[/QUOTE]

Sigh. Where is Underwood Dudley? We need him.

[QUOTE=Wynand]You draw an acute angle, any angle 10 degrees, 20, 35 , doesn't matter. Now you bisect the angle, as usual. Now you extend the bisect line far enough to draw an exact invert of the angle. Now you bisect one of the arms of the inverted angle. If you draw a line back to the initial vertex of the initial angle, it will be exatly one third of the angle.[/QUOTE]No, it won't.

Trisecting an angle.
 

[QUOTE=Kevin]Here's what it says in one of my books......

1. If we can trisect any angle, then we can trisect a 60 degree angle.
2. If we can trisect a 60 degree angle, trig shows us that x^3-3x-1 must have a constructable solution.
3. If the cubic had a constructable solution, then it must have a rational solution.
4. Assume solution c/d where c and d have no common factors.
5. Plugging in and multiplying by d^3 gives us c^3-3cd^2-d^3=0.
6. Rewriting this one way, c(c^2-3d^2)=d^3, so c is a factor of d^3, but since c and d have no common factors, c=+-1.
7. Rewriting another way, we get d(3cd+d^2)=c^3, so similarly d=+-1.
8. We conclude that c/d must be +-1, and plugging both into the cubic fail to solve the equation.
9. Since all steps up to the beginning are (supposedly) valid, we must have erred in out assumption that we can trisect any angle.

So i guess you're thing about trisecting a 60 degree angle must not work. :?[/QUOTE]

:smile:
Kevin your book is right. Note the ruler must be unmarked.
Trisection of an angle in general is impossible
.
The multiples of 60* can be trisected. 90* and multiples also can be trisected.
0* and 180* are very much angles.
Your cubic stems from the formula that cos theta (\0\) = g say and follows from the trig formula. in terms of \0\* divided by 3
Your proof stands for 60* and it is sufficient to exhibit only one angle that cannot be trisected, since, a valid 'general method' would have to cover every single example

Hey the URL given by ixfd64 is a good one. It says angle 27* can be trisected under the usual conditions. I have put on my thinking cap!
Mally :coffee:

[QUOTE=mfgoode]

Hey the URL given by ixfd64 is a good one. It says angle 27* can be trisected under the usual conditions. I have put on my thinking cap!
Mally :coffee:[/QUOTE]

This is trivial. Note that 4*27 = 108 and that 108 - 90 = 18.

We can construct a 108 degree angle because it is the interior angle
of a regular pentagon and 2^2^1 + 1 = 5 is prime. We can construct a
27 degree angle by bisecting it twice. We can construct an 18 degree
angle by subtracting 90 from 108. Thus, we can construct a 9 degree
angle.

All angles that are multiple of 3 degrees can be constructed since:

sin 3° = [tex]\large \frac {\sqrt {8-\sqrt{3}-\sqrt{15}-\sqrt{10-2*\sqrt{5}}}}{4}[/tex]

cos 3° = [tex]\large \frac {\sqrt {8+\sqrt{3}+\sqrt{15}+\sqrt{10-2*\sqrt{5}}}}{4}[/tex]

In general, the only angles that we can construct are multiple of [tex]\frac {3}{17*257*65537}[/tex] degrees, provided that there is not an unknown Fermat prime number.

In general, the only angles that we can construct are multiple of [tex]\frac {3}{17*257*65537}[/tex] degrees, provided that there is not an unknown Fermat prime number.

For example, the cosine of 360/17 degrees is:

[tex]\large \frac {\sqrt{15 + \sqrt{17} - \sqrt{2(17-\sqrt{17})} + \frac {\sqrt{2(34+6\sqrt{17}+\sqrt{2(17-\sqrt{17})} - \sqrt{34(17-\sqrt{17})} + 8\sqrt{2(17+\sqrt{17})}})}{2}}}{4}[/tex]

[QUOTE=alpertron]In general, the only angles that we can construct are multiple of [tex]\frac {3}{17*257*65537}[/tex] degrees, provided that there is not an unknown Fermat prime number.[/QUOTE]

The paragraph quoted above is incorrect, but I cannot edit my own messages, so I write the corrected paragraph below:

In general, the only angles that we can construct are multiple of [tex]\frac {3}{2^n*17*257*65537}[/tex] degrees, provided that there is not an unknown Fermat prime number

Trisecting an angle
 

[QUOTE=R.D. Silverman]This is trivial. Note that 4*27 = 108 and that 108 - 90 = 18.

We can construct a 108 degree angle because it is the interior angle
of a regular pentagon and 2^2^1 + 1 = 5 is prime. We can construct a
27 degree angle by bisecting it twice. We can construct an 18 degree
angle by subtracting 90 from 108. Thus, we can construct a 9 degree
angle.[/QUOTE]
:surrender Thank you Bob for the excellent explanation for not only the 27*angle but also the 9* angle.
Gauss aged 17 yrs. investigated the constructability of regular "p-gons" (polygons with p sides) where p is a prime number. He derived that only if p is a prime "Fermat number" that this is possible.
p = 2^ 2^n +1

Now the first Fermat numbers are 5 , 17 , 257 , 65537 . Euler factorised F(5) and hence proved that it is composite. No further F(n) primes have been found for n>4 so far. May I say that this is also a challenge for GIMPS?
I would include the prime 3 for n=0 . Why has this prime not been included?

Alperton: Thank you for expanding my horizon on this topic. I need to study it further before I can comment on it. :unsure:
As you say that all multiples of 3/expression are the number of angles that can be constructed then knowing the 9* angle can we construct a 3* angle?
Mally :coffee:

In order to actually construct the 3 degree angle, you can start from the well known pentagon and hexagon constructions. These have angles of 360°/5 = 72° and 360°/6 = 60° respectively. Subtract both angles and you get 12°. Bisect it twice and finally you get 12°/2^2 = 3°.

Trisecting an angle
 

:surprised Thank you Alperton. You have to spoon feed me!
Please explain the denominator of 3 in your previous post. I recognise the \Fermat numbers in it but I would imagine the angles will either be huge or very tiny.
Can a relationship be worked out for the number of sides in a regular polygon and the number of angles that can be constructed by unmarked ruler and compass? :confused:
Mally :coffee:

The heptadecagon can be constructed, see for example this [URL=http://en.wikipedia.org/wiki/Heptadecagon]Wikipedia article[/URL].

The angle between two succesive vertex is 360°/17.

Using the 3° angle constructed in my last post, just start drawing a 120-gon, where one of its vertex is a vertex of the 17-gon. You can draw other 16 120-gons using the other 16 vertex of the 17-gon.

