## Weight distribution to four points

## Weight distribution to four points

(OP)

Hello Everyone,

brand new here and need some help.

I am having a problem with our machine building department and all the engineers at our location are process, automation or software guys, so can't work this problem out.

The details are too hilarious to list in the starting thread, and I want to keep it a bit professional, so suffice it to say, there has been a screw-up.

I am looking for feedback and a proper formula for the problem at hand.

Here is thread that was close to what I am looking for: thread404-117973: Weight Distribution on Four Wheels, but not exactly, although it was a help.

Situation description: A rigid rectangular machine (600 kg total weight) on four feet(1 at each corner) needs to be lifted (ca. 3 inches) by lowering the wheels (straight, screw down jack mechanism) that are positioned at each corner. Assume that there is no suspension and that the wheels are steel.

The CG is not in the middle but rather in the rear right quadrant.

The problem: the wheels do not go down all at once, but rather one at a time.

Is there an accepted formula (or procedure) for determining how much weight(how much lifting force) is (needed) on each wheel (jack)?

I know it is not the total weight divided by 4 because the CG is off center. And I think it is more of a total divided by 3 since "lifting" one corner turns it into a 3 pressure point item instead of 4 but again that is just my feeling.

Any help will be appreciated.

Thanks.

brand new here and need some help.

I am having a problem with our machine building department and all the engineers at our location are process, automation or software guys, so can't work this problem out.

The details are too hilarious to list in the starting thread, and I want to keep it a bit professional, so suffice it to say, there has been a screw-up.

I am looking for feedback and a proper formula for the problem at hand.

Here is thread that was close to what I am looking for: thread404-117973: Weight Distribution on Four Wheels, but not exactly, although it was a help.

Situation description: A rigid rectangular machine (600 kg total weight) on four feet(1 at each corner) needs to be lifted (ca. 3 inches) by lowering the wheels (straight, screw down jack mechanism) that are positioned at each corner. Assume that there is no suspension and that the wheels are steel.

The CG is not in the middle but rather in the rear right quadrant.

The problem: the wheels do not go down all at once, but rather one at a time.

Is there an accepted formula (or procedure) for determining how much weight(how much lifting force) is (needed) on each wheel (jack)?

I know it is not the total weight divided by 4 because the CG is off center. And I think it is more of a total divided by 3 since "lifting" one corner turns it into a 3 pressure point item instead of 4 but again that is just my feeling.

Any help will be appreciated.

Thanks.

## RE: Weight distribution to four points

## RE: Weight distribution to four points

Here is the problem in a bit more detail:

The reason I am looking for "the" correct procedure(s) for calculating this is that we can't just go to that department with , "maybe you could, or maybe you should..." because they are at a different business unit (location) and/but this is something they never tackle.

Then they can get the right "size" of jacks ordered.

The problem is, they ALREADY built it wrong.

Yes, that's right, their "calculations" if they did any were WAY, WAY off and the damn unit won't go up.

Actually, their manager slipped and told me that, "there is no way any human being could turn the crank to lower the wheels".

And...the actual regulations here regarding lifting equipment are "Safety factor 2,5 times required capacity compared to calculated need".

So at the minimum, each wheel should be able to lift 375 kg, even if they just divided weight by 4.

## RE: Weight distribution to four points

## RE: Weight distribution to four points