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Torque, Force, and a truck cover

  1. Dec 23, 2003 #1

    zeb

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    Hello...

    I am doing a little retrofitting with my Tonneau cover and my truck, but I need some help. I have cut my cover length-wise down the center, and need to rework the gas springs that hold the doors open. Here is the situation:

    Each door weighs 56 pounds, and is 32" wide. They are hinged at the edges so that the 32 inches extends out into the center of my truck bed. The gas springs (compressed nitrogen gas filled shocks .. i.e. open assist door props) are mounted 5" inside the hinge and 2.75" below the hinge. When the door is open, the angle of the door is ~70° (measured about the hinge), and the angle of the spring is ~78° (measured about the spring hinge point). There are two springs per door. The question is how much force does each gas spring need to exert in order to keep the door open? I understand that Force = Torque / Lever Arm, and that the lever arm is always measure perpendicular to the line of force and through the axis of rotation. So in a case like this, the lever arm becomes really small and the required force become really large. Here is a diagram to help see what I am doing (12k):

    [​IMG]

    I have calculated 83.318 lbs of force, per gas spring, but I don't trust the script I wrote to calculate it, and I can't try the springs before I special order them.

    Any thoughts?
     
  2. jcsd
  3. Dec 23, 2003 #2

    krab

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    Just balance the torques. The torque due to the weight of the door is 56lbs times 16 inches. (I am assuming that the door is evenly weighted so the centre of gravity is halfway between the hinge and the edge.) Since you will use 2 springs, each will need to provide half this, for 448 inch-lbs.

    The torque exerted by the spring is

    [tex]\tau=448\mbox{ in-lbs}=rF\sin\theta[/tex]

    We want F. [tex]\inline{\theta}[/tex] is the angle between F (the spring direction) and r, r being the vector from the hinge pivot to the spring pivot. This can be considered as comprised of 2 angles:

    [tex]\phi_2:\ \tan\phi_2=2.25/16[/tex]

    and

    [tex]\phi_1:\ \tan\phi_1=(2.75-2.25)/(16-5)[/tex]

    r is just given by

    [tex]r^2=16^2+2.25^2[/tex]

    Putting this all together, my answer is F = 150 pounds.
     
  4. Dec 23, 2003 #3

    zeb

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    150 lbs. when the gas spring is at what angle (relative to the closed position)? My script get 150 lbs. when the door is closed and the angle of the spring is nearly parallel to the door (near 0°). The important figure is how much force is needed to hold the door open when the gas spring is at ~78° (from closed).

    Doesn't [tex]\inline{\theta}[/tex] need to be 90° (by definition of "Lever Arm") and r the vector between the pivot point and the intersection of the line of Force at 90°?
     
  5. Dec 23, 2003 #4

    krab

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    Right. I thought you wanted the maximum force. This occurs when the door is at the closed position.
    Right. The angle is [tex]\inline{\phi_1}[/tex], which is 2.6 degrees.
    I can't calculate that force unless I have more info. Need to know the thickness of the door and where its CG is.
    Comes down to the same thing. Your definition of r is my [tex]\inline{r\sin\theta}[/tex].
     
  6. Dec 23, 2003 #5

    zeb

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    okay, good... we're on the same page.
    I'm, not sure of the exact CG, but it's pretty close to the center of mass (i.e. 16" from the hinge point). Also, the thickness of the door varies, but for this application there is plenty of room for error. The gas springs are specified in increments of 10 lbs, and the manufacturer recommends calculating an extra 10% of force for "tiring" over time (leave that out of the calculation). Also, I have designed the mount points of the gas springs to be adjustable so that I can fine tune the amount of force by changing the distances and angles. So, regarding the door thickness, is the value for distance "below" the top of the door (being in line with the hinge point) good enough for this calculation? I can't really get the thickness of the door at that point.
     
  7. Dec 23, 2003 #6

    krab

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    I get something like around 60 pounds.
     
  8. Dec 25, 2003 #7

    zeb

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    hmmm... 60 pounds each doesn't seem right. Looking back at your first post:
    shouldn't r be [tex]r^2=5^2+2.75^2[/tex] in your equation? The spring pivot point should be the point closest to the door hinge (take a gander at the image I linked to in the first post)... would this change your answer?
     
  9. Dec 28, 2003 #8

    krab

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    No. The spring pivot point does not rotate with respect to the hinge, so the torque there is irrelevant.

    BTW, I forgot that you have 2 springs per door.So the answer at fully open is more like 30 pounds per spring.
     