Finally you have 120*17 = 2040 points in the circle. The angle between these points is 3°/17.

Since the 257- and 65537-gons are also constructible, using the same procedure you can construct an angle of 3°/(17*257*65537). Finally you can bisect this angle n times to have an angle of 3°/(2^n*17*257*65537)

Trisecting an angle
 

[QUOTE=alpertron]The heptadecagon can be constructed, see for example this [URL=http://en.wikipedia.org/wiki/Heptadecagon]Wikipedia article[/URL].
:bow:
Thank you for the URL. It was most enlightening. I knew it was possible according to Gauss but did not have a method. But it seems very difficult to divide into 17 parts as The stone mason hired to erect it on Gauss' tomb refused as it was too complicated.
I will try and tabulate the different angles starting from 3* angles up to 90* angles which can be constructed geometrically. Do you by chance have a ready reckoner for these constructions? It will be valuable information for the forum.
Mally :coffee:

[QUOTE=mfgoode]
I will try and tabulate the different angles starting from 3* angles up to 90* angles which can be constructed geometrically. Do you by chance have a ready reckoner for these constructions? It will be valuable information for the forum.
Mally :coffee:[/QUOTE]

I see nothing wrong with your doing so for your own education and
entertainment. In fact, I applaud such efforts.

However:

(1) Unless you restrict to angles that are an integral number of degrees, your
list will be infinite. (e.g. bisecting a 45 degree angle etc.)

(2) Caiming that the result will be "valuable information" is ludicrous beyond
words. This information is *useless*. And it is easily reconstructed.

[QUOTE=R.D. Silverman]Sigh. Where is Underwood Dudley? We need him.[/QUOTE]

I had to google for him (which lead immediately to wikipedia).
Author of /The Trisectors/, it appears.

Thanks Bob - my Christmas pressy wish list has grown by one!

Phil

[QUOTE=R.D. Silverman]I see nothing wrong with your doing so for your own education and
entertainment. In fact, I applaud such efforts.
...
(2) Caiming that the result will be "valuable information" is ludicrous beyond
words. This information is *useless*. And it is easily reconstructed.[/QUOTE]Which information?

I agree that a list of angles would be useless.

On the other hand, a nicely laid out exposition of [b]how[/b] the angles may be constructed with compass and unmarked straight edge would be quite educational for others to read and would be worth the effort of posting. I'd like to see a geometric construction of a 257-gon, for instance. I know it's possible but I've never seen such a construction. The analogous proof of constructibility of a 65537-gon may be too much of a challenge...

Don't you like seeing new and elegant proofs of a theorem that has been proved long before by other means?

It's possible that information about a "ready-reckoner" may also be useful to me if it leads me to a tool which I find useful and which is more convenient to use than, say bc(1) or Pari/gp.


Paul

[QUOTE=mfgoode]I will try and tabulate the different angles starting from 3* angles up to 90* angles which can be constructed geometrically. Do you by chance have a ready reckoner for these constructions? It will be valuable information for the forum.[/QUOTE]

After constructing the 3 degree angle it is trivial to construct any multiple of it.

Maybe you want the formulas for sines and cosines of multiples of 3 degrees.

sin 0º = cos 90º = 0

sin 3º = cos 87º = [tex]\sqrt{-\sqrt{5/128-\sqrt{5}/128}-\sqrt{15}/16-\sqrt{3}/16+1/2}[/tex]

sin 6º = cos 84º = [tex]\sqrt{15/32-3*\sqrt{5}/32}-\sqrt{5}/8-1/8[/tex]

sin 9º = cos 81º = [tex]-\sqrt{5/16-\sqrt{5}/16}+\sqrt{10}/8+\sqrt{2}/8[/tex]

sin 12º = cos 78º = [tex]\sqrt{\sqrt{5}/32+5/32}-\sqrt{15}/8+\sqrt{3}/8[/tex]

sin 15º = cos 75º = [tex]\sqrt{6}/4-\sqrt{2}/4[/tex]

sin 18º = cos 72º = [tex]\sqrt{5}/4-1/4[/tex]

sin 21º = cos 69º = [tex]\sqrt{-\sqrt{\sqrt{5}/128+5/128}-\sqrt{15}/16+\sqrt{3}/16+1/2}[/tex]

sin 24º = cos 66º = [tex]-\sqrt{5/32-\sqrt{5}/32}+\sqrt{15}/8+\sqrt{3}/8[/tex]

sin 27º = cos 63º = [tex]\sqrt{\sqrt{5}/16+5/16}-\sqrt{10}/8+\sqrt{2}/8[/tex]

sin 30º = cos 60º = [tex]1/2[/tex]

sin 33º = cos 57º = [tex]\sqrt{\sqrt{5/128-\sqrt{5}/128}-\sqrt{15}/16-\sqrt{3}/16+1/2}[/tex]

sin 36º = cos 54º = [tex]\sqrt{5/8-\sqrt{5}/8}[/tex]

sin 39º = cos 51º = [tex]\sqrt{-\sqrt{\sqrt{5}/128+5/128}+\sqrt{15}/16-\sqrt{3}/16[/tex]+1/2}

sin 42º = cos 48º = [tex]\sqrt{3*\sqrt{5}/32+15/32}-\sqrt{5}/8+1/8[/tex]

sin 45º = cos 45º = [tex]\sqrt{2}/2[/tex]

sin 48º = cos 42º = [tex]\sqrt{\sqrt{5}/32+5/32}+\sqrt{15}/8-\sqrt{3}/8[/tex]

sin 51º = cos 39º = [tex]\sqrt{\sqrt{\sqrt{5}/128+5/128}-\sqrt{15}/16+\sqrt{3}/16+1/2}[/tex]

sin 54º = cos 36º = [tex]\sqrt{5}/4+1/4[/tex]

sin 57º = cos 33º = [tex]\sqrt{-\sqrt{5/128-\sqrt{5}/128}+\sqrt{15}/16+\sqrt{3}/16+1/2}[/tex]

sin 60º = cos 30º = [tex]\sqrt{3}/2[/tex]

sin 63º = cos 27º = [tex]\sqrt{\sqrt{5}/16+5/16}+\sqrt{10}/8-\sqrt{2}/8[/tex]

sin 66º = cos 24º = [tex]\sqrt{15/32-3*\sqrt{5}/32}+\sqrt{5}/8+1/8[/tex]

sin 69º = cos 21º = [tex]\sqrt{\sqrt{\sqrt{5}/128+5/128}+\sqrt{15}/16-\sqrt{3}/16+1/2}[/tex]

