Brain can't work out the Maths

Weeeellll, if your loads are evenly distributed, I reccon the ball weight will be just under 180kg, but that seems VERY high, so my confidence in the answer is low. Maybe someone with a calculator will have a go?

Cheers,
Peter

I agree with the answer of just under 180kg. Regards Graham

Mac

It is a lever so this will work

Theory of operation

The principle of the lever tells us that the above is in static equilibrium, with all forces balancing, if F1D1 = F2D2.The principle of leverage can be derived using Newton's laws of motion, and modern statics. It is important to note that the amount of work done is given by force times distance. For instance, to use a lever to lift a certain unit of weight with a force of half a unit, the distance from the fulcrum to the spot where force is applied must be twice the distance between the weight and the fulcrum. For example, to cut in half the force required to lift a weight resting 1 meter from the fulcrum, we would need to apply force 2 meters from the other side of the fulcrum. The amount of work done is always the same and independent of the dimensions of the lever (in an ideal lever). The lever only allows to trade force for distance.

Archimedes was the first to explain the principle of the lever, stating:

"(equal) weights at equal distances are in equilibrium, and equal weights at unequal distances are not in equilibrium but incline towards the weight which is at the greater distance."

Archimedes once famously remarked: "?a ß? ?a? ?a??st???? ta? ?a? ????s? pasa?." ("Give me a place to stand and with a lever I will move the whole world.")

The point where you apply the force is called the effort. The effect of applying this force is called the load. The load arm and the effort arm are the names given to the distances from the fulcrum to the load and effort, respectively. Using these definitions, the Law of the Lever is:

Load arm X load force = effort arm X effort force. If, for example, a 1 gram feather were balanced by a one kilogram rock, the feather would be 1000 times further from the fulcrum than the rock; if a 1 kilogram rock were balanced by another 1 kilogram rock, the fulcrum would be in the middle.

Now that didn't work ... LOL


Lever




Someone called me an idiot today ,, LOL now I can see why

Cheers

Richard

Would the guys who arrived at a ball weight of 180 kg mind enlightening me as to how you reached that conclusion ?

Thanks. KK

cut the crap...just give us the formulae...:)))))

There is not a formulae as such. Use the pretty picture from Richard.
Work out the net effect of the 2 loads using half the lengths to allow even spread of the load (rear = 0.6 x 500 pushing the ball up), (front 1.15 x 800 pushing the ball down)
Ballanced the answer by the ball (divide by 3.5).

Cheers,
Peter

Macy

Perhaps if you read this you will ensure that you get the right calculations. In any event the ball download should NOT exceed 10-15% of the Gross Trailer Mass. Hope this helps.
Cheers
Pete

Thanks for your help fellas, the 180kg ball weight is about what I got and can ujust the balance with the placement of the water tank.
Cheers Mac

IMHO all of the calculations above are incorrect and misleading UNLESS your loads are TOTAL verticle loads only touching the trailer at one point ( 1.2m to the rear and 2.3m to the front.)

In a real world the load would be spread out ( presumably evenly) and would make all of the diagrams and calculations above inadequate and WRONG. Nice high school classroom maths - but not related to the real word.

For example, if the load at the rear is 500kg across the distance of the axle to the 1.2m and it is an even load then it is really only 0.6m from the axle for these calculations ( the average distance of the load). Similarly the front would be 1.15m, again assuming an even distribution which is highly unlikley. There is simply not enough info to get this right. Also does it include the tare wieght?

The actual correct calculation is quite complex using the integral of the weight from 0 to 1.2m and the integral of the weight from 0 to 2.3m. Not something that can be worked out here.

I strongly suggest that you measure the towball weight, after it will be quicker and have the bonus of being right! Otherwise it could be dangerous!!!

A bathroom scale, piece of wood and brick should do the trick to half the load. And given that the loads are single points, you can even use the calculations suggested above :-)

If in doubt, take the trailer (loaded) to a weigh bridge; disconnect the trailer. With the jockey wheel down, place the trailer on the weigh bridge with the trailers wheels just off the bridge. check the weight. That will give you a pretty accurate towball weight. Yes I realise that the jockey wheel is further back than the towball hitch, but this is a reasonable method to check your maths.

still cant work it out...am i dumb or what?

I admire the aptitude of all those mathemeticians who have responded with various formulae etc and I certainly don't intend to comment on any of that techo stuff.

However, the thing that baffles me is what this trailer is going to look like, given the dimensions provided:

The box is 3.5 meters long (what's that, about 11+ feet?) and the "A" frame is going to be 1.2 meters (approx 4 feet?).

The axle is located approximately 1/3 of the way from the back to the front of the box.

To me, this is much too far towards the rear of the box section of the trailer. This will "look" wrong and would explain why the proposed ball weight will be quite high @ around 180 kg (assuming that is a fairly accurate calculation, which I have no reason to doubt).

Is it just me?........... or is this proposed trailer design just plain wrong?

Roachie

Im still try

Image Could Not Be Found

hope it works

Cheers

Richard