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

[Geordielad]

TL;DR Summary

I am trying to calculate the hand pressure required to close a Hatch door which is held in the open position by 2 gas springs. I have done all the trig and geometry and force calculations except for the Hand Pressure, there I am confused.

 

[russ_watters]

Please show your calculations. Also I don't see the force applied by the springs.

 

[Geordielad]

Hi Russ,
Thanks for the response. The formula I have to balance forces is F*E = S*J from which I get F = (S*J)/E
weight of door 41 kg 410N COG is 0.804 (M) Moment arm E is 0.448 (M)
F = (410 * 0.804)/0.448 735.8 (N) total 367.9 per spring extended spring.
F = (410 * 1.163)/0.477 999.6 (N) 499.8 compressed spring

 

[erobz]

TL;DR Summary: I am trying to calculate the hand pressure required to close a Hatch door which is held in the open position by 2 gas springs. I have done all the trig and geometry and force calculations except for the Hand Pressure, there I am confused.

View attachment 332339

Maybe you could just get rid of the circles and replace with dimensions with increase font size all over?

The gas springs ( if they are truly gas springs), are going to apply a variable force, because 1) the gas acts as a non-linear spring and 2) the geometry of the applied force is changing as the door closes. Are they really gas springs or are they gas dampers?

The force is going to vary as a function of the angle. Also how the force is applied; is it always normal to the door, or more likely does it start out vertical and become more "normal" as the door approaches the closed position?

 

[Geordielad]

Thanks Erobz,
I will update my sketch. Yes they are Gas Springs. I need to calculate hand force required to open and to close the hatch door.

 

[erobz]

Thanks Erobz,
I will update my sketch. Yes they are Gas Springs. I need to calculate hand force required to open and to close the hatch door.

Ok. And think about how the hand is applying the force (in which direction). I think the hand will more or less apply a force vertically throughout the range of motion.

 

[Geordielad]

 

[erobz]

Can you balance torques about the hinge using these variables I have labeled?

Please see LaTeX Guide for learning how to post mathematics on the site.

 

[Geordielad]

Thanks Erobz,
I think it is something like Torque=2Fs*(Moment)+Fh*e - 0.5(e cos theta)W=0
but I cannot get it to balance.

 

[erobz]

Thanks Erobz,
I think it is something like Torque=2Fs*(Moment)+Fh*e - 0.5(e cos theta)W=0
but I cannot get it to balance.

You should resolve all forces into components perpendicular to plate to examine the sum of torques, not just the weight of the door.

Also please try to use LaTeX Guide to write your equations. Its clean format that makes communicating the math efficient

 

[Geordielad]

Please excuse my ignorance but I dont understand "perpendicular to plate"

 

[erobz]

Please excuse my ignorance but I dont understand "perpendicular to plate"

"perpendicular to door" I should have said. The torque looks at components of forces perpendicular to the moment arm of each force. You did it correctly with the weight, but did not do it for the other forces acting on the door.

 

[erobz]

Just in case you have gotten stuck; The dashed components of the forces are what contribute to the torque.

 

[Geordielad]

Thanks,
I think I should have written 2 ((c cos β) Fs) + (e cos θ) FH – 0.5 (e cos θ)W = 0

 

[erobz]

Thanks,
I think I should have written 2 ((c cos β) Fs) + (e cos θ) FH – 0.5 (e cos θ)W = 0

Not quite...but close. Check which trig function defines the desired spring force component, and check your signs. You've defined a counterclockwise torque as positive (I think...). Are your torques consistent with that convention?

 

[erobz]

Also, use Latex to reply with math, it's going to become indispensable for communicating the upcoming steps.

Follow the link: LaTeX Guide

 

[Geordielad]

Thanks I think this is what you are asking for 0.5 (e cos θ)W + (e cos θ) FH – 2 C(cos β) Fs = 0
I don't have Latex I used word.

 

[erobz]

Thanks I think this is what you are asking for 0.5 (e cos θ)W + (e cos θ) FH – 2 C(cos β) Fs = 0
I don't have Latex I used word.

Latex is a part of the site.

Like if you want to say "$\sin \beta$"

instead of " cos β " ..hint,hint

You type the code:

## \sin \beta ##

if you want to show a nicely formatted equation in the center of the reply all by itself: You type the code:

 $$ e F_H \cos \theta + \frac{1}{2} e W \cos \theta - 2 c F_s \sin \beta = 0 $$

And you get:

$$ e F_H \cos \theta + \frac{1}{2} e W \cos \theta - 2 c F_s \sin \beta = 0 $$

as the formatted math text.

Like I said, follow the directions in the LaTeX Guide I've now linked three times...

or you can find the link in the bottom left corner of the reply box

 

[Geordielad]

I used 0.5W because I was taking the weight force from the COG. which would be half of e. Am I missing something?
Apologies I do not know how to load Latex and MathJax.

 

[erobz]

I used 0.5W because I was taking the weight force from the COG. which would be half of e.

No, I flopped the variables on accident. My bad, I edited the equation.

Am I missing something?

What you are missing is that the component of the spring forces perpendicular to the door is $2F_s \sin \beta$

Apologies I do not know how to load Latex and MathJax.

You type the code directly into your reply here, the website does the rest for you. Hit reply to this message. You will see exactly what I typed, to get the formatted math you see.

 

[Geordielad]

Erobz thanks a million. you have been very patient.

 

[erobz]

Erobz thanks a million. you have been very patient.

Hold on, that is just the tip of the iceberg. We haven't even really begun to untangle this. There are a few significant challenges ahead if you are looking to find the force $F_H$ as a function of $x$?

Even if you are satisfied to do it numerically using a scale drawing to find all the angles at dozen or so points, we still have to talk about the spring force.

 

[Geordielad]

Yes I need Fh but I am really struggling to use Latex. I must spend some time reading the guide.

 

[erobz]

Yes I need Fh but I am really struggling to use Latex. I must spend some time reading the guide.

Well, If you get stuck\ need help, just talk me through the issues you are having.

Don't be afraid to test it out with a few extraneous posts.
For an example of how it works, go to post #18, and copy entirely what is inside either box labeled code and paste it in to a new reply. Hit preview in the top right of the formatting header. It will render the code. Hit post reply and the code will be rendered in the new reply.

 

[Geordielad]

$$ e F_H \cos \theta + \frac{1}{2} e W \cos \theta - 2 c F_s \sin \beta = 0 $$

 

[Geordielad]

Okay I think I understand I write the equation as code and then preview it to get a proper math style format.

 

[Geordielad]

$$ e F_H cos \theta + \frac 1 2 e W cos \theta -2 c F_s sin \beta =0 $$
is this correct ?

 

[erobz]

$$ e F_H cos \theta + \frac 1 2 e W cos \theta -2 c F_s sin \beta =0 $$
is this correct ?

Correct equation.

For the latex code with the trig functions its like

\sin

you forgot the "\" in front.

 

[erobz]

A next step may be to just solve it for $F_H$.

After that, finding $\sin \beta$ as a function of $x$ is pretty straight forward using the Law of Cosines. It's when you get to finding $\cos \theta$ as a function of $x$ it gets a bit involved... Finally, we would find the spring force as a function of $x$.

Just waiting on you to tell me when you get stuck, and show me what you've figured out. Here is a diagram to get you started on the trig.

 

[Geordielad]

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

 

[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.