sin 72º = cos 18º = [tex]\sqrt{\sqrt{5}/8+5/8}[/tex]

sin 75º = cos 15º = [tex]\sqrt{6}/4+\sqrt{2}/4[/tex]

sin 78º = cos 12º = [tex]\sqrt{3*\sqrt{5}/32+15/32}+\sqrt{5}/8-1/8[/tex]

sin 81º = cos 9º = [tex]\sqrt{5/16-\sqrt{5}/16}+\sqrt{10}/8+\sqrt{2}/8[/tex]

sin 84º = cos 6º = [tex]\sqrt{5/32-\sqrt{5}/32}+\sqrt{15}/8+\sqrt{3}/8[/tex]

sin 87º = cos 3º = [tex]\sqrt{\sqrt{5/128-\sqrt{5}/128}+\sqrt{15}/16+\sqrt{3}/16+1/2}[/tex]

sin 90º = cos 0º = 1

Trisecting an angle
 

[QUOTE=R.D. Silverman]I see nothing wrong with your doing so for your own education and
entertainment. In fact, I applaud such efforts.

However:

(1) Unless you restrict to angles that are an integral number of degrees, your
list will be infinite. (e.g. bisecting a 45 degree angle etc.)
[/QUOTE]]
:redface: Yes Bob, I agree that restricting the angles to integral numbers is a fine suggestion. Thanks for the tip..I appreciate your direct and straight from the shoulder pointers which are concise, precise and to the point

Quote Fatphil[I had to google for him (which lead immediately to wikipedia).Author of /The Trisectors/, it appears.
Thanks Bob - my Christmas pressy wish list has grown by one!
Phil ]
:smile: Following your example I checked on Dudley and found it interesting. I may also order one of his books

Quote Xilman [Don't you like seeing new and elegant proofs of a theorem that has been proved long before by other means?

It's possible that information about a "ready-reckoner" may also be useful to me if it leads me to a tool which I find useful and which is more convenient to use than, say bc(1) or Pari/gp.]
:unsure: Thanks Paul for the encouragement I will start work on this ready reckoner.
The fact is that all constructable numbers are algebraic.

Quote Alpertron: [Maybe you want the formulas for sines and cosines of multiples of 3 degrees.

sin 0º = cos 90º = 0

sin 3º = cos 87º = ]
:rolleyes: Thanks Alpertron for the table of sines and cosines.
Please give me an example say sin 3* angle as to how you arrive at the value
in the sq.rt.form?

:smile: I have come across a wonderful method to prove that a heptagon inscribed in a unit circle cannot be constructed by unmarked ruler and compass.
It is yours for the asking. Also you can come across it in "What is Mathematics?"
By Courant and Robbins
Please excuse my method of quoting as I still dont have the hang of it
Mally :coffee:

[QUOTE=mfgoode]]
Please give me an example say sin 3* angle as to how you arrive at the value
in the sq.rt.form?[/QUOTE]
In order to compute sin 3º we have to first compute sin 72º and cos 72º (equivalent to constructing the pentagon) and then sin 60º and cos 60º (equivalent to constructing the hexagon). Using the difference of angle formulas we can find sin (72º - 60º) = sin 12º and cos (72º - 60º) = cos 12º, and then using the formulas for halving angles twice we can find sin 3º and cos 3º.

First step: computing sin 72º and cos 72º:

We can start with the formula for multiple angles:

[tex]cos\, 5t\, =\, (cos\,t)^5\, -\, 10 (cos\, t)^3\, (sin\, t)^2\, +\, 5 (cos\, t) (sin\, t)^4[/tex].

When t = 18º we get cos 5t = cos 90º = 0

Let c = cos t, s = sin t.

[tex]\large 0 = c^5 - 10 c^3 s^2 + 5 cs^4[/tex]

c is not zero so we can divide both sides by c.

[tex]\large 0 = c^4 - 10 c^2 s^2 + 5 s^4[/tex]

Since (cos t)^2 + (sin t)^2 = 1: (1)

[tex]\large 0 = (1-s^2)^2 - 10 s^2 (1-s^2) + 5 s^4[/tex]

[tex]\large 0 = 1 - 12s^2 + 16s^4[/tex]

This is a biquadratic equation that gives four solutions, one of them is sin 18º = cos 72º = [tex](\sqrt 5 - 1)/4[/tex]

From (1) we get the value of sin 72º.

The cos 60º and sin 60º are even easier to compute.

Then we use the following identities to compute sin 12º and cos 12º:

sin (a+b) = sin a cos b + cos a sin b
cos (a+b) = cos a cos b - sin a sin b

Let a = 72º and b = -60º, and of course, cos (-a) = cos a and sin (-a) = -sin a.

Finally use the formulas:

[tex]\large \cos\left(\frac{x}{2}\right) = \pm\, \sqrt{\frac{1 + \cos(x)}{2}}[/tex]

[tex]\large \sin\left(\frac{x}{2}\right) = \pm\, \sqrt{\frac{1 - \cos(x)}{2}}[/tex]

twice to find first sin 6º and cos 6º and then sin 3º and cos 3º

Note that you can only get an approximation but using the same techniques that allow for 3 degrees divided by any power of 2 could be used to create other angles inbetween. 3/2+3/4=1.5+0.75=2.25 -> That's 2.25 degrees from combining the trisection of two different harmonics of 3 degrees. I use the word 'harmonics' here in the radio frequency way - k/n or k*n are both harmonics for k=constant and n=nth harmonic. Actually, the same techniques I've seen here for going from 12 degrees to 3 degrees are also used to determine interference from two radio stations. There's a reason that FM stations tend to be near 3-digit primes.

Imagine frequencies A and B of a pure carrier wave (no signal).
Interference examples:
A+B
A-B
A/6+B
mA+nB; m and n = +/- harmonics, A and B = carrier frequency/wavelength.
Note that wavelength and frequency are inversely related using that simple little constant 'c' but I'm sure most of the people reading this know that! :P Sorry if my sense of humor is dry.

It can be argued that since the carriers are waves, then you can use trigonometry to determine their interactions. Of course it does, since we're talking circles=cycles. If A's wavelength is shorter than B, then from A's persepective B is always lagged behind. That is, for every A you will never see a complete copy of B since A ends too fast. A way to draw this, is to assume that 360 degrees is A, and B's angle is greater then 360 degrees since it's slower.

Does anyone have any diagrams explaining this in a visual way?

