# Building a cabinet need gas spring lbs selection

1. Apr 3, 2006

### thecrow

hi i hope this post is in the correct forum as i am new here. i am doing a home project and building custom cabinet/trunk that opens like a car hood. i am in the process of buying gas springs to ease in the closing and opening of the door/lid (and to stay open when at the top position). i realize that if the gas spring/shock is too powerful it can result in the bending of the material (while closing or preloaded), or if it is too weak then there is no assistance while opening and the lid cannot remain in the opened position by itself. what formula do i need to calculate this?

the lid will be approximately 50 lbs, but i am considering other materials that may weigh less or more. so an equation and explanation would help me a lot! i am hoping to learn from the calculation process instead of getting just a straight answer as i plan to build different cabinets of all shapes and sizes with different gas springs. the door/lid has TWO hinges so i will be using TWO gas springs in total. the lid will NOT open to 90 degrees upright.

here is the idea and some measurements.

http://ca.geocities.com/thecrow@rogers.com/cabinet.GIF [Broken]

what force lbs gas spring should be used in this project?

Last edited by a moderator: May 2, 2017
2. Apr 3, 2006

### Cliff_J

Ok, this is physics but would also fall under a area of civil engineering called statics. You may want to review trig so this makes more sense.

First, you would likely want the nitrogen charged shocks to be flipped around and pointed towards the open end and not the hinge end of the lid. The way you have them laid out is similar to most SUV hatches, but why will be clear as I type.

You need to find the X and Y component of the weight of the lid. The weight can be thought of as a single point at the center of mass - halfway up the lid if its a solid material of all one type of construction. If its 44 inches long, then this is 22 inches from the pivot. Lets say it is 50lbs.

Ok, now use trig to find the Y or vertical component of the weight of the lid based on the angle its open. When closed, its 0 degrees from horizontal, so cos(0) = 1 * 50lbs = 50lbs.

At 45 degrees, its cos(45) = .71 * 50lbs = 35.5lbs vertical.

cos(90) = 0 so there's no vertical load when open 90 degrees.

Ok, so now its similar with the shock. Lets say its 15 inches from the hinge to the base of the shock, and then with a 10 inch shock its 25 inches to where the shock attaches to the lid.

We need to find the angle of the shock so we can determine how much of its force is vertical and horizontal. At 0 degrees, its all horizontal and zero vertical, so the lid will initiall be quite heavy. You may want to lower its pivot so it is at an angle when closed to supply the vertical component strongest when closed and then loose its advantage as it goes up. For now, I'll assume its like a car hood where its very little assistance until a certain angle and then plenty until it is open.

Lets use 50 degrees open to find out what force is needed at the shock for equilibrium. So cos(50) = .64 * 25 inches = 16 inches of horizontal displacement - 15 inches for the base for 1 inch of displacement for the top of the shock from vertical.

Then sin(50) = .77* 25 = 19.25 tall at the top of the shock.

19.25/1 = 19.25 and we take the inverse tan to get around 87 degrees for the angle of the shock.

Our lid at 50 degrees, cos(50) = .64 * 50lbs = 32lbs

Since that's at 22" and our shock is at 25", then 25/22 = 1.14 and then 32/1.14 = 28lbs vertical is needed at that point.

sin(87) = .998 and then 28/.998 = 28.01lbs of force needed at the shock.

If you have really good calculus skills you could graph this out and find the optimum shock, or a nice CAD program like Solidworks would do too. But if you know enough trig (or can look it up) and a little common sense, you could make a table of every 5 degrees and see what the shock is doing in terms of net weight assistance and find out what the weight of the lid would be at each point. A little better would be to use an excel table, you could get an idea at each degree and graph it out - you just need to convert degrees to radians by multiplying by pi and dividing by 180 and then plug it into the respective function.

You'll also note that in my example, the shock is likely not even going to allow 50 degrees of lid opening because it is likely already past its open length.

