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.
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.
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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