Here's one I made real quick. :)

Trisecting an angle
 

[QUOTE=alpertron]In order to compute sin 3º we have to first compute sin 72º and cos 72º (equivalent to constructing the pentagon) and then sin 60º and cos 60º (equivalent to constructing the hexagon). Using the difference of angle formulas we can find sin (72º - 60º) = sin 12º and cos (72º - 60º) = cos 12º, and then using the formulas for halving angles twice we can find sin 3º and cos 3º. ....etc./ Unquote.
:bow:
Thanks Alpertron for the very valuable information. Im sorry I did not reply earlier.
Mally :coffee:

Trisecting an angle.
 

[QUOTE=nibble4bits]Note that you can only get an approximation but using the same techniques that allow for 3 degrees divided by any power of 2 could be used to create other angles in between. 3/2+3/4=1.5+0.75=2.25 -> That's 2.25 degrees from combining the trisection of two different harmonics of 3 degrees. I use the word 'harmonics' here in the radio frequency way - k/n or k*n are both harmonics for k=constant and n=nth harmonic. Actually, the same techniques I've seen here for going from 12 degrees to 3 degrees are also used to determine interference from two radio stations. There's a reason that FM stations tend to be near 3-digit primes. ...etc. Unquote/
:smile: Thank you Niddles4bits for your interest.

In simple language I understand you are saying that the process can be carried on indefinitely of halving a given angle.

Please read Silverman's post n. 21 point no. (1). Its useful to know the integral values of angles. Most angle values in cos and sin are approximate except for 0* and 2 pi and a few other particular trig signs like tan 45*.

The rest of your post though irrelevant is interesting. Its right up Alpertrons street and I leave it to him to answer.

Your point of FM stations tending to be near 3-digit primes is very interesting.
However the stations marked and which can be tuned into are mostly divisible by 3 on my FM radio.. Kindly clarify on this point

The thumb nail sketch resembles SHM of the beginning of a sinusoidal wave and if two waves are out of phase then interference can occur.

Regards the term 'c' its not as simple as it seems. I am labouring on a post for Xilman to get a thorough meaning of the term esp. in Relativity.
Mally :coffee:

If I took ANY binary number and used the technique I showed you, then yes adding a series of halves+4ths+8ths+16ths+...(a_n)/(2^n) would generate approximations.
k/f(x) -> k=integer, f(x)=sum of (a_0)+(a_1)/2+...(a_n)/(2^n)
This formula should work. :)

I was thinking of stations such as 92.9 and 94.3 which are close to each other - the closer ones are the ones I'd worry about. 98.1, 98.5, and 98.9 are also deliberately arranged that way also.

I showed angles A and B in my diagram as coinciding every 8th (or 9th depending on point of view) cycle. In reallity they would take a lot longer period of time to even come close to alignment. I was lazy - it's was easy to draw. :P

For most people working with radio waves, c can be considered constant. If I'm making an antenna for 20 MHz then I know I need a length of twice the wavelength. "All else being equal" applies here - most people need to just simplify the process to make it effecient. Yes, I know c as a constant is misleading when you take account of nonuniform frames of reference.

<EDIT> Actually an even simpler formula is k*f(x)*2^-n where k and f(x) are integers for a constant and a sum, and n is the number of times to halve that constant.

By the way, I found that the polynomial

[tex]65536X^{16}-262144X^{14}+430080X^{12}-372736X^{10}+182784X^8-50176X^6+7040X^4-384X^2+1[/tex]

has the roots sin 3°, sin 21°, sin 33°, sin 39°, sin 51°, sin 57°, sin 69°, sin 87° and their negatives.

In binary:
int(log(101010110100001010101011110101))=29 in decimal=11101

101010110100001010101011110101/(10^11101)=
101010110100001010101011110101/100000000000000000000000000000=
1.01010110100001010101011110101=~1.33797214366495609283447265625

To get the original string '101010110100001010101011110101' from '1.33797214366495609283447265625' is not too hard. The value of k can be constant=1 or whatever you need to estimate your solution. Too bad that any arbitary increase in binary digits will just end up making it slightly less wrong. :/ The point is not to make a table, but to make an algorythm that creates tables on the fly.

Trisecting an angle
 

[QUOTE=alpertron]By the way, I found that the polynomial

[tex]65536X^{16}-262144X^{14}+430080X^{12}-372736X^{10}+182784X^8-50176X^6+7040X^4-384X^2+1[/tex]

has the roots sin 3°, sin 21°, sin 33°, sin 39°, sin 51°, sin 57°, sin 69°, sin 87° and their negatives.[/QUOTE]
:bow:This is remarkable, and excellent work Alpertron
This is a characteristic polynomial of degree 16 with eigenvalues sin3*,sin21*etc.
If you could give us the 16 x 16 matrix and trace of the matrix it will be very valuable. Thank you.
Mally :coffee:

Trisecting an angle
 

[QUOTE=nibble4bits]If I took ANY binary number and used the technique I showed you, then yes adding a series of halves+4ths+8ths+16ths+...(a_n)/(2^n) would generate approximations.
k/f(x) -> k=integer, f(x)=sum of (a_0)+(a_1)/2+...(a_n)/(2^n)
This formula should work. :)

I was thinking of stations such as 92.9 and 94.3 which are close to each other - the closer ones are the ones I'd worry about. 98.1, 98.5, and 98.9 are also deliberately arranged that way also.

I showed angles A and B in my diagram as coinciding every 8th (or 9th depending on point of view) cycle. In reallity they would take a lot longer period of time to even come close to alignment. I was lazy - it's was easy to draw. :P

For most people working with radio waves, c can be considered constant. If I'm making an antenna for 20 MHz then I know I need a length of twice the wavelength. "All else being equal" applies here - most people need to just simplify the process to make it effecient. Yes, I know c as a constant is misleading when you take account of nonuniform frames of reference.

<EDIT> Actually an even simpler formula is k*f(x)*2^-n where k and f(x) are integers for a constant and a sum, and n is the number of times to halve that constant.[/QUOTE]
:smile: Quote/ "Yes, I know c as a constant is misleading when you take account of nonuniform frames of reference." /unquote
'c' is a constant regardless of the frame of reference uniform or non-uniform.
Mally :coffee:

[QUOTE=mfgoode]:bow:This is remarkable, and excellent work Alpertron
This is a characteristic polynomial of degree 16 with eigenvalues sin3*,sin21*etc.
If you could give us the 16 x 16 matrix and trace of the matrix it will be very valuable. Thank you.
Mally :coffee:[/QUOTE]

Thanks Mally, but there is no need for using matrices here.

If we want to find a polynomial that has a root cos 3°, just use the relation:

cos (20 * 3°) = cos 60° = 1/2

If x = cos 3°, we find that T[sub]20[/sub](x) = 1/2 where T[sub]n[/sub] is the [URL=http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html]Chebyshev polynomial[/URL].

