Rincewind Posted April 25, 2005 Report Share Posted April 25, 2005 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°). Quote Link to comment Share on other sites More sharing options...
C1ay Posted April 25, 2005 Report Share Posted April 25, 2005 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. Quote Link to comment Share on other sites More sharing options...
tom Posted April 25, 2005 Report Share Posted April 25, 2005 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). Quote Link to comment Share on other sites More sharing options...
C1ay Posted April 25, 2005 Report Share Posted April 25, 2005 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. Quote Link to comment Share on other sites More sharing options...
Robust Posted April 25, 2005 Author Report Share Posted April 25, 2005 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. Quote Link to comment Share on other sites More sharing options...
C1ay Posted April 25, 2005 Report Share Posted April 25, 2005 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. Quote Link to comment Share on other sites More sharing options...
Robust Posted April 26, 2005 Author Report Share Posted April 26, 2005 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 Quote Link to comment Share on other sites More sharing options...
Rincewind Posted April 26, 2005 Report Share Posted April 26, 2005 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." - PythagorasEh? 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? Quote Link to comment Share on other sites More sharing options...
C1ay Posted April 26, 2005 Report Share Posted April 26, 2005 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." - PythagorasIt's beginning to look like we have another candidate for the Strange Claims Forum. Quote Link to comment Share on other sites More sharing options...
Robust Posted April 27, 2005 Author Report Share Posted April 27, 2005 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 Quote Link to comment Share on other sites More sharing options...
Rincewind Posted April 27, 2005 Report Share Posted April 27, 2005 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? Quote Link to comment Share on other sites More sharing options...
maddog Posted April 27, 2005 Report Share Posted April 27, 2005 What I marvel at here is how so many are speaking the same stuff back to eachother and then saying the other has it "wrong" or "backwards". Guys, the Greekshad this figured out thousands of years ago ! ;) Heheh... ;) maddog Quote Link to comment Share on other sites More sharing options...
Robust Posted April 27, 2005 Author Report Share Posted April 27, 2005 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. Quote Link to comment Share on other sites More sharing options...
Rincewind Posted April 27, 2005 Report Share Posted April 27, 2005 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!" Quote Link to comment Share on other sites More sharing options...
Robust Posted April 27, 2005 Author Report Share Posted April 27, 2005 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. Quote Link to comment Share on other sites More sharing options...
Rincewind Posted April 28, 2005 Report Share Posted April 28, 2005 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. Quote Link to comment Share on other sites More sharing options...
Robust Posted April 29, 2005 Author Report Share Posted April 29, 2005 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. Quote Link to comment Share on other sites More sharing options...
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