Just another how Heavy is my 4x4 thread

[Daan]

solidworks Daan,

just some toys some friends are knocking up - more bling than its humanly possible to gaze upon without going blind

Im still using Autocad and crayons - Ive spotted a problem though, the wax from the crayon is difficult to rub off the screen

I'll give it a try- you will be surprised what I can face.

Try tipex.

Daan

[02GF74]

A bit of camber or a dip in the surface can throw the corner weights right out, as could uneven tyre pressures across an axle.

just curious but how can tyre pressure affect the weight? (only way I can see this is one tyre is so low that the whole thing is tipping over to one side but the effect on wieght would still be minimal?) off to scratch my head andthink about it.

... whilst you're at it, some explain in simple terms how spring rate/adjustment can affect wheel weight .

[LR90]

Good result Andy.

If you raise one end while its on the scales you can calc where the COG is (inc how high).

[DaveSIIA]

just curious but how can tyre pressure affect the weight? (only way I can see this is one tyre is so low that the whole thing is tipping over to one side but the effect on wieght would still be minimal?) off to scratch my head andthink about it.

The sidewall of a underinflated tyre will compress more than one where the pressure is correct, causing the wheel to sit slightly lower at that corner. This will cause the spring at that corner to be slightly less compressed (or the other three to be more compressed), reducing the measured wheel weight at that corner.

Agreed, the effect is secondary but it is measurable with the sort of kit MogLite used.

[pugwash]

Good result Andy.

If you raise one end while its on the scales you can calc where the COG is (inc how high).

whats the calculation for that then? can you work this out with a hi lift and a weighbridge?

[b101uk]

the difference in the L/R mass on each axle is the mass of the diff and its offset on the axle, if you look at the image of the mass on the digital readout, the front R and the rear L have the diff nearest to them (beam axles and 4 point scales don't work to well for cross axle mass due the the high unsprung mass of the axle its self)

[Turbocharger]

The sprung/unsprung balance won't show up in a static test AFAICS. Much more likely it's because Andy found his springs lying in a hedge somewhere.

Andy - very tidy result, and I find the 25/25/25/25% (+/- 5%) weight distribution more impressive than the low overall weight. It's a testament to a well thought-out design, good result.

How does a stock Ninety or your Ibex fare on similar scales (for comparison)?

[MogLite]

The sprung/unsprung balance won't show up in a static test AFAICS. Much more likely it's because Andy found his springs lying in a hedge somewhere.

Andy - very tidy result, and I find the 25/25/25/25% (+/- 5%) weight distribution more impressive than the low overall weight. It's a testament to a well thought-out design, good result.

How does a stock Ninety or your Ibex fare on similar scales (for comparison)?

Thanks John

I haven't had chance to weigh the Ibex or the RaRo classic or P38a I have at my disposal. But if I find a spare hour or two after Seven Sisters I will.

I'd be very interested in the calculations for centre of gravity. I'd love to prove that a portal vehicle when designed with a clean sheet can rival a modified vehicle.

[Tonk]

i'd be interested to weigh a vehicle before and after a playday, just to see how much mud can weigh down a vehicle, might get your scales dirty though

[dollythelw]

heres an idea Andy,

if I book the manitou we can have a rollover angle session?

[LR90]

Thanks John

I haven't had chance to weigh the Ibex or the RaRo classic or P38a I have at my disposal. But if I find a spare hour or two after Seven Sisters I will.

I'd be very interested in the calculations for centre of gravity. I'd love to prove that a portal vehicle when designed with a clean sheet can rival a modified vehicle.

Look here for info on how to calc COG height.

[Mark90]

Look here for info on how to calc COG height.

Ohh A-level maths flash backs

[b101uk]

“I'd be very interested in the calculations for centre of gravity”

Well the fact the front/back and left/right 4 point loads are near the same says that the CoG is ~central to the wheelbase (Y) and ~central wheel track (X) the only thing you don’t know it the height of the CoG from the ground (Z)

If the CoG is at 50% point in the X,Y and Z plane then for every 1deg of angle 1.1111111111111111111111111111111R% of the mass is transfer to the lower contact points from the upper contact points, if a it tips over at 45.0deg then that means the CoG height (Z) = the distance from the outer contact point to the centre point in the plane the object is being tipped

If the rise in mass per 1deg of tilt is >1.1111111111111111111111111111111R% then the CoG is higher and it will tip over at <45deg, if the rise is lower than <1.1111111111111111111111111111111R% then the CoG is lower and it will tip over at >45deg

E.G1

Now lets say you get 1.5474% rise in mass per 1 deg then the CoG would cause the object to fall over at 50 / 1.5474 = 32.31deg if you then take 90deg / 32.31deg = 2.7855153203342618384401114206128,

So 100 / 2.7855153203342618384401114206128 = 35.90%

So if the outer contact points were 1.85m apart then 1.85m – 35.90% = 1.18585m witch is the height of the CoG in the Z plane.