Thus p(x) = 2*T[sub]20[/sub](x) - 1 is a degree 20 polynomial that has the requested root.

We can notice that cos(20 * 15°) = cos(20 * 75°) = 1/2.

The four numbers cos(15°), cos (-75°) and their negatives are roots of the polynomial q(x) = 16x[sup]4[/sup] - 16x[sup]2[/sup] + 1. The polynomial I wrote in my last post is just p(x) / q(x).

Trisecting an angle
 

[QUOTE=alpertron]Thanks Mally, but there is no need for using matrices here.

If we want to find a polynomial that has a root cos 3°, just use the relation:

cos (20 * 3°) = cos 60° = 1/2

If x = cos 3°, we find that T[sub]20[/sub](x) = 1/2 where T[sub]n[/sub] is the [URL=http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html]Chebyshev polynomial[/URL]./Unquote
:bow:
Thank you very much Alpertron, for introducing me to the wonders of the Chebyshev polynomials.
Kindly bear with me, and tolerate my questions, so that I can get a better idea of its properties and please clarify.
Now given the first few Chebyshev polynomials of the first kind, I find that the ‘first’ coefficients are powers of 2 such that say T (6) =2^(6-1) -…. Giving 32.
I also find a relationship to successive terms as an integration of the previous terms thus
T3 (x) = 4* x^3 -3*x +
T4 (x) = 8* x^4 -20x^2 +1.
Kindly explain to me how exactly the coefficients are obtained? Such as the ‘2nds’ 3 , 20, etc.
Is there an elementary rule for obtaining these coefficients or does the process require a direct integration of the Contour Integral?
In other words given T (n) can we find T (n+1) or T (n-1) ? by elementary integration or have we to integrate the Contour Integral every time? And follow a simple rule for the coefficients? Sloane's A008310 did not have your coefficients and I couldnt open it further.
Thank you,
Mally :coffee:

Using identity (21): T[sub]n+1[/sub] (x) = 2 x T[sub]n[/sub] (x) - T[sub]n-1[/sub] (x) I coded the following program in UBASIC (this interpreted language can operate with polynomials):

[CODE]
10 input "Degree";N
20 PrevPoly=1
30 ActualPoly=_X
40 for I=2 to N
50 NewPoly=2*_X*ActualPoly-PrevPoly
60 PrevPoly=ActualPoly
70 ActualPoly=NewPoly
80 next I
90 print ActualPoly
[/CODE]

In our example, if I enter 20 I get:

524288*X^20 - 2621440*X^18 + 5570560*X^16 - 6553600*X^14 + 4659200*X^12 - 2050048*X^10 + 549120*X^8 - 84480*X^6 + 6600*X^4 - 200*X^2 + 1

As an exercise for you, you could try to find a 48-degree polynomial that has the root sin 1° among others.

[QUOTE=mfgoode]
'c' is a constant regardless of the frame of reference uniform or non-uniform.
[/QUOTE]

Actually what I was refering to is what if I told most people that several galaxies are moving away from us at 60% of light speed. They'de think that the any two in opposite directions must be moving away from each other at 1.2c. Talking about this sort of stuff is disgusting, LOL. :yucky:

Trisecting an angle
 

[QUOTE=alpertron]Using identity (21): T[sub]n+1[/sub] (x) = 2 x T[sub]n[/sub] (x) - T[sub]n-1[/sub] (x) I coded the following program in UBASIC (this interpreted language can operate with polynomials):

[CODE]
10 input "Degree";N
20 PrevPoly=1
30 ActualPoly=_X
40 for I=2 to N
50 NewPoly=2*_X*ActualPoly-PrevPoly
60 PrevPoly=ActualPoly
70 ActualPoly=NewPoly
80 next I
90 print ActualPoly
[/CODE]

In our example, if I enter 20 I get:

524288*X^20 - 2621440*X^18 + 5570560*X^16 - 6553600*X^14 + 4659200*X^12 - 2050048*X^10 + 549120*X^8 - 84480*X^6 + 6600*X^4 - 200*X^2 + 1

As an exercise for you, you could try to find a 48-degree polynomial that has the root sin 1° among others.[/QUOTE]
:surprised Thanks Alpertron for identity (21).
I have to confess that I'm nil in programming and its a great drawback for me.
However I must get some one knowledgeable here who can work it out for me.
Mally :coffee:

Trisecting an angle
 

[QUOTE=nibble4bits]Actually what I was refering to is what if I told most people that several galaxies are moving away from us at 60% of light speed. They'de think that the any two in opposite directions must be moving away from each other at 1.2c. Talking about this sort of stuff is disgusting, LOL. :yucky:[/QUOTE]
:yawn:
Come to think of it nibble4bits how would you explain it to the people, never mind what they think.! Id like to hear from you. Can anyone explain this to me?
Mally :coffee:

[QUOTE=mfgoode]:yawn:
Come to think of it nibble4bits how would you explain it to the people, never mind what they think.! Id like to hear from you. Can anyone explain this to me?
Mally :coffee:[/QUOTE]
Here's what I say when this comes up:

The Michelson-Morley experiments show that the universe is a stanger place than was expected. When you try to figure out a way for the universe to behave consistently with this experimental observation, this is what you end up with.

[QUOTE=nibble4bits]Actually what I was refering to is what if I told most people that several galaxies are moving away from us at 60% of light speed. They'de think that the any two in opposite directions must be moving away from each other at 1.2c.[/QUOTE]Well, they [i]are[/i] moving away from each other at 1.2c, in [i]our[/i] frame of reference.

But they're not moving away from each other at 1.2c in either of their own frames of reference.

In a manner of speaking yes I think they are as no light from one ever reaches the other. I'm no expert so anyone can explain why I am wrong I'd appreciate it.

[QUOTE=garo]In a manner of speaking yes I think they are as no light from one ever reaches the other. I'm no expert so anyone can explain why I am wrong I'd appreciate it.[/QUOTE]As far as observers in each galaxy are concerned, the other galaxy is travelling at less than the speed of light so they do indeed see light from the other galaxy.

Try a thought experiment: forget about the galaxies for a moment and think only of a couple of observers on a trajectory such that they only just miss each other, together with a third observer right next to the point of closest approach. The third observer first sees the other two approaching at 0.6c and then, after they pass, receding at 0.6c in opposite directions.

What do the other two see? They first see their opposite number approaching at less than the speed of light. For a brief instant, light from one travels to the other at right angles to the direction of flight. Then they recede from each other at less than the speed of light.


