B How to Calculate Hand Force to Close a Hatch Door Against a Gas Spring?

[erobz]

As I understand it the terms are interchangeable.

Maybe the words are used interchangeably, but the design of the one I had in mind is very much unlike the other in how they respond to displacement. The "strut" is the one that fits the application as a door opening assist.

 

[Geordielad]

Erobz, I have figured out how to write the $f (x)$you mention in post 29 using the law of cosines and since I know all 3 sides. $f(x)=\cos\left(\frac {x^2+a^2-b^2}{2xa}\right)$ This gives me the angle designated $sin beta$ in post 29. for all the intermediate positions of the cylinder.
Instead of finding \cos\theta as a $f(x)$ could I use $\cos angle BOA$ as a $f(x)$?
I am still working on $F_s$ as a $f(x)$.

 

[erobz]

Erobz, I have figured out how to write the $f (x)$you mention in post 29 using the law of cosines and since I know all 3 sides. $f(x)=\cos\left(\frac {x^2+a^2-b^2}{2xa}\right)$ This gives me the angle designated $sin beta$ in post 29. for all the intermediate positions of the cylinder.
Instead of finding \cos\theta as a $f(x)$ could I use $\cos angle BOA$ as a $f(x)$?

You should be looking for $\cos \beta$ as a function of $x$:

$$ b^2 = c^2 + x^2 - 2cx \cos \beta $$

$$ f(x) = \cos \beta = ? $$We are trying to get rid of $\sin \beta$ in the numerator of the eq found in post #31, by replacing it with an expression of the variable $x$.

I am still working on $F_s$ as a $f(x)$.

We are going to assume that it is a gas strut (in the application it makes the most sense), and that the force $F_s$ is approximately constant in magnitude over the stroke. If you want to examine how it changes, and how that effects the force we can do that after as a refined model. The challenging part is getting the geometry worked out.