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High School Maths 07/06

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HIGH SCHOOL MATHS REVISITED

Some time ago I came across another 60’s rule which makes load calculations simple.

The force required to move a weight up a slope with an angle B is derived by the formula:-

Force A required to move 2.5T  =  Weight X Sin (Angle B)

Fire up the trustee computer and behold you can calculate the following for a loaded up Pajero

            Force required = (2400 kgs X sin (15 degrees)  =  621 kgs

The problem with this formula is it requires a computer or a scientific calculator and generally neither of which we take away on the weekend

Now there is a very simple 60’s formula which gives a sufficiently accurate result and requires nothing more then a mobile telephone for the calculator and a half empty wine bottle.

            Force required = Weight   X   (Angle B   /   60)

Now as all the Maths students will tell you the formula can be rewritten as

Force required = (Weight   /   60)   X   Angle B

If your NM weighs approx 2400kgs loaded up for the weekend away

The results are then

Physics

 

60’s Rule

 

2400 * Sin(15)

621 kg

2400 / 60 * 15 degrees

     600 kg

2400 * Sin(25)

1014 kg

2400 / 60 * 25 degrees

    1000 kg

2400 * Sin(30)

1200 kg

2400 / 60 * 30 degrees

1200 kg

2400 * Sin(40)

1543 kg

2400 / 60 * 40 degrees

1600 kg

2400 * Sin(50)

1838 kg

2400 / 60 * 50 degrees

2000 kg

 

It starts to get a bit out at the high numbers but would only cause to us to err on the side of caution so I think it is acceptable.

Now all you need to know is how far your eyes are from the ground so when you hold up your half empty wine bottle (no, we don’t drink and drive) as a level you can guess the distance to get the angle.

The angle of the slope requires careful consideration as the world is neither flat nor comprised of even slopes. For example a well known rock shelf at Brooklyn is a combination of individual slopes of 15 and 45 and 70 degrees each of which need to be considered in the overall exercise. Next month we will look at the condition of the surface and its effect on the load.

The foregoing was calculated by an accountant and not an engineer. Accordingly it is offered for comment and discussion but no warranty as to the reliability or accuracy of the information is given or implied. 

The earth isn’t flat and the surface also changes to add to our amusement. Sometimes it is bitumen, sometimes grass, and for 4WDers usually dirt but also often sand or mud. Ask anyone who has been bogged and they will soon tell you it’s a lot harder to move a vehicle if it is bogged in mud than it is to move one on dry ground. So we need to make some allowance for the condition of the surface in our calculations of the load.

The following I believe is commonly used by a large organisation which specialises in drab green or camouflage trucks around Australia. The process is to add to the calculation from last month a fraction of the total weight to be moved.

Therefore for our 2400kg Pajero we can get the following:-

Surface

Bitumen

Grass

Gravel

Sand

Mud

Factor

1/25th

1/7th

1/5th

1/4th

1/2

Weight 2400 kg

100 kg

345 kg

480 kg

600 kg

1200 kg

By combining the table from last month and this one we get the following calculation of the potential load:-

 

Bitumen

Grass

Gravel

Sand

Mud

15 degrees

721 kg

966 kg

1101 kg

1221 kg

1821 kg

25 degrees

1014 kg

1359 kg

1494 kg

1614 kg

2214 kg

30 degrees

1300 kg

1545 kg

1680 kg

1800 kg

2400 kg

40 degrees

1643 kg

1888 kg

2023 kg

2143 kg

2743 kg

50 degrees

1938 kg

2183 kg

2318 kg

2438 kg

3038 kg

These are basic numbers and the practice of adding 10% as a safety margin has considerable merit. All of the calculations in this exercise have been done in kilograms as the winch data normally refers to kgs per layer on the drum. The maximum load for the last two layers on my 4000kg winch is 3115kg and 3333kg. Both of which are below the 10% margin applied to the Mud column at 50degrees

A panel beater has also suggested all loadings should be applied down the centre line of an NM/NP vehicle level with the floor pan/“chassis structure”. He suggests the vehicle may suffer “chassis” distortion when loadings as low as 1500kgs are either not down the centre line or one wheel causes resistance to recovery. His suggestion was to watch the gaps around doors if in doubt but I suspect by the time the gap visibly changes it is too late. It begs the question should we always use equaliser straps with NM/NP vehicles if the recovery point is to one side?

All the calculations have been based on a stationery recovery vehicle. The dynamics change dramatically if the recovery vehicle is moving such as the basic snatch strap recovery.

The foregoing was calculated by an accountant and not an engineer. Accordingly it is offered for comment and discussion but no warranty as to the reliability or accuracy of the information is given or implied.

Now your test …

Out on your weekend away you come across a Nissan Patrol loaded to the hilt GVM 3040kgs with a 1200kg camper on the back stuck half way up a muddy hill with a slope of 30degrees. What is the potential load?? First correct answer at the meeting wins the prize. Just to get in ahead of some, no you cannot leave it there!

Bill Parker


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Last Modified on Tuesday, 23 September 2008
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