E.G2

Now lets say you get 0.674822222222222222222222222222% rise in mass per 1 deg then the CoG would cause the object to fall over at

50 / 0.674822222222222222222222222222 = 74.09 deg

if you then take 90deg / 74.09deg = 1.214738831151302469968956674315

So 100 / 1.214738831151302469968956674315 = 82.32%

So if the outer contact points were 1.85m apart then 1.85m – 82.32% = 0.32708m witch is the height of the CoG in the Z plane.

E.G3

Now lets say you get 1.1111111111111111111111111111111% rise in mass per 1 deg then the CoG would cause the object to fall over at

50 / 1.1111111111111111111111111111111 = 45 deg

if you then take 90deg / 45deg = 2

So 100 / 2 = 50%

So if the outer contact points were 1.85m apart then 1.85m – 50% = 0.925m witch is the height of the CoG in the Z plane.

In the above you will see the numbers 50 90deg & 100 witch are constants, in other words using 100 can be equivalent to using 90deg or 100% of mass or using 50 to represent 45deg as 45deg is 50% of 100, 50 is also the point at ware once the mass of the CoG goes beyond this point it will fall over, etc, etc. and the above can be used in X or Y vs. Z CoG calcs

The point being is gravity is a constant as is the mass of the object and to a relative degree the dimensions of the wheels other than tyre deflection and suspension travel or design.

100% mass / 90deg = 1.1111111111111111111111111111111R

75% mass / 67.5deg = 1.1111111111111111111111111111111R

50% mass / 45deg = 1.1111111111111111111111111111111R

25% mass / 22.5deg = 1.1111111111111111111111111111111R

12.5%mass / 11.25deg = 1.1111111111111111111111111111111R

6.25% mass / 5.625deg = 1.1111111111111111111111111111111R

3.125% mass / 2.8125deg = 1.1111111111111111111111111111111R

1.5625% mass / 1.40625deg = 1.1111111111111111111111111111111R

0.78125% mass / 0.703125deg = 1.1111111111111111111111111111111R

0.390625% mass / 0.3515625deg = 1.1111111111111111111111111111111R

Of cause this is b101uk’s dyslexic simpleton dumb idiots version devoid of needlessly complex equation witch you wont find on the net or in a textbook, witch should mean you wouldn’t need an A level or O level or even a CSE, cos I don’t!!!!! Besides anyone with a winch on the 4x4 should know all this anyway, because for every 1 deg of slope 1.1111111111111111111111111111111R% mass is put as load onto the winch in addition to all the other loads.

b101uk

[pugwash]

B101 thanks for the reply- my only query is that your equations look like they rely on a symetrical linear mass distribution which i don't have on my vehicle! could be a bit of an issue.

think the jeep way will work better for off road beasts.

cheers

Jim

[Mark90]

Or Jez's way

[dollythelw]

I like the mindless use of machinery to find definitives, it works for bus and forklift manufacturers

[Mark90]

Aye, as well as being much more fun than doing sums you know for sure how far it'll go before falling over

[Lara]

Well if you ask me, the front / rear and side to side weights are better than most race cars as a percentage of total And certainly nothing that you will feel in this trucks usage

Well done Andy, a lovely looking machine and the weight is very good for such a vehicle.

Look forwards to seeing it in the flesh.

Lara.

[b101uk]

“B101 thanks for the reply- my only query is that your equations look like they rely on a symetrical linear mass distribution which i don't have on my vehicle! could be a bit of an issue.”

well if you equate “linear” to equal something that is “constant” and repeatable then what you say is true, BUT Gravity is a “constant” at ~9.81m/s/s, angle is a “constant” that ramps up at a “linear” rate between 0 and 90deg, something will always fall over as soon as the CoG passes lower outer contact point so transfers >50% of mass beyond this point, the ratio of 0% to 100% vs. 0deg to 90deg is 1.1111111111111111111111111111111R:1

witch is a “constant” that ramps up at a “linear” rate, and say for slight changes in geometry from suspension type and how it moves and tyre deflection the dimensions of the 4x4 remain in essence the same (wheel base, track vs. CoG height) and finally the mass of the 4x4 remain in essence the same (accepting fuel use etc and obvious changes)

So if everything is a “linear and/or constant” and only 1 thing changes then the only thing that can happen is a linear answer that changes at a constant rate vs. other inputs?

BUT

If you apply the same formula that I used but rather than calculate the CoG height vs. the “X” plane (left/right) but you apply it in the “Y” plane (wheel base front/back) it tells you the % of distance between the wheels in the “Y” plane the CoG is situated, as the same laws apply front to back as left to right

&

If you repeat the same formula for left & right, front & back to get the 4 different results and you plot them on a diagrams (top/side/back) then draw lines from the relevant points depicting the outer contact points to the 4 relevant CoG placements giving the CEP (circular error of precision) and the point the lines intersect is the true CoG without error as each angle etc will have been done twice left & right, front & back and each formula yields a CoG height and % offset of wheel base or tyre outer edges.

I don’t think I am doing to badly for cooking up a formula on the fly!

But as said “I like the mindless use of machinery to find definitives, it works for bus and forklift manufacturers” is best

b101uk