3. Aug 14, 2006

### xcor635

Cliff,

Your logic is awesome and is helping me determine the weight and need for shocks, but I can figure out one calculation you are using.

How are you getting a 10 in shock, and where does this measurement come from with the picture indicated?

4. Mar 19, 2008

### Sibuna

There is an easier way to design gas springs into applications than all that trig. You can use moments of inertia and the calculations are much easier. Check out this website www.gassprings.110mb.com[/URL] it's all explained there, and you can download a spreadsheet to do all the calculations for you.

Mike

Last edited by a moderator: Apr 23, 2017
5. Apr 9, 2008

### hxlgasspring

here is a gas spring manufacturer,China HXL Gas Spring, who supply many types of cabinet gas spring ,the website is http://www.gas-springs.cn. About calculating the F1 of the gas spring, you can visit the page you will find the solution: http://cabinet-gas-spring.spring-manufacturer.com/f1.html [Broken]

Last edited by a moderator: May 3, 2017
6. Dec 7, 2010

### XYZ313

Cliff_J,

Your answer is amazing, but I did not get some points. Although a long time passed I hope you will find time and will to answer my questions; what you mean by base of the shock? I don't doubt your calculations, but if shock base is vertical distance from the hinge to the shock's lower attached point and 25" is the distance along the lid from the hinge to the point where the shock attached, the shock cannot be 10" long, as it should be sqrt(15^2 + 25^2) = 29.15" (approx.)

Also I could not understand how you get 87 degree for the angle of the shock?

Please let me know what I am getting wrong.

I would like to design spreadsheet for this purposes but I need to understand this points first.

Thanks a lot.

Last edited: Dec 7, 2010
7. Dec 8, 2010

### nvn

XYZ313: Your questions are well-founded, because the above posts are not only very confusing, but also yield the wrong answer. The correct answer follows.

Let the hinge, point A, be at the origin; therefore, the (x, y) coordinates of point A are A(0, 0). Let the base of the gas spring (the lower end) be point B, having coordinates B(bx, by). The lid centroid is point C, located at radius a1 from the origin. The upper end of the gas spring, attached to the lid midplane, is point D, located at radius a2 from the origin. Let us use the following nomenclature. The following solution assumes the hinge pins are virtually frictionless (well greased), and lie on the lid midplane.

bx = x coordinate of gas spring base.
by = y coordinate of gas spring base.
a1 = distance (radius) from hinge axis of rotation to lid centroid.
a2 = distance (radius) from hinge axis of rotation to gas spring upper end.
theta = lid slope angle, measured from the +x (horizontal) axis.
phi = gas spring slope angle, measured from a horizontal axis.
L1 = gas spring current length.
W = lid weight.
P1 = gas spring axial force, applied by lid to gas spring at point D due to gravity (with no human pushing up nor down on the lid). Positive P1 means gas spring in tension; negative P1 means gas spring in compression. If there are two gas springs, divide P1 by 2.

L1 = {[a2*cos(theta) - bx]^2 + [a2*sin(theta) - by]^2}^0.5.

phi = acos{[a2*cos(theta) - bx]/L1}.

P1 = -W*a1*cos(theta)/[a2*sin(phi - theta)].

Now, subtract P1 from the spring force in the gas spring, to obtain the net gas spring axial force applied at point D.

You will need to guard against a singularity in the above solution. I.e., do not allow phi = theta; otherwise, P1 goes to infinity. The domain of the above solution is 0 ≤ theta ≤ 180 deg, 0 ≤ phi ≤ 180 deg, and theta ≠ phi.

Try computing L1, phi, and P1 for the given example (bx = 15, by = 0, a1 = 22, a2 = 25, theta = 50 deg, W = 50), and we can check your answers.

Last edited: Dec 9, 2010
8. Dec 9, 2010

### XYZ313

NVN,

Thank you for your reply. Terrific. wow, I was looking for phi calculation. I will definately check this. Do you have spreadsheet or program (not CAD, simple)? If not I want to develop one.
I hope its OK if I contact you in case of further question.

