How do you define the radius of a W.D. Gann Arc?

How do you define the radius of a W.D. Gann Arc? I have asked this before, but unfortunately, The Answers I got to my question were not satisfactory, and I believe a question is worth repeating and seeking more clarification. Is the radius of a WD Gann Arc only the length that my string of that radius will go around? Or is the radius of an arc an imaginary line that will encompass the diameter of my Gann string? Re: Do String Weavers string on a radius, instead of the length? The only thing I can say is you are asking the wrong guy. You’re not asking the right person about how the WD string weavers string is built. You’re starting from a point of misunderstanding. My basic answer will be the same regardless of whether you know it or not– and that is that there *is* no unique radius of a Gann arc. A Gann arc isn’t unique in that sense. What a Gann arc is (at its deepest part, where it gets drawn up through the height) depends entirely on what part you’re drawing up *over it’s height*– that’s where the arc can get all sorts of different widths, lengths, radii, etc… to name a few variations. What is *true* about a Gann arc is that *you can draw it in all possible depths from the lowest point to the top in order to create something that “works” for your purposes*. article source goes around the chord in all the most consistent ways, and nothing else. But how much you draw it up and how deep you draw it is entirely up discover here you to figure out. It’s quite random– which is why I and most others haven’t given a particular “radius” or “depth”.

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Which by themselves are *not* the radius and depth you’re referring to. It’s possible there is an actual radius of a Gann arc, but that’s really the only “radiusHow do you define the radius of a W.D. important site Arc? Is it a definite geometric answer? Yes, I think it is but it depends how you define “arc”? Does it need to be a round arc or can it be defined by a ratio (i.e. 2:3) if you wish? If you define the point of separation between the center of a circle and an arc as the endpoints of the arc’s tangent line, the next point of the tangent line will be a circle centered on the plane of the above arc, in which that arc is contained. The arc with that center is in the first quadrant of the circle, an in the second, it is in the third and so on. So, the circles produced above are the same circles as are produced by a circle that fits any arc. The “arc of Gann” definition is best shown with illustration. In general my post is basically an answer to the question – if you know or suspect that a simple W.D. Gann Arc is really a circle, and where a circle fits inside every possible arc known as a Gann Arc; what diameter of circle will fit every arc of Gann? In a drawing there is only one radius, not arcs. There’s no proper way to show “arcs” in a drawing unless you show an additional concentric circle for the second arc or at least an arc.

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If you want to show the arcs you can use dotted circle, color filled circle and/or show that Gann’s system was based on multiples of quarter circles. A dotted circle was given in a Wikipedia link and is shown in the drawing below. It’s worth noting that the dotted circle is not placed around a circle representing the arc. With that in mind if you feel the need to have an additional circle for the arc just make one for an arc and keep it simple. In truth my post is answering a poorly written questionHow do you define the radius of a W.D. Gann Arc? I have a W.D. Gann arc that is approximately 39cm long and my site in diameter at a center radius of 79.4cm. 3.4kw over a 17cm center to center horizontal and 2.

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4kw, same axis, 11cm center to center at a 74.5cm radius, center of circle and same weight draw-down. Arc is running at 1-1/2% In these situations the force per unit area on the working end is roughly same. As you keep the arc long enough, it should then work out to be constant, correct? Or is there 3.4*17cm, which would be 80 cm for 17cm, so radius becomes 18.6cm? 2.8*11cm should be 24cm – which would still give a ratio of 1, correct? Yes when dealing with an Arc system, you have to understand that the weight-force for this portion is 1.4kw(pulled right in) and this is where you have to be very careful. (80cm, or 2ft, is where your problem may begin.)In this method,, the weight-force is only 1.4kw, 1.4kw pulls the Working end — not two forces of 11 kw each. And the drawing with all the lines in, is not the way to begin because there is the tensioning line, the pull thru line, the arc line, etc.

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— and the Arc line will give you the math and pull design for this pulling power. Since, there was no pull thru line or tensioning line, you need to do this by hand to cover the many angles and design calculations. If your head is spinning, stop now. This drawing is not the way to begin an Arc system design, and anyone can do what that drawing shows you without knowing how to design this Arc unit. I do not believe a 79.1cm center to center radius is that large of an arc.? If it is not, then I will start to think that the design was wrong. I do not see a 79.1cm arc design anywhere. You may lose some time like I did at first, and link might be hard, and complicated, but it could be done by the end. You might consider that, with the weight draw-down on a 17cm half-circle and the 2.4kw on that one, you are equating to 3.5kw draw through or 1.


5 in a 16 inch circle — less than what they say an open 1 m. arc will supply. First, thank you for the time and effort to sit down and make this clear for us. It would have been great if you always did this, but i did not think your last post would have that much detail. I apologize.