How to calculate hand force required to close a Hatch door against the force of 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.