Posted by John Collins (194.164.232.108) on October 31, 2003 at 06:56:51:
In Reply to: Re: Another one for John - apologia drawing posted by Jonathan on October 31, 2003 at 04:49:14:
Sorry Jonathan,
I should read what I write more carefully. I will try to explain what I mean more accurately.
Right, a triangle with a corner of 120 degrees has two other corners each of 30 degrees. Assuming that you have drawn a circle and divided it into four equal segments by drawing two diameters at right angles to one another,look at your drawing. Let's say that one of the two diameters you've drawn runs from 9 o'clock across to 3 o'clock. Place a protractor on the horizontal line at the point where the line touches the rim at say 9 o'clock, so that you can draw an angle which comes out from that point into the middle of the circle. Mark off 30 degrees (on the right side). Draw the new 30 degree line from the rim into the circle to about half way across to the adjacent radius. Repeat for the adjacent radius and then do the same thing for the other radii. Then repeat the whole process for the under sides of all the radii. That means you've drawn eight 30 degree angles. Now place the point of a pair of compasses in the centre and draw a circle through the four points where, within each segment, the two 30 degree angles meet in each case. If you've got it right it looks like a four pointed star inside a circle. You should now have what I believe to be the essential working area of the mechanisms for a one way wheel. They operate within a 120 degree segment placed four times, each within a right angle inside the circle.
It's possible that such a diagram was intended also to work with just three mechanisms and weights and it is easy to do the same excercise just dividing the circle into three segments instread of four. But I tend towards the four weight design because of the 90 degree inclusion in the Apologia drawing.
I'll post a diagram to Scott to elucidate this.
John Collins
: I'm sorry, but I still don't understand. Your second sentence starts out by surmising that 120+60+60=180, and nothing past that made a lot of sense. I suppose you mean 120+30+30=180, but then I get confused as to how the angle of 120 at the rim manages to be off center even though it should be 60 on both sides of the line it's made from, thereby making it symetrical.
: : OK. Draw a circle. Divide it into four equal segments through the centre. A triangle with one corner of 120 degrees has two other equal ones of 60 degrees - right? So at the point where the each of the four lines touches the rim, measure a 30 degree angle from the rim across at a tangent towards the centre (60-90=30). Do the same from the other side of the segment and where they meet is a 120 degree angle, but positioned off to one side of the centre of the circle. Do the same for each of the other three segments and you get a circle divided into four and within each segment is a 120 degree angle. Draw a circle through each of the 120 degree angles and you have a small inner circle. I suggest that Bessler intended to show that his four mechanisms operated through a 120 degree angle within each right angle of the wheel.
: : John Collins
: :
: : : What are you talking about? How can a pie wedge have 90 and 120 degree angles?
: : : : Hello again Michael,
: : : : There is a good drawing of the Worcester wheel by Dircks, (I'll have to check that it was he) which shows exactly what the wheel probably looked like. In my opinion if the drawing is correct then the wheel won't turn. There is no other documentary evidence about the Worcester wheel and no proof that it ever worked other than the surmise written in your quotation.
: : : : So in answer to your question, I personally don't think that the white pie sections are arrows. I have spent a lot of time analysing all of the drawings and I've come to some conclusions and made some discoveries on all of them. I believe that in the Apologia drawing the JB is telling us that there are four sections, because of the approximately 90 degree pie sections, but that each 90 degree segment has an internal angle of approximately 120 degrees (90 + 30). If you draw this on paper you will d