  10. Dec 29, 2003 #9

    zeb

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    I'm positive that 30lbs/spring is not right. Before I started this thread, I had calculated 30lbs/spring and bought 4 of 'em... They hardly affected the weight of the doors, and they certainly didn't hold them open. So I returned the springs and started to rethink my calculations... which is why I looked here for help.
    You are absolutely right, BUT there is an imaginary point that exists which marks the intersection of the line of force and the line that runs perpendicular to the line of force and goes through the door hinge (the axis of rotation). Lets call this point P. The lever arm in the equation for force is the distance from the door hinge to the point P. The tricky thing is that as the door opens, the angle of the force (relative to the door) changes, thus the imaginary point P gets further from the door hinge as the door opens until the door reaches an angle of 49° after which the imaginary point P starts to get closer to the door hinge. At this point where the door is at 49°, the needed force per spring (if my calculation are correct) is 78.514 lbs. As the door opens to 70°, the needed force per spring increases (since the lever arm is now getting smaller) to ~83.318 lbs. All of this is based on the concept that the lever arm must always be perpendicular to the (constantly changing) line of force.

    Does this make sense?
     
  11. Dec 29, 2003 #10

    krab

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    That's probably because you need 150 pounds per spring to open the door.
    Well, maybe your diagram doesn't accurately reflect the situation. Anyway, could you draw an improved diagram? The 5 inches at the bottom appears as long as the 2.75 inches above it. When the scale is wrong, the angles will be wrong, and it is hard to make intuitive guesses.
    Look, Zeb, we went through this. This thing you call "lever arm" is identical with my [tex]\inline{r\sin\theta}[/tex]. The advantage in my technique is that you don't have to draw extra lines, drop perpendiculars, or anything; just find the distance from the hinge point to the force point, call that r and find the angle between r and the force (spring) direction. It would be a good exercise for you to go ahead and prove this for yourself: Draw my r ([tex]\inline{=\sqrt{16^2+2.25^2}}[/tex] from hinge pivot to free end of spring pivot, make a right triangle with this r and the point you call P, and you will see that from P to the hinge pivot is [tex]\inline{r\sin\theta}[/tex].
    You seem to be forgetting that as the door opens, gravity has a smaller and smaller "lever arm" about the door hinge: at 90 degrees, the door would be balanced i.e. the force required to keep it open is zero. If you find this last statement to be incorrect, please give me a better diagram, because it means the centre of mass is not where I think it is.
     
  12. Dec 29, 2003 #11

    NateTG

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    This is good fun and all, but wouldn't it be easier to look up the replacement part, and continue from there?
     
  13. Dec 30, 2003 #12

    zeb

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    Re: NateTG
    um... what replacement parts? The door was originally one piece with only two gas springs needed to open it. Once I cut it down the center, the cover became a whole different beast. The two cover pieces now hinge on the side of the truck bed instead of the front of the truck bed, there now needs to be a total of four springs instead of two, and the new springs are positioned at a much different angle than they were originally. Too many factors have changed.

    Re: krab
    There's no need to get upset, there is nothing personal about anything I have said, it's just a simple debate... and since my money and time is on the line, you can't blame me for wanting to get this right (not to mention wanting to understand why it works). I am not criticizing you in any way, I am just one of those people who needs to understand things and doesn't like to take anything for granted or blindly accept other peoples calculations. I sought out this forum to get some help understanding the math involved. I appreciate your help and time.

    I understand that the needed force for the springs is very large when the doors are closed (or near closed), but when I unlatch the doors, I will be lifting at the end of the 32" pushing straight up which doesn't take very much force because of the mechanical advantage. As I open the door more, the springs start to take over for me.

    Yes, the diagram is not to scale but I am not deriving any angles from that diagram, nor am I letting any intuition get factored in.

    Yes, I see how you get your r and I like your method better. I am not sure why you comprised two angles for theta though. I understand the need for [tex]\phi_2[/tex], but why [tex]\phi_1[/tex]?

    I have not forgotten about gravity, I was hoping to get past the mechanics first and then factor in gravity; so I should have mentioned that without gravity, the needed force for the springs would increase past 49°. So, to factor in gravity, would we figure the weight of the door to be an function of the CG and the open angle?

    The door is made of fiberglass covering an aluminum frame. I would estimate (I can try to get an accurate figure tomorrow, but I think it will be difficult) that the CG is about 10-14 (call it 12) inches from the hinge and about 3/4" below the hinge. The door at it's thickest is probably just over 2". Also, the hinge is located at a "corner" of the doors' cross-section, meaning that the door would be balanced when the open angle is just a little over 90°. But obviously, the door won't open past 70°.
     
  14. Dec 30, 2003 #13

    krab

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    I was using the halfway mark of 16 inches. That makes the forces even smaller. I get 113 lb. per spring to open the door, 25.13 lb. to hold at 70 degrees. Here are the formulas.

