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Re: [PSUBS-MAILIST] dome hatch lift force
Thanks Sean,
Have printed out your email & will decipher it over the
comming days.
Alan
----- Original Message -----
From: "Sean T. Stevenson" <cast55@telus.net>
To: <personal_submersibles@psubs.org>
Sent: Thursday, November 05, 2009 1:59 PM
Subject: Re: [PSUBS-MAILIST] dome hatch lift force
Just sum all of the forces acting on the dome. Examining the forces in
the vertical direction only (since the horizontal forces will cancel),
gives you the expression:
F (sum) = F1 (dome weight) + F2 (force from external water pressure) +
F3 (buoyant force) + F4 (force from internal air pressure).
Assuming a sign convention of positive for the down direction, forces F3
and F4 above will be negative.
F1 [N] = m [kg] * g [m/s^2], where m is the mass of the acrylic dome,
and g is 9.80665 m/s^2.
F2 [N] = P [Pa] * A [m^2], where P is the external water pressure and A
is the projected area of the dome in the horizontal plane (i.e. pi *
dome base radius^2).
P, as used in the above equation, may be calculated as follows:
P [Pa] = rho [kg/m^3] * g [m/s^2] * h [m], where rho is the water
density (1000 kg/m^3 or 1025 kg/m^3 for fresh water or seawater,
respectively), g is 9.80665 m/s^2, and h is the height of the water
column above the base of the dome. Technically, you would need to add 1
atmosphere equivalent of pressure to this value of F2 (approximately 101
300 Pa) to get the absolute pressure acting on the dome, since the
atmospheric air adds pressure over and above the water column weight;
however, since you also have one atmosphere of pressure acting on the
inside of the dome (assuming a one atmosphere submersible), the two
forces cancel each other and need not be considered.
F3 [N] = V [m^3] * rho [kg/m^3] * g [m/s^2], where V is the volume of
water displaced by the dome, rho is the water density, and g is the
acceleration due to gravity (9.80665 m/s^2).
F4, as mentioned above, need not be considered if you disregarded the
atmospheric pressure when calculating F2. If you do need to consider it
(necessitating adding it to your calculation of F2), then:
F4 [N] = P [Pa] * A [m^2], where P is the internal pressure acting on
the dome, and A is the projected area of the dome in the horizontal
plane (i.e. pi * dome base radius^2). Calculating F4 (and adding 1 ATA
to F2) will be necessary for an ambient pressure submersible, since the
interior cabin pressure will be equal to the ambient water pressure at
the cabin waterline, and thus greater than the ambient air pressure when
the boat is surfaced.
Simply add up all of the forces, taking care to note that they all have
correct signs (dome weight and force due to external pressure acting
down, dome buoyancy and force due to internal pressure acting up), and
the result is the net force acting on the dome.
Clear as mud?
-Sean
Alan James wrote:
Thanks Sean,
What I'm after is just the lifting force of the acrylic dome
wich will be mounted in the hatch; to gauge what sought
of lifting pressure it will initially exert on the attachments
to the hatch.
I was looking at the weight of a column of water directly
above the dome & assuming that weight would be forcing
the dome on to the hatch & opposing the flotation in the
dome. However after playing round with stuff in the kitchen
sink & spa pool (at 3AM) I concluded I've got something wrong.
Regards Alan
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