Again, many thanks

9. Dec 9, 2010

### nvn

Hi, XYZ313. Normally when you reply to a post, please press the "New Reply" button at the top or bottom of the page, not the "Quote" button. Please do not waste PF bandwidth by quoting posts for no reason. See the PF Rules link at the top of each page:

"Replying to Posts: Please use the correct buttons! ... When you quote from a post, please delete large sections that are not directly relevant to your response, to make reading easier."

10. Dec 25, 2010

### XYZ313

Sorry, NVN. I think this is forgivable as in other forums it's OK to reply quoting. Would you let me know what kind of formula

P1 = -W*a1*cos(theta)/[a2*sin(phi - theta)]

is in physics? I mean when this formula is applied in physics? Which branch of physics? I am pretty good at math, but unfortunately horrible in physics... I think P1 should be calculated like

P1=W*a1*cos(theta)/[sin(pi/2-phi)*(bx^2+by^2) ^ 0.5]

or maybe some modification of this formula.

Again this is thoughts of person who is horrible at physics...

Please let me know what you think and guide me to that branch of physics...

Thanks

Last edited: Dec 25, 2010
11. Dec 25, 2010

### nvn

XYZ313: That is a branch of physics called Vector Mechanics: Statics.

See the last sentence of post 7. The formula you posted gives a completely different answer, and is incorrect.

12. Dec 25, 2010

### XYZ313

Thanks for your reply. I attached a picture of gas spring attached to the lid.

In correct installation the correct inputs would be (speaking of the drawing and on existing example):

Hinge coordinates - (0,0)
G=W - center mass - assume evenly distributed - 50lbs; coordinates (a1=25,0)
A(ax,ay) - coordinates of the point, spring attached to the lid; assume ay=0; ax=a2= 22;
B(bx,by) - coordinates of the point, spring attached to the base; assume bx=0; by = -15;
alpha=theta(max)=50deg
beta=phi(max)-gas spring slope angle +x
Assume L1 is gas spring's length when theta=0; fully compressed;
Assume L2 is gas spring's length when theta=theta(max)=50; fully extended;

L1=[(ax-bx)^2 + (ay-by)^2]^0.5= 27 (rounded)
L2={[a2*cos(theta) - bx]^2 + [a2*sin(theta) - by]^2}^0.5=[(22*0.6428)^2 + (22*0.7660+15)^2]^0.5 = 35 (rounded)

We are mainly interested at the weight applied at attached point of the lid when the gas spring fully extended, therefore at extended state, where alpha=theta(max)=50deg, let's calculate beta=phi(max):

beta = acos{[a2*cos(theta) - bx]/L2} = acos{[a2*cos(50deg)/L2} = 1.1530 (radians)=66 degrees

P1 = -W*a1*cos(alpha)/[a2*sin(beta - alpha)]=-50*25*cos(50deg)/[22*sin(16deg)]=-133 (rounded);

but I don't know if using "-" sign is correct, because we do not speak about the direction of vectors, if has something to do with vector direction.

Thanks

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13. Dec 25, 2010

### nvn

XYZ313: The sign on P1 is explained in post 7. Reread the explanation in post 7. Your negative sign on P1 is correct.

Your example is different from the previous example in this thread (e.g., you reversed the values for a1 and a2), but that is OK.

For your example, your answers are correct. Nice work. The correct answer in your example is actually P1 = -132 (rounded), but you got -133 probably because you rounded so much.

Notice, as stated in post 7, my solution in post 7 requires that points A and D in post 7 must lie on the lid midplane. (But you used all different nomenclature, except for point B, which makes the thread somewhat confusing.)

14. Dec 25, 2010

### XYZ313

nvn: yes, I used different nomenclature, but therefore I posted that drawing. One more thing, assuming that point A (D in your post) is close enough to midplane, calculations still true. Moreover it is located at substantial distance from the midplane trigonometry could be of help )
And I reversed values a1 and a2 intentionally, as attaching gas springs to the lid at the point between hinge and center mass makes more sense, and corrected (bx=15,by=0) to (bx=0,by=-15) which makes more sense... In gas spring design works it is accepted to add 10-15% to P1 value to assist with weight... And yes, those are rough calculations...

