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

[erobz]

Okay solving for F_H
$$0.5 (e cos \theta) W + (e cos \theta) F_H - 2(c sin\beta) F_s=0 $$
$$0.5 (e cos \theta) W + (e cos \theta) F_H = 2(c sin\beta) F_s $$
$$ (e cos \theta) [ 0.5 W + F_H ] = 2(c sin \ beta) F_S $$
$$ [ 0.5 W + F_H ] = 2(c sin \ beta) F_S / (e cos \theta) $$
$$ F_H = (2(c sin \ beta) F_S /(e cos \theta)) - 0.5 W $$

You have forgotten the ""\"" in front of the trig functions( don't put spaces between "\" and "cos", or "\" and "beta", as you can see when you do that teh code does not render the symbol. And its better to use the " \frac{}{}" latex function for fractions. Also remove parathesis that aren't needed, it just clutters up the code. I will fix them for you... Pleases hit reply to this so you can see the changes.

$$0.5 e \cos \theta W + e \cos \theta F_H - 2 c \sin\beta F_s=0 $$
$$0.5 e \cos \theta W + e \cos \theta F_H = 2c \sin\beta F_s $$
$$ e \cos \theta [ 0.5 W + F_H ] = 2c \sin \beta F_s $$
$$ [ 0.5 W + F_H ] = \frac{2c \sin \beta F_s} { e \cos \theta } $$
$$ F_H = \frac{ 2 c \sin \beta }{e \cos \theta}F_s - 0.5 W $$

 

[Geordielad]

Thanks Erobz

 

[Geordielad]

I have create a spreadsheet to compute all the trig functions to establish the fixed and moving points so that I can make iterations for the moveable mounting bracket (also different gas spring lengths). When I insert the FH formula below I get a reasonable figure to close an open door but assisting a closed door seems very high?
$$F_H=(2c\sin\beta)/(e\cos\theta)F_S-\frac {1}{2}W$$

 

[erobz]

I have create a spreadsheet to compute all the trig functions to establish the fixed and moving points so that I can make iterations for the moveable mounting bracket (also different gas spring lengths). When I insert the FH formula below I get a reasonable figure to close an open door but assisting a closed door seems very high?
$$F_H=\frac{ 2c\sin\beta}{e\cos\theta}F_S-\frac {1}{2}W$$

You can solve this analytically $\sin \beta = f(x), \cos \theta = g(x), F_s = q(x)$, did you try and get stuck, or just opt to go for a numerical approximation for this exact geometry at specific points?

Latex tip if you want a fraction ( like the first term on the right hand side)

\frac {"the numerator"}{ "the denominator" } , it works with anything, not just numbers.

To be more clear, I can see that you can use the geometry in you drawing to find $\sin \beta$ when the door is closed $\theta \approx 0, \cos \theta \approx 1$, but its not clear how you have defined the spring force $F_s$ in a particular position?

 

[Geordielad]

I will rewrite using \frac for the $2c \sin\beta$section.
The gas spring manufacturer gives the formula $F*E = S*J$ where F is force, E is the force moment arm, S is door weight, J is weight moment arm. Then $F=\frac {S*J} {E}$ with this I used the spreadsheet to calculate F for increments of 10 degrees. However I suspect I am missing something as the results for FH are much higher than I expected.

 

[Geordielad]

$$\frac {2c\sin\beta} {e\cos\theta}$$

 

[erobz]

I will rewrite using \frac for the $2c \sin\beta$section.
The gas spring manufacturer gives the formula $F*E = S*J$ where F is force, E is the force moment arm, S is door weight, J is weight moment arm. Then $F=\frac {S*J} {E}$ with this I used the spreadsheet to calculate F for increments of 10 degrees. However I suspect I am missing something as the results for FH are much higher than I expected.

Yeah...that is sum of the torques = 0. Basically, that is a highly naive version of what we have done so far. Just forget about that at this point...It's not describing what you thought/think it is.

The force from the cylinder is supplied by the gas inside the barrel being compressed by the piston as $x$ decreases from its initial length.

What you could find from the manufacturer that would be of benefit is the force supplied by the cylinder as a function of the displacement, given an initial pressure. As it stands now, I was just going to make up a generic model for that using some basic constraints for the rod and barrel geometry and air as an ideal gas being adiabatically compressed.

 

[Geordielad]

So if I understand what you are saying I should be using $F=P*A$ where F is cylinder force, P is pressure in cylinder, A is cross sectional area of piston, $F_1$ is extending force,$F_2$ is compressing force, $V_1$ is cylinder volume extended, and $V_2$ is cylinder volume compressed.
Then $F_1=P*A$ and $F_2=F_1*\frac{v1} {v2}$

 

[erobz]

So if I understand what you are saying I should be using $F=P*A$ where F is cylinder force, P is pressure in cylinder, A is cross sectional area of piston, $F_1$ is extending force,$F_2$ is compressing force, $V_1$ is cylinder volume extended, and $V_2$ is cylinder volume compressed.
Then $F_1=P*A$ and $F_2=F_1*\frac{v1} {v2}$

I'm not sure what you are talking about with the extending/compressing force. The force will just be a function of $x$, it doesn't matter whether the hatch is opening or closing. It's always applying a force in the direction of the blue vector in post #8.

