Tom:
I like your suggestion. I was going to try something with chords of a
circle. Mainly as an exercise. But you used the Phythaurous method. Great.
May I suggest that if you use the extra equation that B (boom length) = Br
(length from reflector to centre) + Bd (length from director to centre then
the iterations need not be done.
Br = (D^2-R^2+(2B)^2)/8B
If you want a 3 element yagi at 28,8 mHz with D=
455/28.8=15.7986, R=500/28.8=17.36, spacing 0.1l then B=
0.2*2*468/28.8=6.5. Calculations make centre 2.253 from R and total radius
of rotation 8.967.
Is this correct?
Chris opr VE7HCB
At 09:17 AM 2002-10-14 -0600, n4kg@juno.com wrote:
>Recently, there was a recommendation made
>to the effect of running a string from the reflector
>tip diagonally across the boom to the end director
>tip, claiming (if I remember correctly) that the
>point at which this string crossed the boom was
>the Rotational Center.
>
>I believe that recommendation to NOT be correct.
>
>My thinking is that as the director is shortened,
>the string will move closer to the boom and the
>crossover will move TOWARD the Director which
>is clearly not the desired result.
>
>There are two approaches that will lead to determining
>the center of rotation.
>
>An iterative measurement approach (either actual or on
>a scaled drawing). Select a point on the boom, measure
>from that point to the tip of the reflector and to the tip of
>the end director. If they are not equal, select another
>point which will reduce the longer dimension and lengthen
>the shorter dimension. Repeat this process until a point
>is found on the Boom which is equidistant from the
>reflector tip and the end director tip. THAT is the
>Rotational Center.
>
>Mathematically, calculate the rotational radius to
>the reflector tip and the director tip from points along
>the boom until they are equal.
>
>(R/2)^2 + Br^2 = Rr
>where R = Reflector Length,
>Br = distance on boom from reflector to test point
>
>(D/2)^2 + Bd^2 = Rd
>where D = Director Length,
>Bd = distance on boom from end director to test point
>
>When you find Br and Bd such that Rr = Rd,
>you have found the rotational center.
>
>I like the scaled drawing approach best :-)
>
>Tom N4KG
>
>
>
>
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