I appreciate you help. I am going to study Vector Mechanics: Statics... Is there any book on your mind that could be useful to beginners?

Many thanks

15. Dec 27, 2010

### nvn

I don't have a book in mind to refer you to. Unlike post 7, in the following solution, the hinge and gas spring upper end are no longer required to lie on the lid midplane. Below, I use the same nomenclature I used in post 7, except I changed the names bx and by to xb and yb.

Let the hinge axis of rotation, point A, be at the origin; therefore, the (x, y) coordinates of point A are A(0, 0). Let the base of the gas spring (the lower end pin) be point B, having coordinates B(xb, yb). The lid centroid is point C. The upper end of the gas spring is point D. Let us use the following nomenclature. The following solution assumes the hinge pins are virtually frictionless (well greased).

• xb = x coordinate of gas spring base pin.
• yb = y coordinate of gas spring base pin.
• a1 = x coordinate of lid centroid when theta = 0.
• a2 = x coordinate of gas spring upper end pin when theta = 0.
• a3 = y coordinate of lid midplane when theta = 0.
• a4 = perpendicular distance from lid midplane to gas spring upper end pin (point D); positive downward. Positive a4 means point D is below lid midplane; negative a4 means point D is above lid midplane.
• theta = lid slope angle, measured from the +x (horizontal) axis.
• phi = gas spring slope angle, measured from a horizontal axis.
• W = lid weight (positive downward).
• xd = x coordinate of point D.
• yd = y coordinate of point D.
• L1 = gas spring current length.
• P1 = gas spring axial force, applied by lid to gas spring at point D due to gravity (with no human pushing up nor down on the lid). Positive P1 means gas spring in tension; negative P1 means gas spring in compression. If there are two gas springs, divide P1 by 2.

xd = a2*cos(theta) + (a4 - a3)*sin(theta).
yd = a2*sin(theta) - (a4 - a3)*cos(theta).

L1 = [(xd - xb)^2 + (yd - yb)^2]^0.5.

phi = acos[(xd - xb)/L1].

P1 = W*[a1*cos(theta) - a3*sin(theta)]/[yd*cos(phi) - xd*sin(phi)].

Now, subtract P1 from the spring force in the gas spring, to obtain the net gas spring axial force applied at point D.

Important: The above solution is valid only if yd - yb ≥ 0. The domain of the above solution is -90 ≤ theta ≤ 180 deg, 0 ≤ phi ≤ 180 deg, and yd - yb ≥ 0.

As an example, let's say a lid is 1270 mm long, the lid midplane is 10 mm above the hinge axis (when theta = 0), point D is 100 mm below the lid midplane (when theta = 0), and the x coordinate of point D when theta = 0 is 560 mm. Therefore, a1 = 635 mm, a2 = 560 mm, a3 = +10 mm, and a4 = +100 mm. Let's say xb = 120 mm, yb = -380 mm, W = 220 N, and theta = 50 deg. Therefore, xd = 428.905, yd = 371.134. Notice yd - yb = 751.13 mm ≥ 0, which is within the valid range. Therefore, using the remaining equations, above, gives L1 = 812.173 mm, phi = 67.645 deg, and P1 = 220*400.510/(-255.512) = -344.85 N.

16. Dec 28, 2010

### XYZ313

Thanks, I found a book: Vector Mechanics for Engineers: Statics and Dynamics by F.P. Beer. Love it! It is pure Math

17. Dec 28, 2010

### XYZ313

btw: nice work!

is it possible for you to plot forces (vectors) involved?

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18. Dec 30, 2010

### XYZ313

Done,

nvn: Many thanks for your great help.

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19. Jan 2, 2011

### nvn

Nice work, XYZ313.