And considering that the vast majority of forklifts I train upon are pretty much rated between 4000 and 6500 lbs, 1"=approx 100lbs., sounds good.

I think that if I was brought into litigation, I would have plenty of support from the industry to back up my teachings.

What I am concerned about, though, are students not paying attention, even though they were instructed, when acquiring these lengthy pallets (loads), and end up hurting themselves, or wrecking the forklift.

Knock on wood, all's well!

Dan

Appreciate all you say as I too train operators.

In regard to the steer wheel lift off I explain to operators that if that is happening they are lifting 40% to much - 1400 kg when they should only be lifting 1000 kg. And that the result of overloading is that the forklift will tipover even more readily.

Consider the 78" carpet prong mentioned - would probably have a 40" load centre. The 100 lb per inch rule would say reduce the capacity by 1600 lb.

So for a 2000 lb forklift the predicted capacity would be 400 lb. The actual capacity would be over 1400 lb.

But for a 10000 lb forklift the predicted capacity would be 8400 lb but the actual capacity would be only 7100 lb. At 8400 lb it would be around 20% overloaded.

My suggestion for you if you beleive that you need a "X lb per inch" would be to relate it to truck size say:
2000 lb - 3500 lb 50 lb/in
4000 lb - 6500 lb 100 lb/in
7000 lb - 8000 lb 150 lb/in
9000 lb - 12000 lb 200 lb/in...

You'd have to do your own risk assessment and develop your own values for the rules keeping in mind that if there was a serious incident that resulted from someone using your rules you might be drawn into litigation.

John,

It all depends on which 'people' you are quoting who should have a complete understanding of this concept.

I am not quite sure of the educational levels of the operators you are training in Australia however, many of the operators that I train here in Canada are doing this work for one main reason...they could not tolerate sitting in class all day when they were attending school. Several of them have not even completed high school (secondary), let alone pass their math classes.

When I ask how much is 5000-600=, some cannot even provide the correct answer. This does not apply to all my students, but a fair number of them.

When attempting to deliver this information in the lowest common denominator, the K.I.S.S. principle is mandatory. Adding, multiplying and dividing goes beyond the intelligence level of many. They need a relatively quick and easy means of approximating the weight of a load based upon different load centers. If an instructor is only going to confuse the **** out of them, than what's the point of teaching these materials. As instructors, we are there to teach them forklift dynamics, not math. And I can only vouch for my students, that 7 of 10 would fail to comprehend your formula, thereby, not putting these calcualtions to work when an oversized load is being acquired.

State all the laws and principles you want, and I fully wholeheartedly am a zero tolerance guy when it comes to forklift safety, but try teaching the 'book' to a bunch of guys working in some dumpy warehouse in eastern Kentucky. Chances are, they will pick the load up regardless, and when the rear end starts to lift, then they will put the load down.

I am not in arguement with you at all. I am simply stating that if this forklift safety stuff becomes too complex for them, they will just continue to do their own thing, their way, and the time spent educating thse operators would be for not.

Now, if these 'people' you talk about are on the management level, that is a different story. Management generally possesses a better comprehension for these matters, and math, but how many times do you end up training the operators, and a shirt/suit/tie is nowhere to be found?

So as I said earlier...5000x24/36=HUH?

RPM7 - excellent

Forklift safety demands that people understand the issues completely.

The degradation in capacity of the forklift tynes (forks) is faster than for the forklift because the formula only considers the distance from the fork heel. A fork rated at 2500 kg @ 600 mm will be rated at 2500 * 600/900 = 1670 kg.

However for a counterbalance forklift with a distance of 400 mm from the centre of the drive axle to the fork face of 400 mm, the formula is 2500 * (600+400)/(900+400) = 1920 kg.

For that reason some forklift manufacturers actually fit forks with a higher rating than the forklift. For example if they fit forks rated at 3000 kg @ 600 mm in the case above, then the fork rating at 900 mm is 3000 * 600/900 = 2000 kg which is close to the forklift rating as a whole.

Finally in regard to the 100 lb/in rule that started this discussion. That rule of thumb only applies to forks with ratings of 3000 - 4000 lb (with an error in the range of up to 20%), or coounterbalance forklift trucks with a distance from the centre of the drive axle to the fork face of 16 inches (400 mm) with ratings of 4000 - 5500 lb (with an error in the range of up to 20%). SO MY RECOMMENDATION IS TO NEVER USE SUCH A RULE.

If you want to be conservative you use the fork strength related rule of "rated load" x "rated distance" / "larger centre of mass distance".

BUT ALWAYS KEEP IN MIND THAT AT FULL HEIGHT, ESPECIALLY FOR LIFT HEIGHTS OVER 4500 mm THE SITUATION MAY BE MORE COMPLEX AND YOU SHOULD CONTACT AND GET ADVICE FROM THE MANUFACTURER/ SUPPLIER.

I find that the rating of a forklift has to take three aspects into account.
1st: the strength of the forks
2nd: the strength of the mast & finally the counterweight of the forklift itself.

If the forks are rated for 2500kg @ 600mm (the formula for them will be (2500 x 600)/(new load centre).
The mast is rated similarly.
For the amount the forklift can counterweight using moments the formula is the same as some guys have mentioned above.
For example. A forklift rated at 3000kg @ 600mm Load centre with front overhang (dist of front axle to front of fork) of 200mm. Has to lift @ 1000mm Load centre the formula is.
(3000*(600+200))/(1000+200).