But yes $F = PA$ . You have to think about the length of the rod, and the length of the barrel and how the volume of air inside the barrel changes with the total cylinder length $x$, (the air can't be compressed to zero volume at $x = x_{min}$, it could however be over pressurized ( in a sense that its physically withheld from applying $F = PA$ because the piston is up against the end caps. You must decide what will be the initial condition. I assume the springs hold the door in a certain position on their own without being fully extended. How you choose the initial condition and the parameters involving the cylinder are how you set the initial charge $P_o, V_o$. You have to be careful in your selection of cylinder characteristics so everything fits.

Furthermore, I would use an adiabatic compression model:

$$ P_o V_o ^n = P_x V_x^n $$

Where for air undergoing an adiabatic process $n = 1.4$

 

[Geordielad]

What I was trying to say was that when the cylinder is compressed there is less volume therefore a higher pressure than when extended.
Your suggestion to solve Fs analytically sound good but I don't understand $\sin\beta=function (x),\cos\theta=g (x), F_s=q (x)$ could you please explain how to find Fs.

 

[erobz]

What I was trying to say was that when the cylinder is compressed there is less volume therefore a higher pressure than when extended.

ok.

Your suggestion to solve Fs analytically sound good but I don't understand $\sin\beta=function (x),\cos\theta=g (x), F_s=q (x)$ could you please explain how to find Fs.

I mean all of those quantities $\sin \beta, \cos \theta,$ via trigonometric relations, identities, etc... are functions of $x$ and $F_s$ is a function of $x$ via some model like "The Ideal Gas Law".

The idea is that you could plot your $F_H$ values as continuous function of $x$ over some specified range ( i.e. from open to closed ). Provided you can find the correct relationships.

 

[erobz]

Just out of curiosity what is this for...work, school, play?

 

[Geordielad]

It is to help a friend who has even less math knowledge than me. I am 74 years young and my math is rusty to say the least. My apologies but I do not understand function of (x). and The ideal gas law is way over my head. Could I ask you to recap how to find Fs we got so far and then I lost the plot.

 

[erobz]

It is to help a friend who has even less math knowledge than me. I am 74 years young and my math is rusty to say the least. My apologies but I do not understand function of (x). and The ideal gas law is way over my head. Could I ask you to recap how to find Fs we got so far and then I lost the plot.

Basically, A function of $x$ means there is a relationship that is determined by the variable $x$, and a value of $x$ produces a single value for $f(x)$

For example, a function of $x$ may be $f(x) = k$. If you plotted it for all $x$ you get a horizontal line that at the value $k$, if $f(x) = x$ you get a line through the origin at a slope of 45 degrees. A function maps some $x$ value to some other value. Informally, its mathematical notation for solving for one variable in terms of another.

see: https://www.mathsisfun.com/sets/function.html

 

[Baluncore]

The ideal gas law is way over my head.

Don't let that worry you.

With a piston in a cylinder, like a bicycle pump, if you move the piston to 50% of the stroke, you double the pressure and force, as you move it to 75%, it doubles again to four times the force and pressure, to 87.5% takes it to eight times the force and pressure. That is horrible and non-linear, not good for the average application.

A gas strut is different. It has no piston or seal, just a perforated guide to keep it straight. As you push the rod in, the gas volume only changes by the volume of the rod, so the pressure changes much less than with a piston. The gas strut also starts with a high internal pressure, so it offers a good pressure from 0% to 100%. The gas strut has a large diameter compared with the rod, so the force and pressure ratio from 0% to 100% changes by less than about 20%, I call that constant. You can change the force by charging the strut with gas through the rod-seal, to the force and pressure you require.

 

[Geordielad]

Is there an easier way for me to calculate Fs?

 

[erobz]

Is there an easier way for me to calculate Fs?

Do you have any manufacturer specs on the gas cylinder being used ( besides that equation you quoted earlier)? You want to be looking for a force vs displacement curve ( or set of curves ) for some specified initial conditions. If you can’t find that, I’ll give you a basic model. Since you don’t seem to be up to the challenge in the mathematical solution, you can just spot check some points using your CAD model. If your model is parametric, it won’t be much effort at all.

 

[jrmichler]

A gas strut is different. It has no piston or seal, just a perforated guide to keep it straight. As you push the rod in, the gas volume only changes by the volume of the rod, so the pressure changes much less than with a piston.

When I designed the gas strut lift system for the topper for my previous truck, the auto parts store made me a copy of the catalog page for gas struts. Almost all the important information was there - extended length, compressed length, and force. The one piece of information not there was the effect of temperature. The gas fill follows the idea gas law, where the pressure is proportional to the absolute temperature. The force is proportional to the pressure. I neglected to include the effect of temperature on the force, and the resulting design would not hold the top up at temperatures below 30 deg F. Other than that, it worked very well.

When designing with gas struts, you can safely assume that the force is constant over the entire stroke because the variation will be small compared to other variables. But you do need to keep the effect of temperature in mind if your door will be used on cold weather.

 

[erobz]

@Geordielad , is it a gas spring or a gas strut? @Baluncore, @jrmichler seem to have doubts (justifiable)that it is truly a gas spring. I am currently taking you at your word that it is a gas spring, which (if it's not) may be taking you places you don't need to go?

 

[Geordielad]

As I understand it the terms are interchangeable.

 

[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.