Load cells and weighing multihull builds

You confused feet with inches, so the setoff is 12 times greater.
But you are right, in the case of a very small angle the difference regarding the weighing process is neglectable.
I did the sketch to show this principle and in order to produce a readable sketch I had to take a larger angle and a greater ratio between VCG and boat breadth. This suggests a far larger difference, as it would be in reality. Sorry, it was not my intention to misdirect you.

Nevertheless, my point, that weighing all "corners", one after the other, will not give you a sum which is the boats weight, stands, because the reasons are different. If you look at my table weighing examples above, aren't they convincing?

 

A load cell compresses only microns, so deflection has a huge impact on the measurements. Weighing half the boat, while lifting it from the outermost edge, is the only accurate and simple method. Also, while lifting, the boat has to be supported from the opposite outermost edge. Otherwise, complicated corrections for the overhangs will need to be introduced.

 

How come this isn't like measuring corner weights on a car? At ride height (dwl for a boat) you place the scales under each wheel. I would think he'd want to know the corner weights at dwl since that tells him the weight distribution.

 

I disagree. Focusing on the "outermost edges" may be is fine for a small boat if it provides points of action at these edges to support its weight or link it to the scale like a bow eye or a rub rail.
A boat wheighing a few tons will mostly not provide such points because they would have to bear large loads without deform, break out or something similar.
And there is no need doing corrections about overhangs.
You have to arrange two parallel horizontal "lines" perpendicular to the center line of the boat. To avoid damage caused by to much pressure from the supports, these "lines" must have some width to provide sufficient area. So the lines are slim rectangulars in reality. You don't know exactly, where the pivot line ist within the mentioned line width. The critical point is not to alter at all this arrangement by moving the scale from line A to line B. Not an easy thing to do. E. g. as the scale has some thickness, it has to be replaced with a dummy of same thickness and vice versa.
Of course these measurement lines should be as far apart as possible, but this is limited by the shape and construction of the boat and may be other cirumstances.

 

Load cells like piezoelectric, the most common type, work by generating a voltage when they get deformed. The strain is measured in microns, so a minimal deflection of the structure will give false readings. There is no way you can move a load cell from one location to the other without disturbing it. The only setup that will work is to have multiple cells, or to lift half the boat from the extremes. A boat this size should be relatively easy. Fallguy says it will be a total of only abou 8,000 lbs. That is well within the limits of two small chainfalls. A simple scale built from a beam can be used to lift and weigh the boat. You guys are getting really complicated. Beam scales are very accurate and easy to build in the field. Unequal-arm beam balance (steelyard) https://www.europeana.eu/en/item/2020801/dmglib_handler_mcdsc_3645025

 

thanks for catching my math error, I did the correction and then the correction factor would be 96/2000 or 4.8%..

the test you did is not the same as the boat test, but it is similar and shows a flaw in my thinking, to a point

when you lift the table leg 2", the opposing (short side) leg is supported by the scale, and there is no correction factor applied for the Vg change the long way

in order to test the table weight; you'd need to support the structure on the other side and then apply some correction factor for the Vg

First, let's make sense of the scale. It is clearly weighing most, but not all of the other side. If we divide the table into sections, each 1/4 is 2.7kg. When we lift the table and weighed it, we'd expect to get 5.4kg, except the other legs are doing some of the work, how much is calculable.

Making some crude assumptions, let's say table is 24x48,Vg is 32. Skipping the math, we get a long way correction of 14%. So, the other side takes 14% of the weight off the scale. Here that would mean 5.4kg•86% or 4.64kg calc vs 4.72 measured. And I am guessing the measures; not the weights.

what happens when you weigh the table with only the short side supported at the same height as the scale? I will make a guess that the scale reads 14% less than 2.7kg or roughly 2.32kg.

As for the bucket, some of the bucket's load is getting carried by the other two table legs as well. Same 14% could be applied, but that would be erroneous because the Vg does not move based on the table height. But also calculable. It is rather late and I am a bit tired. Maybe if you are willing to test my theory or supply real measures, I'll make an attempt at why the bucket doesn't give its full weight. If we use 14% less, we'd get 7.042kg on the scale, but I believe that 14% is a bit off. However 7.042 calc vs 7.16 measured ain't awful far off considering all the estimates here

 

As you lifted the leg of the table and put it on the scale, one of the other legs will have come off the floor. Ie as I mentioned before, 3 point contact establishes a plane that is not coplanar with the floor.

