Suction mounts on a canopy

Why is the maximum possible force Pi*D*14.7 ?



"Matt Herron Jr." wrote in message
ups.com...
Just so we all agree on the laws of physics, any force exerted on the
canopy by a suction cup is due to the differential in pressure on one
side of the plex vs. the other. trapped air bubbles in the plex would
not have an effect. air pressure at sea level is about 14.7 PSI.
Assuming the suction cup pulled a perfect vacuum (unlikely) and it had
a diameter of 2", then the maximum possible force would be:
Pi*D*14.7=92.4 lbs of force. Not insignificant. Another way to look
at this is that it would require 95 lbs of force to pull the suction
cup off the canopy.

At 6000' atmospheric pressure drops to about 12 psi, yielding 75lbs of
force. Still pretty high. Note that deflection of the canopy in this
area would be pretty small, but stress internal to the material would
be high. Low temperatures, UV exposure, etc would exacerbate the
issue.

In reality, a suction cup probably doesn't come anywhere close to
pulling a perfect vacuum, so the numbers would be much lower, but I
couldn't guess how much. I can't offer any analysis on skin/nipple
distances.

Matt (jr)

On Apr 13, 4:03 am, Simon Taylor
wrote:

Back to gliding, which always tends to distance itself
from the world of skin and nipples, this does imply
that any minute pockets of trapped air in the canopy
might pull- sorry, might cause the very insidemost
parts of the canopy to be PUSHED in towards the low
pressure within the suction cup, potentially damaging
the canopy. However, I presume such pockets don't exist;
there would be visible depressions in the canopy where
these pockets had cooled after the forming of the canopy,
and any such pockets would be just/almost as prone
to deforming the canopy during a high wave flight.

Without the existance of air pockets, I reckon the
situation would be just as I described before - complete
with disclaimer..

the canopy. It may or may not be a visible deformation.