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Degree Distasnce


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#18 Rincewind

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Posted 25 April 2005 - 12:28 AM

You got it just backwards, Clay. The distance between each angular degree on the circumference is given by radius/radian (as the quickest result).

Can you give us an example calculation showing how the two versions of the formula work, please C1ay and Robust?

Let's say using a radius of 7 cm and an angle of 0.48 rad (27.502°).

#19 C1ay

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Posted 25 April 2005 - 06:48 AM

Can you give us an example calculation showing how the two versions of the formula work, please C1ay and Robust?

Let's say using a radius of 7 cm and an angle of 0.48 rad (27.502°).

Sure rince, 7*.48 = 3.36. Here's a chart I threw together to show the distance for each 10 degree marker around a unit circle, r=1.

Degrees	Radians	r*rad	r/radian
10	 0.1745	 0.1745	 5.7296
20	 0.3491	 0.3491	 2.8648
30	 0.5236	 0.5236	 1.9099
40	 0.6981	 0.6981	 1.4324
50	 0.8727	 0.8727	 1.1459
60	 1.0472	 1.0472	 0.9549
70	 1.2217	 1.2217	 0.8185
80	 1.3963	 1.3963	 0.7162
90 	 1.5708	 1.5708	 0.6366
100	1.7453	1.7453	0.5730
110	1.9199	1.9199	0.5209
120	2.0944	2.0944	0.4775
130	2.2689	2.2689	0.4407
140	2.4435	2.4435	0.4093
150	2.6180	2.6180	0.3820
160	2.7925	2.7925	0.3581
170	2.9671	2.9671	0.3370
180	3.1416	3.1416	0.3183
190	3.3161	3.3161	0.3016
200	3.4907	3.4907	0.2865
210	3.6652	3.6652	0.2728
220	3.8397	3.8397	0.2604
230	4.0143	4.0143	0.2491
240	4.1888	4.1888	0.2387
250	4.3633	4.3633	0.2292
260	4.5379	4.5379	0.2204
270	4.7124	4.7124	0.2122
280	4.8869	4.8869	0.2046
290	5.0615	5.0615	0.1976
300	5.2360	5.2360	0.1910
310	5.4105	5.4105	0.1848
320	5.5851	5.5851	0.1790
330	5.7596	5.7596	0.1736
340	5.9341	5.9341	0.1685
350	6.1087	6.1087	0.1637
360	6.2832	6.2832	0.1592


Notice that for r*radians the distances grow as the angle grows and for r/radians the distance shrinks as the angle grows. Perhaps robust can explain why the distance should be smaller for larger angles.

#20 tom

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Posted 25 April 2005 - 01:10 PM

L = angle ( in radians ) * radius

angle ( in radians) = angle ( in degrees ) / 360 * 2 pi

L = angle ( in degrees ) / 360 * 2 pi * radius

We're talking about 1 degree so

L = 1 / 360 * 2 pi * radius

L = 2 pi / 360 * radius

We all agree on this, right?

The only problem is caused by robsts' notation. He calls 360 / 2pi radian. This is why his formula is

radius / radian

No, Rincewind, the radian = 360-degrees/2pi. It is not determined by linear length (as is the radius).



#21 C1ay

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Posted 25 April 2005 - 02:02 PM

No, Rincewind, the radian = 360-degrees/2pi. It is not determined by linear length (as is the radius).


The only problem is caused by robsts' notation. He calls 360 / 2pi radian. This is why his formula is radius / radian


It looks as if he is intentionally reversing things. We all know a circular arc is the radius times the angle in radians and not their quotient. We also know that radians are equal to 2π times θ/360 and not the reverse. I hope any freshmen passing by here are ignoring this gibberish.

#22 Robust

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Posted 25 April 2005 - 04:12 PM

It looks as if he is intentionally reversing things. We all know a circular arc is the radius times the angle in radians and not their quotient. We also know that radians are equal to 2π times θ/360 and not the reverse. I hope any freshmen passing by here are ignoring this gibberish.


First off, Clay, The title of this posting is "Degree Distance" - the distance between each angular degree on the circumference. It is readily found by either the formula radius/radian or, as Pythagoras might have it: pi/40.

#23 C1ay

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Posted 25 April 2005 - 04:34 PM

First off, Clay, The title of this posting is "Degree Distance" - the distance between each angular degree on the circumference. It is readily found by either the formula radius/radian or, as Pythagoras might have it: pi/40.

So. The distance is still the radius times the angle in radians even if the angle is only one degree or radius*(2π/360) or even if you regroup it you have (radius*2π)/360.

And what is pi/40 supposed to be? That would be 4.5° increments so it is certainly not the distance of one degree.