The two legs that will be in contact with the floor will be the leg on the scale and the one diagonally across from the scale, (hereinafter called the diagonal leg)
and a third leg which will carry some weight. The third leg will be the one that the center of gravity is slightly toward and off the line from the scale leg to the diagonal leg. Ie most of the weight will be carried by the scale leg and the diagonal leg.
The table is "teetering" around this line. The third leg is only carrying a very small piece of the weight to counterbalance the moment arm that the CG makes with the diagonal line.

So from your calculations (units kg)
The table weighs 10.80
The scale reads the weight of the scale leg at 4.72. (ignore the very slight transfer of weight from the scale leg to the diagonal leg due to the height difference for now)
The diagonal leg scale , if the table is relatively weight symmetric should read 4.72.
This brings the total to 9.44 and the third leg carrying weight to 10.8 - 9.44 = 1.3

So the 1.3 kg, is the force required to keep the table from rotating around the diagonal line. Ie the CG is not precisely over top of the diagonal line.

So as a check. Grab a bathroom scale and adjust the height as your new scale and put it under the diagonal leg. This will eliminate any weight transference due to the height differential between the scale leg and the diagonal leg.
Take a 1 kg moveable weight and move it around the table until two legs come free, ie the third leg is not carrying the weight of the of the influence of the center of gravity
The scale weights should be then the table weight (10.80 + 1 kg)/ 2 = 5.9 each

 

I completely agree.
I had to leave the office yesterday after I took, prepared and posted the pics. So there was no time for me to explain or interpret the results.
My goal with this post was to shatter the idea it would be possible to find the weight of a boat by weighing 4 "corners" separatly and simply add up the 4 weights.

 

I always assumed there would be a correction factor.

 

Interesting, but this would not be possible with a large boat; raising the diagonal to the scale would result in all sort of unpredicatables @gonzo has said.

What confused me was a 3 leg supported table reading 0. That would mean the 3 legs were actually lifting the table off the scale and were too high I believe.

Gonzo's beam test is surely the best, but still requires a correction factor as the Vg moves. The beam is hard to come by is all and the boat needs to get up quite high and then balance gently on the scale. Not likely easy.

I may have to rent three additional scales to get an accurate weight and then I can determine corrections and Vg. Return the 3 scales and understand the impact using a single scale for the remaining work. Or get closer to done then weigh which goes against the general concern. Or suck it up and spend more and more money.

 

If your beaching keel is strong enough, you could take your single scale and try to orient the scale on a floor jack and position it along one keel. Slowly jack it up, and eventually the boat will become "loose" in it supports. If it is the front one, then move your jack back. Move the jack around until it appears that the one hull is only supported by the jack. This will weigh the one side. If you think that the beaching keel is not strong enough, you could add in say a 6 foot length of 6 x6 to
spread the load
Do the other side the same way and you will have the overall weight

For the CG
Again if the beaching keel is strong enough, you could just place a 6 x 6 beam crossways under the keels at a strategic location, and move this beam until the boat is balanced. If you think that the width of the 6x6 is to large WHERE IT CONTACTS the ground, you could reduce this width to make it narrower to gain a bit more accuracy. Ie make it able to tilt easier

 

After lifting the diagonal leg to the same height as the weighbridge and supporting it in this place, I was trying to show how depending the scale reading is in relation of support of the other two legs. I supportet one of them with a more or less fitting wood piece and used a wedge to be able to adjust the height of the last support. To my surprise the first attempt leads to a reading of 0.00 kg at the scale, which means the remaining force on it equals to less then 10 gram mass. Then I took the pic. Pulling the wedge a tiny bit would have lead to some reading (up to 5 kg).

 

followed the first bit; the keel is only 4" off the ground, so the jack, scale and a 6x6 will be about 8-9" total or say about 5" more; maybe doable

the second bit makes me nervous because the sacrificial timber on the keel is a bit narrower (this in hindsight is less than ideal), but the keel is say 4,5" wide and the sacrificial is say 3.5" wide...so a 6x6 at 5.5" wide is only about 20 square inches...5000 pounds is 250 psi...too much for me

 

You only need to lift the boat a fraction of a inch. There isn't enough tilting to make a significant difference in the measurement.

 

please explain what you mean, thanks
Do you mean for measuring the aft section with a beam across or across the beaching keel?