In a nutshell what I'm saying is that the most a forklift can lift will be what the forks are rated to and after that you can use the moments formula as long as it's values are less than the forks.

There are two ways to look at this.
1) Lose 100lbs./inch
2) (Capacity x LC)/new lc

Now, which formula do you think many operators will find easier to figure out? Load center gets removed by 6"=600lbs. lost, from 5000=4400lbs. The guys I train even freak at these simple formulas, nevermind having them figure out 5000x24, divided by 30= HUH?

The usual "de-rating" formula is; A truck with a 4000lb capacity @24" load centre is lifting a load with a 30" load centre -
Multiply 4000x24 (our capacity x load centre) =9600. Divide 9600 by new load centre (30) = 3200lbs our new capacity

Thanks for the info, this helps to clarify the issue as most people only talk about counter balance load centers.

I ran across a salesman that had a factory downrating graph fo the specific model that allowed me to see when I should consult the factory for a revised spec plate.

IMPORTANT NOTE: As previously stated these caculations should only be applied to situations where the load is not being handled at anywhere near full height as their are other factors that need to be considered near full height including side stability, the fact that the floor may not be flat and level (a slope of up to 2% is allowed), et cetera.

Firstly you need to read and reflect on all the issues mentioned above in respect of counterbalance forklifts, and take heed of advice about getting a rerating from the manufacturer because ratings are complex.

For a reach truck you need to measure the load distance L from the outrigger wheel centres to the rated load centre with the forks fully extended - the overturning moment related to the load is then the rated load by L.

Then determine the new distance L distance = L1 at 30 or 36 inches load centre.

The adjusted load is then the RL * L / L1.

As an example a Clark NPR 22 is a single stroke reach truck and has a rated load of 2000 kg at 24 inches. The distance from the load centre to the centre of the outrigger wheels when the forks are fully extended is 28.6 in.

For a 30 in load centre that distance becomes 34.6 ins and the rated load = 2000*28.6/34.6 ~= 1650 kg.

And for a 36 in load centre that distance becomes 40.6 ins and the rated load = 2000*28.6/40.6 ~= 1400 kg.

Similarly for a Clark NPR 15 D, a double stroke reach truck (and because of the greater fork extension has a lower rating even though the reach truck is heavier) has a rated load of 1350 kg at 24 inches. The distance from the load centre to the centre of the outrigger wheels when the forks are fully extended is 50.67 in.

For a 30 in load centre that distance becomes 56.67 ins and the rated load = 1350*50.67/56.67 ~= 1200 kg.

And for a 36 in load centre that distance becomes 62.67 ins and the rated load = 1350*50.67/62.67 ~= 1090 kg.

Does anyone know the formula to calculate the downrating on a 30-36 inch load center for a reach truck? I would apreciate your help.

Sorry, I forgot. Go to rightline dot com and you will find one.

Most of the attachment manufacturers have easy to use online capacity calculators. You just plug in the numbers. I've used them all and they all come out the same. Go to www.rightline.com and there is one right on the home page. I use it a lot and it is pretty accurate.

You can roughly do it in your head, 2000x24 divided by 78.This gives a slightly lower number than the true capacity.You should always work on the capacity at the maximum lift height.This also gives a safety margin.

Johnr-j is correct in his advice.

However if you are only going to carry the load at a low height - that is much lower than the maximum lift of the mast - then the formula is as follows:

Rated load x (rated distance plus the distance from the fork face to the centre of the drive axle) = modified rating x (modified distance plus the distance from the fork face to the centre of the drive axle).

For example a 2000 kg @ 24 in forklift plus 16 in from fork face to centre of drive axle would have a revised rating of 2000 x (24 + 16) = revised rating x ( 78 + 16) giving a rating of 850 kg.

Note that this revised rating includes the effect of the carpet prong on stability. Assuming the prong is supported on the forks, and the prong weighs 100 kg then conservatively the device can carry a roll of carpet 850 - 100 = 750 kg.

It is still recommended that the manufacturer be contacted re a new rating as they can give the rating at full mast height.

There are many factors that determine the rated capacity of any lift trucks beyond simple mathematical calculations of inch pound ratings, etc., etc. Other factors that determine the capacity are:

1. Amount of mast tilt
2. Physical limits of: a.) mast & carriage bearings b). Tilt cylinder strength, c.) Tire compounds & tire sizes d.) fork section (W &T) as well as length. & type (std. taper or full taper) used, e.) limitations of any attachment, f.) lift cylinder column strength.

Best advise let the manufacturer do the work & the lawyers won't get more $$$$.

4 metre carpet boom.......LC now 78"
Truck rated capacity 2000lbs@24"

LC extended by 54"

reduce capacity 100lbs/per inch.....5400lbs

Its only a 2000lbs rated truck to start with!

A fork truck is basically a ratio & proportion machine. Therefore , I don't see the formula you offer working very well. It can reasonable be said that each time you double the Load Center the capacity is reduced by about 50%. This is not perfect, but, it is close enough for rough calculations. One must also take into consideration the extra weight of any special forks or attachment above the weight of normal forks. As said above, this is not perfect but it will help you from making a gross capacity error.