Calculation help!!!! Physics was too long ago.

[fbaly]

Alright everyone. I am trying to figure out how much down force is on my rear hatch.
I drew up a diagram of what I am dealing with.
Anyone good at this?



<img src="http://farm6.static.flickr.com/5260/5551326003_f44e5764dd.jpg" width="434" height="500" alt="188937_774999038661_11011175_42169161_597968_n" /></a>

[mvperini]

why

mike

[asianflava]

Calculation schmalculation, just use the hit and miss method.

Start with say a 100lb lift and see if it is enough or too much. It also depends on where they are mounted. 100 could be enough mounted one way, not enough mounted another.

I got mine from NAPA and asked if I could try different sizes. He was fine and let me return the ones that didn't work out.

[Corwin C]

I'm missing where you want to apply the force. Is this for hatch struts or force on the hinge or ???

On this type of calculation all of your vertical forces times the horizontal distance from a common reference point HAVE TO SUM UP TO ZERO if it remains stationary. Keep track of your positives and negatives in both force and distance for this to work (upward positive, downward negative etc.)

Just from your diagram, if it takes 40 lb (upward) to hold the hatch stationary, and the hatch weighs 70 lb (downward), then the hinge is applying a force of 30 lb (upward) to balance everything.

Now some assumptions ... (all the engineers are collectively shaking their heads and saying "shame, shame")

If the 40 lb force applied to the hatch is lets say 50" horizontally from the hinge you can find the center of gravity of the hatch in that position by using algebra (if you move the hatch you will need to re-calculate because the horizontal distance will change). I'm going to use the hinge as my reference point.

(Force at hinge times distance from hinge) plus (force at end of hatch times distance from hinge) plus (force (weight) of hatch times center of gravity distance from hinge) equals ZERO.

(30lbs * 0") + (40lbs * 50") + (-70lbs * x") = 0

Solve for x (I had a professor in college that would write "ATAMH" on the board ... And Then A Miracle Happens)

x = (40 * 50) / 70

Meaning that the center of gravity of the hatch in this position is about 28.57" from the hinge.

If you apply other forces (say a pneumatic hatch strut), then you have to include them in your calculation. I know that this doesn't answer your question, but hopefully it will jog your memory or at the very least help you give us the information we need to find the number you are looking for.

[Yota Bill]

if you are trying to calculate the right weight rating of a strut, you could do all the math...or you could do it in less time and much simpler, with a stick and a bathroom scale.
Place the stick in the position you want the strut to be in, so that it holds up the hatch, while pushing against the scale. That will tell you the total weight being applied at that angle. Divide in half if you are using two (though that seems to make them a bit too strong, I'd suggest going about 5-10% less then half)

Ive done that on auto hatches and various doors on heavy equipment, and it gets you pretty damn close. May not sit stationary or balanced at a center point, but does that really matter? I dont see a need for it to be anything other then opened or closed.

[eamarquardt]

if you are trying to calculate the right weight rating of a strut, you could do all the math...or you could do it in less time and much simpler, with a stick and a bathroom scale.
Place the stick in the position you want the strut to be in, so that it holds up the hatch, while pushing against the scale. That will tell you the total weight being applied at that angle. Divide in half if you are using two (though that seems to make them a bit too strong, I'd suggest going about 5-10% less then half)

Ive done that on auto hatches and various doors on heavy equipment, and it gets you pretty damn close. May not sit stationary or balanced at a center point, but does that really matter? I dont see a need for it to be anything other then opened or closed.

I like the above technique!

The stick should be perpendicular to the suface of the scale to get a good reading.

Cheers,

Gus

[john warren]

man you guys think too hard

[Rvankeur]

I printed this out and left it for Ed. I'm attaching what he left on the computer this morning.

It's been 25 years since statics class for me (and I never did earn that EE degree afterall), so I'm just the messenger.

[myoung]

I too am unclear about why you want to know how much downward weight is at the edge of the hatch.

If you are considering a mechanical device to raise and lower the hatch, then the task might be unnecessary. I had a similar problem in calculating the force required to raise the pop-up top on my TTT.

The mathematics is really simple as others have shown but in searching for a linear actuator to accomplish the raising and lowering task, there were only to models to choose from: one was fast and could lift 40 pounds and the other was slower and could lift 150 pounds. The 40 pound one would have been marginally acceptable, but to be on the safe side, I opted for the 150 pound version. Problem solved.

[fbaly]

Alright so does this make more sense.
I fixed some errors.
And with the help of Rvankeur's friend I did some changes.

<img src="http://farm6.static.flickr.com/5264/5556022507_15e1862ba5.jpg" width="434" height="500" alt="door2" />

Thoughts?
I would say picking a mounting point between 12 and 24 inches would be a good idea.
Now to find one that fits!

[fbaly]

I didn't mean hindge points I was trying to say gas struts points.

[Yota Bill]

you know, with a stick and a bathroom scale, you would have had your answer in less time then you spent drawing that picture.

[Mightydog]

Two approaches to math:

If you want to find the height of the building using a barometer you can:

1) Stand on the roof of the building. Throw the barometer to a point exactly on the horizon. Measure the distance from the bottom of the building to the barometer. This gives the horizon distance at the top of the building, thus giving its height above the ground.

2) Find the owner of the building and say, "...hey, I'll give you this expensive barometer if you tell me how tall your building is."

The stick and bathroom scale is my way of doing math--kind of like method #2. Less chance for error.

[eamarquardt]

I have two slide rules. One for each hand!

Cheers,

Gus

[fbaly]

Yes true I could have saved myself the head ache and used a bathroom scale. But My door is drying right now and not on the project.
So.... This was the only way I could calculate the force.

I also don't have a place to buy gas struts where I could return them.
So I need to make sure I calculate them correctly so I dont have to return to mcmasters.