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Re: Re: [PSUBS-MAILIST] dome hatch lift force




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 
----- Original Message -----
Sent: Thursday, November 05, 2009 8:33 AM
Subject: Re: Re: [PSUBS-MAILIST] dome hatch lift force

I should elaborate a bit on the counterbalance spring forces, since most hatch counterbalance springs are torsion springs implemented as an intended moment acting on the hatch hinge.  

 

This moment

 

M [N * m] =  k [N * m] * theta [radians]

 

is typically applied via short spring ends acting at an effective distance from the actual axis,  so you can work back to the distance you are interested in (presumably at the hatch center of mass),

since F_1 * r_1 = M = F_2 * r_2.  Ideally, you want the required counterbalance force to develop over a spring twist of 90 degrees (pi/2 radians) so that you have full counterbalance (for safety reasons, slightly less than the actual force required to open the hatch) with the hatch closed, and zero counterbalance with the hatch open.  Note that the relationship is not precisely linear, as the counterbalance force in this case varies linearly with hatch open angle, but the force acting to close the hatch (due to gravity) varies with the cosine of the open angle.

 

-Sean

 

 

On Nov 4, 2009, Sean T. Stevenson <cast55@telus.net> wrote:

1)  For force due to material weight:

 

F [N] = m [kg] * a [m/s^2]

 

Where m is the mass and a in this case is acceleration due to gravity: 9.80665 [m/s^2 ].

 

2)  For force due to pressure:

 

F [N] = P [Pa] * A [m^2]

 

where A is the projected area of the hatch in the plane under consideration.  If you have a dome hatch, you only need consider the forces in the direction of the axis, since the radial forces cancel each other.

 

3)  For force due to counterbalance spring:

 

F [N] = k [N/m] * x [m], where k is the linear spring constant and x is the compression / extension distance, OR

F [N] = k [N] x theta [unitless], where k is the torsion spring constant and theta is the twist angle, expressed in radians.

 

4)  For force due to buoyancy, equal to the force due to weight of the displaced water (using equation 1 above):

 

To figure the mass, m [kg] = rho [kg/m^3] * V [m^3].  Use densities of rho=1000 kg/m^3 for fresh water, or rho=1025 kg/m^3 for salt.

 

-Sean

 

 

 

On Nov 4, 2009, Alan James <alanjames@xtra.co.nz> wrote:

Hi psubbers,
I'm trying to calculate what the lifting force will be on the
dome in my hatch when it is just submerged; but aren't
quite sure what I have to take into consideration to calculate this. Any thoughts on this?
Thanks,  Alan 
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