#24 Robust

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Posted 26 April 2005 - 04:11 AM

So. The distance is still the radius times the angle in radians even if the angle is only one degree or radius*(2π/360) or even if you regroup it you have (radius*2π)/360.

And what is pi/40 supposed to be? That would be 4.5° increments so it is certainly not the distance of one degree.

Pi/40 gives the least possible distance between 2 adjacent angular degrees on the circumference. I derive the formula from the perfect ratios of Pythagoras.

"All things number and harmony." - Pythagoras

#25 Rincewind

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Posted 26 April 2005 - 04:45 AM

Pi/40 gives the least possible distance between 2 adjacent angular degrees on the circumference. I derive the formula from the perfect ratios of Pythagoras.

"All things number and harmony." - Pythagoras

Eh? Are you saying that the least possible arclength subtended by an angle of 1° is 0.078539816339744830961566084581988? 0.078539816339744830961566084581988 what? Metres? Miles? Light years?

Why can't it be 0.05 µm?

#26 C1ay

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Posted 26 April 2005 - 05:35 AM

Pi/40 gives the least possible distance between 2 adjacent angular degrees on the circumference. I derive the formula from the perfect ratios of Pythagoras.

"All things number and harmony." - Pythagoras

It's beginning to look like we have another candidate for the Strange Claims Forum.

#27 Robust

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Posted 26 April 2005 - 06:20 PM

It's beginning to look like we have another candidate for the Strange Claims Forum.

Why do you protest so, Clay? Are you contesting Pythagoras now? He's the one who gave the Western world it's system of mathematics. Pi/40 is derived from his perfect ratios. Are you saying that it does not define the least possible distance between 2 adjacent angular degrees on the circumference?

"All things number and harmony." - Pythagoras

#28 Rincewind

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Posted 27 April 2005 - 12:00 AM

Are you contesting Pythagoras now?

No, just your interpretation of, and way of using, his theories.

Are you going to answer my queries regarding your "least possible distance between 2 adjacent angular degrees on the circumference," Robust?

#29 maddog

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Posted 27 April 2005 - 10:59 AM

What I marvel at here is how so many are speaking the same stuff back to each
other and then saying the other has it "wrong" or "backwards". Guys, the Greeks
had this figured out thousands of years ago ! ;) Heheh... ;)

maddog

#30 Robust

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Posted 27 April 2005 - 12:17 PM

Eh? Are you saying that the least possible arclength subtended by an angle of 1° is 0.078539816339744830961566084581988? 0.078539816339744830961566084581988 what? Metres? Miles? Light years?

Why can't it be 0.05 µm?

Yes, that's what I'm saying (using the irrational pi); using the rational pi value of 256/81(earliest known) the distance would be 0.07901234567 ad infinitum. I hold to the latter, yet consider the finite of 3.1640625.

#31 Rincewind

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Posted 27 April 2005 - 04:20 PM

Yes, that's what I'm saying (using the irrational pi); using the rational pi value of 256/81(earliest known) the distance would be 0.07901234567 ad infinitum. I hold to the latter, yet consider the finite of 3.1640625.

Yes, but 0.07901234567 of what units? Inches? Metres? Miles? Angstroms? Light years?

Without a unit, a distance is meaningless. That's like saying, "Phew, I just walked forty-seven, and I'm pooped!"

#32 Robust

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Posted 27 April 2005 - 04:43 PM

Yes, but 0.07901234567 of what units? Inches? Metres? Miles? Angstroms? Light years?

Without a unit, a distance is meaningless. That's like saying, "Phew, I just walked forty-seven, and I'm pooped!"

I don't understand the question, Rincewind. It would be whatever units you use to describe the diameter.

#33 Rincewind

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Posted 28 April 2005 - 03:19 AM

I don't understand the question, Rincewind. It would be whatever units you use to describe the diameter.

So if I choose to describe the diameter in metres, then you're saying that the least possible distance between 2 adjacent angular degrees on the circumference is 0.07901234567 m, or 7.901234567 cm -- the equivalent of approximately 3.11 inches. Is that right?

If it is right, then explain why it can't be 5 cm or 1 cm. If it's wrong, then please explain why it's wrong and what you actually mean.

#34 Robust

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Posted 29 April 2005 - 05:36 PM

So if I choose to describe the diameter in metres, then you're saying that the least possible distance between 2 adjacent angular degrees on the circumference is 0.07901234567 m, or 7.901234567 cm -- the equivalent of approximately 3.11 inches. Is that right?

If it is right, then explain why it can't be 5 cm or 1 cm. If it's wrong, then please explain why it's wrong and what you actually mean.

Whatever units you use, Rincewind, it is simply radius/radian or,if you prefer, pi/40, the latter referencing the Base 10 number system.