Brain can't work out the Maths

Submitted: Sunday, Jan 04, 2009 at 19:39
ThreadID: 64814 Views:3348 Replies:13 FollowUps:24
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G'day all,
I am building a camper trailer and forgotten how to work out the weight and balance.
Rear to axle 1.2mt load 500kg
Axle to front of box 2.3mt load 800kg
Axle to hitch length 3.5.
What's the "ball" weight going to be??????????
Cheers Mac
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Reply By: Peter_n_Margaret - Sunday, Jan 04, 2009 at 20:38

Sunday, Jan 04, 2009 at 20:38
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
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Reply By: gke - Sunday, Jan 04, 2009 at 21:07

Sunday, Jan 04, 2009 at 21:07
I agree with the answer of just under 180kg. Regards Graham
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Follow Up By: Lex M - Sunday, Jan 04, 2009 at 22:07

Sunday, Jan 04, 2009 at 22:07
Agree also
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Reply By: Richard Kovac - Sunday, Jan 04, 2009 at 23:18

Sunday, Jan 04, 2009 at 23:18
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
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Follow Up By: Richard Kovac - Sunday, Jan 04, 2009 at 23:48

Sunday, Jan 04, 2009 at 23:48


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Follow Up By: Member - GeeTee (NT) - Monday, Jan 05, 2009 at 08:09

Monday, Jan 05, 2009 at 08:09
Richard, What program did you use to draw your fancy illustrations ?
I am looking for a similar "draw" program.
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Follow Up By: Mudripper - Monday, Jan 05, 2009 at 09:48

Monday, Jan 05, 2009 at 09:48
Just copy & paste from Wikipedia!
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Follow Up By: Member No 1- Monday, Jan 05, 2009 at 11:15

Monday, Jan 05, 2009 at 11:15
GeeTee
what do you want to draw? Plans, electrical diagrams?
Visio is easy to use and is what i use to do above stuff
mm me for more info
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Follow Up By: Richard Kovac - Monday, Jan 05, 2009 at 21:10

Monday, Jan 05, 2009 at 21:10
GeeTee

Sorry I stole it as per Mudripper said.

cheers

Richard
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Follow Up By: Bonz (Vic) - Monday, Jan 05, 2009 at 23:51

Monday, Jan 05, 2009 at 23:51
Powerpoint will do the same thing too, easy to use
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Reply By: Member - Kiwi Kia - Monday, Jan 05, 2009 at 07:24

Monday, Jan 05, 2009 at 07:24
Would the guys who arrived at a ball weight of 180 kg mind enlightening me as to how you reached that conclusion ?

Thanks. KK
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Reply By: Member No 1- Monday, Jan 05, 2009 at 07:45

Monday, Jan 05, 2009 at 07:45
cut the crap...just give us the formulae...:)))))
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Reply By: Peter_n_Margaret - Monday, Jan 05, 2009 at 10:00

Monday, Jan 05, 2009 at 10:00
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
AnswerID: 342679

Follow Up By: Member - Kiwi Kia - Monday, Jan 05, 2009 at 10:18

Monday, Jan 05, 2009 at 10:18
Thanks Peter, I can see what you have done and it's an interesting 'rule of thumb'. Seems to be some interesting assumptions being made, have you found it to be fairly accurate in practice ?

.
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Follow Up By: Steve63 - Monday, Jan 05, 2009 at 11:12

Monday, Jan 05, 2009 at 11:12
It is not actually a rule of thumb, it is the principle of levers. The assumptions are that there is zero ball weight from the empty trailer (the lever has no mass) and that the load mass is a point mass. The point mass is calculated from the centre of gravity of the mass. Peter assumed it was a uniform mass which for this purpose is reasonable. If you could be bothered you could work out the effect for each piece of the trailer and each piece of the load and the effect of any density gradient for each piece you could work it out exactly.

Steve
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Follow Up By: Member - Kiwi Kia - Monday, Jan 05, 2009 at 11:26

Monday, Jan 05, 2009 at 11:26
Steve63, at 3.5 metres from ball to pivot point it will not take much extra wt well forward of the axle to significantly increase the ccw moments about the pivot point. As I said, assumptions have been made and I am interested to know if they actually stand in reality if someone has that experience. I would have been very dead many times over in the past (nearly was once !) if I had made such assumptions for an aircraft weight & balance calculation.
I know that assumptions have to be made but they are not always practical eg. where is the cg of a washer in the horizontal plane ? :-))
.
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Follow Up By: Member No 1- Monday, Jan 05, 2009 at 11:26

Monday, Jan 05, 2009 at 11:26
what do you mean "There is not a formulae as such"