Paul

[QUOTE=cheesehead]Well, they [i]are[/i] moving away from each other at 1.2c, in [i]our[/i] frame of reference.

But they're not moving away from each other at 1.2c in either of their own frames of reference.[/QUOTE]
You are right because the time in those galaxies runs [b]slower[/b] compared with our frame of reference. Thus their relative speed is less than c.

Without going that far, the people who went to the Moon "gained" about 1/10000th of a second (considering that you need five days to reach the Moon, it is not too much).

[QUOTE=garo]In a manner of speaking yes I think they are as no light from one ever reaches the other. I'm no expert so anyone can explain why I am wrong I'd appreciate it.[/QUOTE]

Two people receding from an inertial observer who measures their velocities to be +/-0.6c along a fixed axis would observe each other's velocities to be +/-0.88c along the same axis. Light from one would reach the other one, would be reflected, and then would reach the originator.

This is all derivable just using high-school maths (in fact, maths that a bright 10 year old could grasp) as long as you're prepared to accept a few assumptions. Basically, assume the speed of light is constant, and that distance is measured by bouncing light off the object, measuring how long the reflection takes to return, and scaling that by the known constant speed of light.

In this particular example, draw the space/time cone for the inertial observer, the two lines with gradients +/-0.6 (OK, I've got the axes oriented unconventially) corresponding to the two 'moving' objects. At an arbitrary point in time, send a light beam (line with gradient +/-1) from one object to the other, and when it arrives, bounce it back (line with gradient -/+1). Notice that the originator does indeed get his response. Conclusion - they are not receding from each other at a superluminal speed.

Get some graph paper, and draw it - it's very enlightening.

I think I misunderstood what the OP was saying. I understand how relativistic mechanics will show that the galaxies recede at 0.88c in each others frames of reference.

I thought it had to with the stretching of space-time. There are galaxies that are receding from us at faster than the speed of light. No light from them will ever reach us. Right?

[QUOTE=garo]I think I misunderstood what the OP was saying. I understand how relativistic mechanics will show that the galaxies recede at 0.88c in each others frames of reference.

I thought it had to with the stretching of space-time. There are galaxies that are receding from us at faster than the speed of light. No light from them will ever reach us. Right?[/QUOTE]Correct, on the assumption that such galaxies exist.


Paul

Stipulating that FTL galaxies do exist, this would predict that we should be able to detect galaxies receding from us at a broad range of fractions of C. This could only occur as a result of distortion from space time expansion. Why do we not detect such a range especially approaching C? The local group of galaxies is essentially moving in tandem. More distant galaxies should progressively be moving away from us at faster and faster speeds up to and including the point at which they become invisible because they are receding from us at the speed of light.

This should provoke a few comments! We're pretty far off topic though.

Fusion

[QUOTE=garo & xilman]
[B]g:[/B] There are galaxies that are receding from us at faster than the speed of light. No light from them will ever reach us. Right?
[B]x:[/B] Correct, on the assumption that such galaxies exist.[/QUOTE]

Could you say some more about this? I thought that the hypothesized galaxies would be at some distance "d" from us right "now," and that light leaving such a galaxy would travel towards us at constant speed c, reaching us in due course. Where did I go wrong?

[QUOTE=wblipp]Could you say some more about this? I thought that the hypothesized galaxies would be at some distance "d" from us right "now," and that light leaving such a galaxy would travel towards us at constant speed c, reaching us in due course. Where did I go wrong?[/QUOTE]The problem is that the spacetime between us stretches while the photon is in transit. If the expansion of spacetime is greater than c, the photon never reaches us. The distance between us is not a constant and so t=d/c is not the time a photon takes to reach us.

BTW, what do you mean by "now"? Simultaneity of spatially separated events is not well defined in relativity.


Paul

[QUOTE=xilman]BTW, what do you mean by "now"? Simultaneity of spatially separated events is not well defined in relativity.
Paul[/QUOTE]
Let me rephrase that. Simultaneity is well defined, but the observed temporal ordering of two acausally connected events depends on the observer.


Paul

[QUOTE=xilman]The problem is that the spacetime between us stretches while the photon is in transit.
...
BTW, what do you mean by "now"?[/QUOTE]
I guess the part I missed must be the logic that says the only way for a galaxy to be receeding from us that fast is to be this far away. I thought we were dropping some of the reality constraints, and I didn't see any reason why we wouldn't see light from a nearby galaxy merely because it was receeding at faster than light speed.

I knew "now" was problematical - that's why I quoted it.

[QUOTE=Fusion_power]we should be able to detect galaxies receding from us at a broad range of fractions of C. This could only occur as a result of distortion from space time expansion. Why do we not detect such a range especially approaching C?[/quote]
But we [i]do[/i] observe galaxies receding from us at almost all fractions of c. Recently I've seen announcements of galaxies with measured redshifts greater than 6. A redshift (AKA z-parameter) of 6 corresponds (assuming the redshift is all due to the Doppler effect) to a recession velocity of 0.96c.

See [url="http://hyperphysics.phy-astr.gsu.edu/hbase/astro/redshf2.html"]http://hyperphysics.phy-astr.gsu.edu/hbase/astro/redshf2.html[/url] for the calculation formulas.

See [url="http://www.journals.uchicago.edu/ApJ/journal/issues/ApJ/v578n2/16370/16370.text.html?erFrom=-6724382198883408314Guest"]http://www.journals.uchicago.edu/ApJ/journal/issues/ApJ/v578n2/16370/16370.text.html?erFrom=-6724382198883408314Guest[/url] for a May 2002 article about a z=6.28 quasar.

Trisecting an angle
 

:surprised :question:
So, what happens when a body recedes at 100% of the speed of light?

infinity.

Therefore (1 + z) will be equal to infinity and so will z be equal to infinity.

No body can therefore recede at a speed equal to the speed of light. This is another way of saying that a body at the observable edge of the universe is invisible.
[url]http://www.aqua.co.za/assa_jhb/Canopus/c999evz.htm[/url]
Mally :coffee:

But, Mally, I need to add a correction to my latest posting.

The equations referenced there are valid in a flat universe with cosmological constant = 0.0, if I understand correctly. Since observations (of high-redshift supernovae) have shown that our universe doesn't satisfy those conditions, there need to be corrections to those equations.

