Moderator: Flannel Jesus
obsrvr524 wrote:-
Appears awfully close but two questions -
- Have you checked the math?
- How did you place the 2 vertical lines that are closest to center?
Motor Daddy wrote:The vertical lines closest to the center were arrived at by a series of straight lines until I had enough lines to run them vertically through the centers of the circles they vertically run though the centers of.
obsrvr524 wrote:Motor Daddy wrote:The vertical lines closest to the center were arrived at by a series of straight lines until I had enough lines to run them vertically through the centers of the circles they vertically run though the centers of.
I assumed that you drew the center lines of those circles before the circles. If the lines were drawn after the circles, how did you know where to draw the circles?
Motor Daddy wrote:I then drew a straight line horizontally through the center of the circle. From the intersection of the line and circle edge on both sides I created the circle on the left and right of the center circle, so the intersection was the center of each circle. That gave me more intersections to draw the lines horizontally on top and bottom of the 3 circles, which gave me more intersections to draw more straight lines.
The only lines and circles that were made was when there was enough points to make those intersections, and therefore new straight lines or circle centers.
obsrvr524 wrote:I followed you that far. But I still don't see how the upper 2 and lower 2 circles could be horizontally positioned other than just arbitrary guess.
And if they are not mathematically situated, how did any maths geek calculate the square?
Motor Daddy wrote:He claimed that trig could work out the angles and length of lines. I don't know Trig, so beats the heck out of me. He made an excellent diagram and showed all angles and used Trig to calculate the side lengths.
The circles have a radius of 1 unit, so the outer square is 4 units x 4 units
obsrvr524 wrote:Motor Daddy wrote:He claimed that trig could work out the angles and length of lines. I don't know Trig, so beats the heck out of me. He made an excellent diagram and showed all angles and used Trig to calculate the side lengths.
The circles have a radius of 1 unit, so the outer square is 4 units x 4 units
I imagine I could work out the math now.
Did you come up with that technique yourself?
obsrvr524 wrote:-
They say it is impossible to make it precisely accurate - but I don't know if that is a proven fact or just a highly educated guess.
Either way - it's a brilliant accomplishment.
Motor Daddy wrote:The reason they say it is impossible is because there is no finite number for the square root of Pi, so how can you make the side length of the square a finite size when there isn't a finite square root of Pi?
obsrvr524 wrote:-
Come to think of it - I know \(\sqrt\pi\) can be drawn.
The question is how to do it with straight edge and compass.
Motor Daddy wrote:My argument is that, how can the circle have a finite area if Pi is infinite, for all practical purposes??
obsrvr524 wrote:Motor Daddy wrote:My argument is that, how can the circle have a finite area if Pi is infinite, for all practical purposes??
This gets back to that base-unit issue. It isn't that pi is infinite - but that trying to express pi in a rational base will yield an infinite string of digits.
But what if you use base-pi.
You know that you can -draw a \(2\pi\) length line merely by rolling a radius 1 disc across a paper exactly once.
Divide that in half with a compass and you have exactly \(\pi\)
Duplicate that length vertically from the end of the \(\pi\) length (making a corner or "L").
Draw a line from the tip of each \(\pi\) segment and you have a line exactly \(\pi\sqrt2 \)
But they don't let you use a disc - so the trick is how to do it using only the compass instead of a disc. It doesn't seem to be a math problem as much as a mechanical problem.
obsrvr524 wrote:-
I think you missed my point.
My point was that a length being an irrational number with an infinite string of decimals (such as \(\pi\) or \(\sqrt2\) doesn't stop you from drawing a line of that length - you can even draw a line of the multiplication of two irrational numbers - \(\pi\sqrt2\).
Motor Daddy wrote:obsrvr524 wrote:-
I think you missed my point.
My point was that a length being an irrational number with an infinite string of decimals (such as \(\pi\) or \(\sqrt2\) doesn't stop you from drawing a line of that length - you can even draw a line of the multiplication of two irrational numbers - \(\pi\sqrt2\).
You can draw a line of that length, but how would you differentiate between 3" and 3.14159", when you only have capability to measure to the closest .1".
If all you have is a ruler marked for Hundredths of an inch, how would you know if the line is 3.14159"? The best you could do is say the line is 3.14", and you don't know if it's 3.14159" or 3.14172".
See what I'm saying? You may be able to calculate to 27 decimal places, but with a ruler that only has Hundredths of an inch, the rest is nonsense to you. You can measure to 2 decimal places, and the remaining 25 decimal places may as well be truncated, you have no purpose for them!
obsrvr524 wrote:
To get \(\pi\) - just roll a 1 unit disc across a paper exactly once and you have exactly \(\pi\) length.
Motor Daddy wrote:obsrvr524 wrote:
To get \(\pi\) - just roll a 1 unit disc across a paper exactly once and you have exactly \(\pi\) length.
How long is that line, 6.28 units or 6.28318 units?
obsrvr524 wrote:Motor Daddy wrote:obsrvr524 wrote:
To get \(\pi\) - just roll a 1 unit disc across a paper exactly once and you have exactly \(\pi\) length.
How long is that line, 6.28 units or 6.28318 units?
Neither one.
It is EXACTLY \(\pi\) in length (forgiving the errors of pencil and paper).
Return to Science, Technology, and Math
Users browsing this forum: No registered users