F1D1 = F2D2

But I cant remember what to do when if you need to calculate say D2 when F1, D1 & F2 are known or D1 if the others are known etc etc

Need some one to show me using the info supplied by Macy

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Follow Up By: Steve63 - Monday, Jan 05, 2009 at 12:04

Monday, Jan 05, 2009 at 12:04
Hi Kiwi,
Perfectly correct for an aircraft. They are very sensitive to weight distributions. As I said it is an assumption. I was trying to say that if you knew enough information you can calculate it exactly. You can not guess how the trailer will be loaded. You will have a good idea how a milatary aircraft is loaded though. Water, food and beer consumtion will all change the ball weight. So an assumption of an even distribution gives a guide.

STeve
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Reply By: Patrol22 - Monday, Jan 05, 2009 at 10:40

Monday, Jan 05, 2009 at 10:40
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
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Reply By: macy - Monday, Jan 05, 2009 at 12:42

Monday, Jan 05, 2009 at 12:42
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
AnswerID: 342706

Reply By: Boobook2 - Monday, Jan 05, 2009 at 12:48

Monday, Jan 05, 2009 at 12:48
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 :-)



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Reply By: Member - Fred B (NT) - Monday, Jan 05, 2009 at 15:19

Monday, Jan 05, 2009 at 15:19
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.
Fred B
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Reply By: Member No 1- Monday, Jan 05, 2009 at 17:34

Monday, Jan 05, 2009 at 17:34
still cant work it out...am i dumb or what?
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Follow Up By: equinox - Monday, Jan 05, 2009 at 21:50

Monday, Jan 05, 2009 at 21:50
A little birdie told me (who happened to be a maths expert).

Answer: 177 kgs

F1D1 + F2D2 = F3D3

Where F1=Answer required.

(1.15*800 – 0.6*500)/3.5 where 1.15 and 0.6 are the halfway points from the wheel of evenly distributed masses.


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Reply By: Member - Roachie (SA) - Monday, Jan 05, 2009 at 22:18

Monday, Jan 05, 2009 at 22:18
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
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Follow Up By: Member No 1- Tuesday, Jan 06, 2009 at 07:26

Tuesday, Jan 06, 2009 at 07:26
i assumed that the box was a storage/tool box on the drawbar
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Follow Up By: Member - Roachie (SA) - Tuesday, Jan 06, 2009 at 08:26

Tuesday, Jan 06, 2009 at 08:26
Yeh, you could be right.....I just assumed "to front of box" referred to the front of the "box" as in "box trailer".......

It would be good if the original poster would come back on and explain what they are trying to achieve.....He/she did come back once and say "tanks fellas......etc"

Roachie
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Reply By: Richard Kovac - Monday, Jan 05, 2009 at 23:07

Monday, Jan 05, 2009 at 23:07
Im still try

Image Could Not Be Found

hope it works

Cheers

Richard
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Follow Up By: Richard Kovac - Monday, Jan 05, 2009 at 23:27

Monday, Jan 05, 2009 at 23:27
I will keep adding, I've got a full bottle of Red


Image Could Not Be Found

next the clacs

Richard
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Follow Up By: Richard Kovac - Monday, Jan 05, 2009 at 23:31

Monday, Jan 05, 2009 at 23:31
Whoops I stuffed up

I will be back
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Follow Up By: Bonz (Vic) - Monday, Jan 05, 2009 at 23:54

Monday, Jan 05, 2009 at 23:54
where is "L"
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Follow Up By: Richard Kovac - Tuesday, Jan 06, 2009 at 00:01

Tuesday, Jan 06, 2009 at 00:01
One L coming up

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Richard
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Follow Up By: Richard Kovac - Tuesday, Jan 06, 2009 at 00:04

Tuesday, Jan 06, 2009 at 00:04
I know ,, I know

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Follow Up By: Richard Kovac - Tuesday, Jan 06, 2009 at 00:18

Tuesday, Jan 06, 2009 at 00:18
I give up
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Follow Up By: Member No 1- Tuesday, Jan 06, 2009 at 21:13

Tuesday, Jan 06, 2009 at 21:13
come on...where is the correct diagram?
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Follow Up By: Member No 1- Tuesday, Jan 06, 2009 at 21:16

Tuesday, Jan 06, 2009 at 21:16
and formulae for us nitwits
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Follow Up By: Richard Kovac - Tuesday, Jan 06, 2009 at 21:43

Tuesday, Jan 06, 2009 at 21:43
I wasn't smart enough.. LOL or maybe not smart enough??

Cheers

Richard
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