    [tex]
    F={12\cos\gamma+0.75\sin\gamma\over 2.75\cos\beta+5\sin\beta}\ {56\over 2}\mbox{ pounds}
    [/tex]

    where [tex]\inline{\gamma}[/tex] is the door angle, and [tex]\inline{\beta}[/tex] is the spring angle. They are related by

    [tex]
    \tan\beta={16\sin\gamma-2.25\cos\gamma+2.75\over 16\cos\gamma+2.25\sin\gamma-5}
    [/tex]
     
  15. Dec 31, 2003 #14

    zeb

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    I totally understand this:
    And I understand that [tex]{12\cos\gamma+0.75\sin\gamma}\ {56\over 2}[/tex] is the distance from the hinge to the line of force (gravity) acting on the CG, times the door weight... right? Would this be the "Torque" in the equation for the Force?

    I don't get the bottom part of the formula for F: [tex]2.75\cos\beta+5\sin\beta[/tex]. If my assumption that the top of the formula in the Torque, then this bottom line should be the "Lever Arm"... right? Could you explain this part? I don't see what 2.75 has to do with [tex]\cos\beta[/tex] or what 5 has to do with [tex]\sin\beta[/tex].
     
  16. Dec 31, 2003 #15

    krab

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    Look at the line that joins the bottom pivot of the spring to the hinge. Its length is [tex]\inline{\sqrt{5^2+2.75^2}}[/tex]. The lever arm is this length times the sine of the angle between this line and the spring.
     
  17. Dec 31, 2003 #16

    zeb

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    okay, I get this for the lever arm:
    [tex]{2.75\over \cos\beta}+(5-2.75\tan\beta)\sin\beta[/tex]
    when I work out the math which is the same as your formula, except that yours is simplified (I'm not that savvy with trig).

    Also, now that I have seen your formula for the Force, I have adjusted mine to accommodate the CG and our figures are very close. The question I have now is why didn't it work out right when I used the 30lb. springs?!? That should've been enough to hold the door open at 70°. I have checked over my figures and measurements, and nothing is inaccurate. Oh, and I have a better figure for the CG: 14.25" from the hinge and .5" below the hinge.

    Any ideas on what we have overlooked (except for the dwindling effects of gravity as the door approaches 90° which only hurts our calculations)?
     
    Last edited: Dec 31, 2003
  18. Jan 1, 2004 #17

    zeb

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    Alright, I get it now. Everything makes sense. The discrepancy between our formulas (aside from method) was that as I said before, I wasn't accounting for gravity, which is what I think you were doing by calculating the distance (as a product of the open door angle) from the hinge to the vertical line running through the CG. This line is the same as the horizontal component of the CG (14.25" now, formerly 16" and then 12") when the door is closed, which is why our figures matched at that point. The problem with my calculation was that I didn't change that figure as the door opened. The other differences, like I said, are our methods. I wrote my script in PHP so that I could generate a diagram on the fly of the distances, angles and the force required. In order to get the same calculations to work for both the actual output and the diagram, I needed to work with points, slopes and lines in a coordinate system. Anyway, I have adjusted my Force equation to account for the changing distance from the hinge to the CG.

    Having said that, I have also figured out why the 30lb springs didn't work. Firstly, the open angle might not be exactly 70°, it might be somewhere closer to say, 68° or 67°, (it's hard to measure exactly)... which by my calculations, requires exactly 30lbs per spring for 68° and 31lbs. for 67°. Of course, if the door opened any further, the springs would hold it there. But since the door only gets harder to keep open as it closes. the springs have less and less of an effect on it's weight, hence my experience "feeling" the weight as I released the door prop.

    Actually, what I should have been seeking from the beginning, is the force required to take over the weight of the door after I manually open it a few inches. Let's say I only want to open the door 4" before the springs start to "kick-in". The angle of the door when the edge is opened 4" is 14.5°. At this angle, the required force is ~87.5 lbs per spring. As soon as the door gets opened that far, it's force acting on the springs equals the springs own force and the door begins to open on it's own.

    Sorry for putting you through the hassle. I do like your formulas better and your explanations have helped me understand the concept even more.

    So, here's another question... If I get 90lb springs, how much force will be required to open the door (with my hand) at it's edge? 90lb springs will "kick-in" at 13.25° and when the door as been raised 3.66" at it's edge. From closed to this angle, the force on the springs drops quickly from 150lbs. to 90lbs. but the fact that the spring is there, reduces the amount of force required to "help" open it from the edge. How do I calculate this? Let's assume I'll be pushing straight up for the 3.66". Having this figure will help me decide at what angle I want the door to open on it's own... If I use 90lb. springs, the door will probably be hard to pull closed once opened. So it's probably better to manually open the door further before the springs kick-in. This way, the required force to close it will be less. I need to reach a happy medium between these two "manual" forces.
     
  19. Jan 4, 2004 #18

    zeb

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    Does anyone have any ideas about this? Krab?
     
  20. Jan 6, 2004 #19

    krab

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    Here's a graph.
    Graph
     
  21. Jan 6, 2004 #20

    zeb

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    ???
     
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