Quoting from two other sources:

[quote=http://www.astro.ucla.edu/~wright/doppler.htm]When z is larger than 1 then cz is faster than the speed of light and, while recession velocities faster than light are allowed, this approximation using cz as the recession velocity of an object is no longer valid. Thus for the largest known redshift of z=6.3, the recession velocity is not 6.3*c = 1,890,000 km/sec. It is also not the 285,254 km/sec given by the special relativistic Doppler formula 1+z = sqrt((1+v/c)/(1-v/c)). The actual recession velocity for this object depends on the cosmological parameters, but for an OmegaM=0.3 vacuum-dominated flat model the velocity is 585,611 km/sec. [/quote](So z=6.3 corresponds to an actual recession velocity of 1.95c !)

[quote=http://www.astro.ucla.edu/~wright/cosmology_faq.html#FTL]
[b]Can objects move away from us faster than the speed of light?[/b]

Again, this is a question that depends on which of the <A href="http://www.astro.ucla.edu/~wright/cosmo_02.htm#MD">[color=#0000ff]many distance definitions[/color] one uses. However, if we assume that the distance of an object at time [i]t[/i] is the distance from our position at time [i]t[/i] to the object's position at time [i]t[/i] measured by a set of observers moving with the expansion of the Universe, and all making their observations when they see the Universe as having age [i]t[/i], then the velocity (change in [i]D[/i] per change in [i]t[/i]) can definitely be larger than the speed of light. This is not a contradiction of special [url="http://www.astro.ucla.edu/~wright/relatvty.htm"][color=#0000ff]relativity[/color][/url] because this distance is not the same as the spatial distance used in SR, and the age of the Universe is not the same as the time used in SR. In the special case of the empty Universe, where one can show the model in both [url="http://www.astro.ucla.edu/~wright/cosmo_02.htm#MD"][color=#0000ff]special relativistic and cosmological coordinates[/color][/url], the velocity defined by change in cosmological distance per unit cosmic time is given by [i]v = c ln(1+z), [/i]where [i]z[/i] is the [url="http://www.astro.ucla.edu/~wright/cosmology_faq.html#z"][color=#800080]redshift[/color][/url], which clearly goes to [i]infinity[/i] as the redshift goes to infinity, and is larger than c for [i]z > 1.718[/i]. For the critical density Universe, this velocity is given by [i]v = 2c[1-(1+z)-0.5] [/i]which is larger than c for [i]z > 3 [/i].

For the concordance model based on CMB [Cosmic Microwave Background - cheesehead] [i]data and the [url="http://www.astro.ucla.edu/~wright/cosmology_faq.html#CC"][color=#800080]acceleration[/color][/url] of the expansion measured using supernovae, a [url="http://www.astro.ucla.edu/~wright/cosmology_faq.html#FLAT"][color=#800080]flat Universe[/color][/url] with [url="http://www.astro.ucla.edu/~wright/glossary.html#Omega"][color=#0000ff]Omega[sub]M[/color][/url][/sub] = 0.27, the velocity is greater than c for z > 1.407[/i].[/quote]

Do you mean that in order to trisect an angle one has to work faster than the speed of light? :grin:

I was thinking of splitting this thread but after ppo's last post decided against it :razz:

[QUOTE=garo]I was thinking of splitting this thread but after ppo's last post decided against it.[/QUOTE]

I didn't know that you had that power.

Actually, I had the same thought. I felt that most of the discussion was "off topic" and belonged in a separate thread (probably in some other part of the forum).

But ppo put us straight!

[QUOTE=Wacky]I didn't know that you had that power.

Actually, I had the same thought. I felt that most of the discussion was "off topic" and belonged in a separate thread (probably in some other part of the forum).

But ppo put us straight![/QUOTE]

Heh! Super-mod :banana:

Actually, I don't care very much if ppo's post manages to bring us back on topic. But the way the thread has evolved it would be a shame to split it. The sum of the two parts would certainly be less than the whole.

Trisecting an angle
 

[QUOTE=ppo]Do you mean that in order to trisect an angle one has to work faster than the speed of light? :grin:[/QUOTE]
:smile:
You don't have too! If you open a scissors fast enough (Finite speed) the point of separation of the two blades can move at the speed or greater than the speed of light (Feynman). :surprised
In the good old days we used to call it a 'virtual' velocity. Today anything 'virtual' is connected with cyber- games et al.
To reiterate a mass less object can travel faster than light speed. There is proof of the existence of tachyons which are always faster than the speed of light but never come below the speed light (c). Theoretically this is possible and compatible with both the theories of Relativity.
As a simple exercise imagine a mirror at a distance 'd' from an object. If the mirror (and not the object) is moved at even half the speed of light away from the object, the image will be moving at 'c' Over half 'C' of the mirror the image moves faster than 'c'
If you cant fathom it out ask a high school student to work it out for you :grin:
Mally :coffee:

Trisecting an angle
 

:smile:
How do you add velocities in Special Relativity?
Check this out.
[URL]http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html[URL]
Mally :coffee:

[QUOTE=mfgoode]:smile:
You don't have too! If you open a scissors fast enough (Finite speed) the point of separation of the two blades can move at the speed or greater than the speed of light (Feynman). :surprised
In the good old days we used to call it a 'virtual' velocity. Today anything 'virtual' is connected with cyber- games et al.
To reiterate a mass less object can travel faster than light speed. There is proof of the existence of tachyons which are always faster than the speed of light but never come below the speed light (c). Theoretically this is possible and compatible with both the theories of Relativity.
As a simple exercise imagine a mirror at a distance 'd' from an object. If the mirror (and not the object) is moved at even half the speed of light away from the object, the image will be moving at 'c' Over half 'C' of the mirror the image moves faster than 'c'
If you cant fathom it out ask a high school student to work it out for you :grin:
Mally :coffee:[/QUOTE]
True, but not very useful.

What relativity *really* says is not that things can't travel faster than light but that information can not be conveyed from place to place at faster than a constant value c. Experiment shows, to extremely high accuracy, that light travels at c in vacuo. There's nothing particularly special about light. It is not the only phenomenon with this behaviour. Physicists would be seriously worried (and very excited) if gravitational waves didn't also travel at c for instance. Pump enough energy into just about anything and you can make it travel as close to c as you wish.

None of the phenomena you give above transfer information from point to point along the path taken.

Paul

Mally,

I have to call you on the "proof of the existence of tachyons" statement above. The last I knew of this, there was theoretical probability but no absolute proof. Do you have something that would prove your statement?

What of quantum entangled particles? Does the event propagate at the speed of light when one of the particles is destroyed? What happens to the energy imbued in an entangled particle when its mate is destroyed?

Fusion

Trisecting an angle
 

[QUOTE=xilman]True, but not very useful.

