Re: Velocity of point on a rim
- From: carlfogel@xxxxxxxxxxx
- Date: Mon, 28 Aug 2006 01:19:41 -0600
On Mon, 28 Aug 2006 05:38:47 GMT, Joe Riel <joer@xxxxxxxxxxxx> wrote:
carlfogel@xxxxxxxxxxx writes:
http://www.flickr.com/photos/52339623@N00/226852675
or http://tinyurl.com/j3h5v
Those looks correct. I screwed up my description, the
velocity arrow on the diagram I drew (ascii arg) corresponds
to the velocity at the rim 90 degrees counterclockwise from
the head of the arrow. Let me try again:
Draw a circle of unit radius (okay it looks more like an ellipse; who
would have thought I'd long for Archimedes' stick and sand). Pick a
point on the rim, say P. Draw the radius from P to the C, the center
of the circle. Draw another radius, to H, so PCH form a clockwise
right angle. Now draw the line segment TH. This corresponds to the
velocity vector at P, translated to T.
-------
--/ \--
-/ \-
/ \ H
-/ /--
/ // \
| /-/ |
P / /- | \
/\ /- / \
| -\ /- / |
/ \ /-- / \
| -\ /- / |
| /\ / |
/ /- -\ | \
| /- \ 90 / |
| /- -\ / |
T ---------------C----------------+--
| |
| |
\ /
| |
| |
\ /
| |
\ /
\ /
| |
\ /
-\ /-
\ /
-\ /-
--\ /--
-------
Joe Riel
Dear Joe,
I see your approach now.
Here's my clunky method again:
http://www.flickr.com/photos/52339623@N00/226852675
or http://tinyurl.com/j3h5v
Here's your more elegant approach, mirror-imaged to put P at 45
degrees again (on the right side) and with a radius of 20 units for
easy comparison:
http://www.flickr.com/photos/52339623@N00/226925395/
or http://tinyurl.com/jp9mx
Both methods produce the same result, right?
(Sorry about flipping your drawing left-right. The Roman soldier
didn't stab Archimedes. The soldier came into the room where
Archimedes had just solved that knotty geometry problem, drawing it
out nicely. Facing the soldier over the drawing, the elderly genius
explained his solution and died of apoplexy when the Roman oaf asked,
"Why is it upside-down?")
Cheers,
Carl Fogel
.
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