What relativity *really* says is not that things can't travel faster than light but that information can not be conveyed from place to place at faster than a constant value c. Experiment shows, to extremely high accuracy, that light travels at c in vacuo. There's nothing particularly special about light. It is not the only phenomenon with this behaviour. Physicists would be seriously worried (and very excited) if gravitational waves didn't also travel at c for instance. Pump enough energy into just about anything and you can make it travel as close to c as you wish.

None of the phenomena you give above transfer information from point to point along the path taken.

Paul[/QUOTE]
:rolleyes:
'True but not very useful'.
Well how useful was it conquering Everest? or building the pyramids or Solomons Temple? Because its in the mind of Man and the collective consciousness. The Spirit of man will not die!

I think that 'things' more scientifically should be called mass-less objects or 'minus mass' objects/particles.

I agree that information cannot be transferred faster than c as it is maintained that the 'limiting velocity of interaction' in the universe is c.
Today c is defined by the standard meter and not the other way around.

The particularity of c is that all electromagnetic radiation travels at c. This was theorised by Faraday and Maxwell long before the Special Theory was formulated and due to this 'find' that Lorentz developed his transformation.
At the time one could very well have said 'Not very useful, so what?'

Pumping energy into mass to reach c requires an infinite quantity as the mass will tend to infinity as it approaches c using the B (beta) formula.

Still infinite forces are real as can be seen in several mechanical linkages.
The 'side rod' of a steam engine provides an infinite force for a split second and that's why journals reach metal fatigue and break.
There is every possibility that g-waves travel at c as also postulated by Einstein.
LIGO claimed that they have detected g-waves between observations some 2000 miles apart. :surprised
Mally :coffee:

Trisecting an angle
 

[QUOTE=Fusion_power]Mally,

I have to call you on the "proof of the existence of tachyons" statement above. The last I knew of this, there was theoretical probability but no absolute proof. Do you have something that would prove your statement?

What of quantum entangled particles? Does the event propagate at the speed of light when one of the particles is destroyed? What happens to the energy imbued in an entangled particle when its mate is destroyed?

Fusion[/QUOTE]
:smile:
[url]http://physics.gmu.edu/~e-physics/bob/tachyons.htm[/url]
Tachyons are postulated to be neutrinos and there is firm evidence of these.
Many particles (almost all ) are first theorised and then found in the labs and not the other way around.

Robert Ehrlich, a George Mason University professor of physics, claims to have possible experimental evidence for the existence of tachyons, hypothetical faster-than-light subatomic particles first proposed in 1962 by Bilaniuk, Deshpande, and Sudarshan. The evidence, published in several articles in the Physical Review D during June and October 1999 consists of an explanation of certain properties of the high energy cosmic rays bombarding the Earth from space. Interestingly, such faster-than-light particles seem to be required by current 12 dimensional theories developed by physicists to unify all the forces of nature.
The answer to your question on decay of a particle is in The URL I have given.
Mally :coffee:

[QUOTE=mfgoode]Today c is defined by the standard meter and not the other way around.[/QUOTE]
It is defined the other way. Look at its definition at [URL=http://en.wikipedia.org/wiki/Meter]Wikipedia[/URL]: It is defined since 1983 as the length of the path travelled by light in absolute vacuum during a time interval of 1/299,792,458 of a second.

The second is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom.

A long time has passed from the meter defined as the length of the platinum-iridum bar located in France.

[QUOTE=mfgoode]

The 'side rod' of a steam engine provides an infinite force for a split second
[/QUOTE]
:no: breakdown by metal fatigue is reached well before the force applied is infinite. Perhaps Paul, being a chemist, can explain better than me what happens to the molecular bonds under stress.
You are probably using the wrong assumption that the rigidity of the rod is infinite.

Trisecting an angle
 

:smile:
For your benefit I c+p portion of my thread again.
"Still infinite forces are real as can be seen in several mechanical linkages.
The 'side rod' of a steam engine provides an infinite force for a split second and that's why journals reach metal fatigue and break."

Perhaps my statement needs further explanation. The journals and axles don't break all at one go of the infinite Milli/ nano second force but gradually on each revolution at a particular position when the 'dead centre' is reached.
Repeated administration of this force gives metal fatigue in the long run.

Ever seen a stone crusher at work and the tremendous force generated to crush even large stones at one go? This is due to the near infinite force at work when the linkage is straight and the angle of the linkage is pi and the force is perpendicular at angle pi/2 when F/cos pi/2 gives % (infinity)
No ! No rod can be so 'rigid' as to have an infinite 'rigidity'. If you can discover one Ill hire you as a consultant to improve on something as simple like a sledge hammer. We are talking Strength of Materials here and not Chemistry'!
I am sure Paul can hold his own explanations/ and replies without you advertising for him. I am well aware of his capabilities, that's why I bothered to post as he is one of the most straight forward and intelligent posters in this forum. But if you feed rubbish into a computer what do you expect but rubbish out of it!
Mally :coffee:

Trisecting an angle
 

[QUOTE=alpertron]It is defined the other way. Look at its definition at [URL=http://en.wikipedia.org/wiki/Meter]Wikipedia[/URL]: It is defined since 1983 as the length of the path travelled by light in absolute vacuum during a time interval of 1/299,792,458 of a second.

The second is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom.

A long time has passed from the meter defined as the length of the platinum-iridum bar located in France.[/QUOTE]
:redface:
Thank you Alpertron. I stand corrected. I have put the cart before the horse!
Mally :coffee:

OK so what does this have to do with triangles, again? Oh yeah, both concepts are of geometry. Other then that they're about as related as a keyboard to a mouse. The link's there but not direct or very causal. The both depend on a central way of thinking (digital gates instead of lines/distances and angles).

[QUOTE=mfgoode]
the angle of the linkage is pi and the force is perpendicular at angle pi/2 when F/cos pi/2 gives % (infinity)[/QUOTE]
:ermm: You should not divide, you should multiply.
[QUOTE] But if you feed rubbish into a computer what do you expect but rubbish out of it!:[/QUOTE]
Being Xmas eve, I dont get the meaninig of this; would you, please, be so kind to explain ?

What's funny is that knowing how geometry works, there's a physical law caused by the trisecting rule. :P Not likely to be important but it's still possible there's an effect since this is a level of abstraction (lines and compass with no rulers).

Trisecting an angle
 

:unsure:
Im not quite sure what you mean in your post but its an interesting observation. Yes the proof of the impossibility of trisecting an angle, in general, by UNMARKED ruler and compass follows a mathematical law.
Its the unsolvability of the cubic equation x^3 -3x -1 = 0 in rational roots.
Please read Kevin's post no.2
Mally :coffee: