Calculating axle weights with a tag axle Motorhome (1 Viewer)

[Geo]

The vast majority of replies seems hooked on "Load" and sharing the weight.
Now assuming for one moment DBK.s and the rest are all 100% correct it still doesn't answer the question asked
No one has asked what that load will be, or which axle will carry what , shared or otherwise
Please answer the question
For the purpose of the Formula and Calculations ONLY ------Which axle is used ? (see diagram) and the same question asked by Jim in the original post.

Use anything other than the center of the rear most axle such as the center of the two or the front tandem axle and you will in effect be incorrectly CALCULATING The wrong overhang figure
Over simplified BUT place a 100 Ton down force on the tow ball and the vehicle WILL swivel on its rearmost wheels, and its the distance from the swivel point and the tow bar that is needed for the calculation regardless of what weight it turns out to be or whether its shared or not

 

[Phillybarbour]

This is an interesting post given I have a tag axle, although I have a very large payload.

Been in transport (hgv) many years and always used the mid axle point as the calculation, but never really questioned if that was correct just thought it was.

 

[Jim]

Over simplified BUT place a 100 Ton down force on the tow ball and the vehicle WILL swivel on its rearmost wheels, and its the distance from the swivel point and the tow bar that is needed for the calculation regardless of what weight it turns out to be or whether its shared or not

Even I can understand over-simplified..

I am working on a question and answer forum where the OP or others can vote a post as "Best Answer" For me this is the best answer .........so far

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[Geo]

For loading your probably right
But for the purpose of this particular calculation it has to be the fulcrum point as it works very similar to a torque multiplier
you add another foot to the overhang by going to the middle of the two axles, or two foot to the center of the 2nd axle, and your formula and calculations will be miles off

 

[andy63]

To calculate axel loads precisely you surely need all the centre of gravity. . Ie the unladen vehicles and then an equivalent c of G for the load (which we know can vary ).. whenever you have them it is a simple matter to determine actual axel loads...
The load adder to the rear of the rear axel would figure in determining the overall c of g of the vehicle to determine those axel loads...
But if you want to know the effect of adding that load to the rear axel in terms of safety and overload you use the manufacturer load rating for that axel... In this case the rear tag and as @Geo has described work out the effect of the leverage of that load..
My thoughts anyway... but ill keep looking ...find it interesting..
Andy

 

[hilldweller]

Actual Axle Weights. (at the time)
1. 1750kgs
2. 1380Kgs
3. 1240Kgs.
Jock.

So 1380 and 1240. 140kg difference in 2620kg so not a lot.

Most "tags" I've seen have pretty massive overhangs which minimises any errors in taking the mid point to do calculations.

Otherwise it's now pretty obvious that vehicle loading, suspension design and set up makes an accurate generalised set up impossible.

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[thehutchies]

Rather than arguing about the theory, couldn't we get some real-life measurements and calculate the values with typical loads to see how much difference it makes?
Any owners out there with a handy tape measure?

 

[2657]

Rather than arguing about the theory, couldn't we get some real-life measurements and calculate the values with typical loads to see how much difference it makes?
Any owners out there with a handy tape measure?

And a handy dynamic axle weighbridge, not many publicly accessible ones about.

I was thinking exactly the same thing myself, everything else is just theory.

 

[andy63]

Rather than arguing about the theory, couldn't we get some real-life measurements and calculate the values with typical loads to see how much difference it makes?
Any owners out there with a handy tape measure?

Certainly not arguing lol..but interested and keen to learn...
But calculating center of gravity and their effective location isn't really practical and that's what I think you need to determine axel loads and even more complicated if you are looking at wheel loading...
A weigh bridge would most probably be the answer..
Andy

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[funflair]

To try to convince myself either way I had to forget the fact that this was a vehicle with suspension as that really does complicate the calculations as in, what if the rear axle is essentially along for the ride and has lower rated or no springs then all the load will be on the middle axle so that would then become the fulcrum, so then look at a nose down attitude where the middle axle is bearing more of the load so this must move the effective fulcrum or pivot point forward of the rear axle.

So I then took the suspension out of the equation and looked at it as a supported beam where the two rear axels share the load but as soon as something is hung on the overhang it uses the rear axle as the Fulcrum/pivot point, so my conclusion was that in an ideal scenario it is indeed the centre of the rear axle that should be used for the calculation. But what difference will it make?

As I said I have taken the wheels off and given the motorhome simple supports, but lets still call these axles it is 8 metres long with 1 metre between the rear axles and a 1.5 metre overhang from the rear axle. Ultimately another 0.5 metre when we include the towball load.

A/ Shows the Axle loads with a uniformly didtributed mass and is worked out by rotational moments about the front axle treating the two rear axles as one at a point midway between them CCW moments =4000x4=16000 CW moments have to be equel so 16000/6 (6 is the distance to centre of rear axles)=2666.6kg take this from the original 4000kg and the front suppport loading is 1333.4kg. I have divided the 2666.6 by 2 to get 1333.3 equelly shared by the two axles.

B/ In this scenario we have introduced the towbar load out at 2 metres from the rear axle, it pivots about the rear axle so again a calculation of moments tells us that this has reduced the front support load by 61.5kg which we must transfer to the rear axle along with the extra 200kg from the towbar, in reality this load is shared by the two rear axles through the suspension compliance.

C/ This scenario is assuming that the fulcrum or pivot point is actually in the centre of the two rear axles so increases the overhang figure in the calcs to 2.5 metres and the point of laod for the CCW moments to 6 metres. Now when we do the new moments calculation the front axle load is reduced by 83.3kg and this must transfer to the rear.

So what difference does it make? well in this example about 20kg so not a lot, but if you use the centre of the two rear axles you will be underestimating the front axle load so be carefull. If you are that close to the limit on any axle and then it would be back to the weighbridge for a ticket.

This may not be the definitive answer but congratulations anyway as it is the end, and I wont be asking questions.

@Jim I don't expect any prizes for simple answers but thanks for asking as i have not done this in a long time.

Martin

 

[thehutchies]

I think the original post was concerned with the differences between the three potential measuring points for twin rear axles. Using part c of Funflair's post above, it just means using distances of 2, 2.5 and 3 metres in the calculation.

 

[2657]

I would have thought that the uniformly distributed load could be treated as a constant in all the calculations and ignored as it is the difference caused by the rear overhang load that we are interested in.

I would think also that using the pivot point in 'B' is correct but the calculation takes no effect of some of the load (moment) being shared by the two axles forward of the pivot point( ie the forward rear axle would also be lighter). I do not know without delving into serious deep thought how to work this out(perhaps ratio's).

Perhaps using the midpoint as a pivot and sharing the extra load between the two rear axles will provide an acceptable approximation.

As stated before some experimentation will be needed to provide a definitive answer, unless someone can still do the sums.........not me!!

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[olley]

Imagine you are able to lift and lower each pair of axles separately, as we lift the rear axle the pivot point moves slowly to the next axle as it takes all the weight. As we lower it back, the pivot moves back towards the rear axle, but not all the way as they share the load. So my belief is under normal conditions its around the mid point between the two axles.

Ian

 

[Jim]

We need a weighbridge, a tag axle van, a calculator and a motorbike, then we can work find out precisely how much weight is added and work backwards

 

[funflair]

I would have thought that the uniformly distributed load could be treated as a constant in all the calculations and ignored as it is the difference caused by the rear overhang load that we are interested in.

I would think also that using the pivot point in 'B' is correct but the calculation takes no effect of some of the load (moment) being shared by the two axles forward of the pivot point( ie the forward rear axle would also be lighter). I do not know without delving into serious deep thought how to work this out(perhaps ratio's).

Perhaps using the midpoint as a pivot and sharing the extra load between the two rear axles will provide an acceptable approximation.

As stated before some experimentation will be needed to provide a definitive answer, unless someone can still do the sums.........not me!!

You are correct about the UD load and was possibly for my benefit but it gives some real life figures to work on except in reality the front axle is a bit light without a lump of engine up there.

If you imagine the motorhome as being supported on three points but without the complication of any suspension then as soon as you apply a turning moment on the overhang the pivot point has to be the rearmost axle or the one closest to the load being examined (same thing).

You are right somebody cleverer than me might have a better set of answers.

Martin

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[funflair]

We need a weighbridge, a tag axle van, a calculator and a motorbike, then we can work find out precisely how much weight is added and work backwards

I have a calculator but none of the others.

Martin

 

[funflair]

Imagine you are able to lift and lower each pair of axles separately, as we lift the rear axle the pivot point moves slowly to the next axle as it takes all the weight. As we lower it back, the pivot moves back towards the rear axle, but not all the way as they share the load. So my belief is under normal conditions its around the mid point between the two axles.

Ian

That was really my first thought which is why I looked at the axles as simple supports without the suspension, but this could well be right, I think it was Geo that said we are not looking at load sharing as the calculation is turning moments which will be about a point closest to the load affecting the turning.

Martin

 

[DBK]

You are correct about the UD load and was possibly for my benefit but it gives some real life figures to work on except in reality the front axle is a bit light without a lump of engine up there.

If you imagine the motorhome as being supported on three points but without the complication of any suspension then as soon as you apply a turning moment on the overhang the pivot point has to be the rearmost axle or the one closest to the load being examined (same thing).

You are right somebody cleverer than me might have a better set of answers.

Martin

You can't assume it pivots around the rear axle, it really pivots around the centre of gravity. The best way to imagine this is to assume another support under the c of g and then jack this up a bit so it takes the full weight of the MH, assuming no suspension for the moment.

With the MH lifted like this then imagine adding some more weight to the rear of the MH which will then lower the back and both rear axles will come under load.

Of course there isn't a support under the c of g but it might help to understand what is happening when the hypothetical scooter is added, it just moves the c of g rearwards a bit and the load on the rear axles increases and the load on the front axle reduces.

Anyway, I've had a think about this and have realised my earlier thoughts were wrong because I ignored the suspension, assuming the MH was a rigid beam (no deflection) supported on fixed points. This theoretical idea ignores the real world!

If both tag axles have identical springs and both are compressed equally then both axles must be taking the same load and the mid-point between them can be used for estimating the effect of adding a scooter to the rear.

Of course, the MH may not be level but I still don't think the effect wont be significant.

But thanks for Jim for an excellent opportunity to reawaken some very old synaptic connections - or whatever they are called.

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[funflair]

You can't assume it pivots around the rear axle, it really pivots around the centre of gravity. The best way to imagine this is to assume another support under the c of g and then jack this up a bit so it takes the full weight of the MH, assuming no suspension for the moment.

With the MH lifted like this then imagine adding some more weight to the rear of the MH which will then lower the back and both rear axles will come under load.

Of course there isn't a support under the c of g but it might help to understand what is happening when the hypothetical scooter is added, it just moves the c of g rearwards a bit and the load on the rear axles increases and the load on the front axle reduces.

Anyway, I've had a think about this and have realised my earlier thoughts were wrong because I ignored the suspension, assuming the MH was a rigid beam (no deflection) supported on fixed points. This theoretical idea ignores the real world!

If both tag axles have identical springs and both are compressed equally then both axles must be taking the same load and the mid-point between them can be used for estimating the effect of adding a scooter to the rear.

Of course, the MH may not be level but I still don't think the effect wont be significant.

But thanks for Jim for an excellent opportunity to reawaken some very old synaptic connections - or whatever they are called.

But a pivot has to be a point that can be measured to make the calculation, if we are looking at overhang loads for a single axle there is no argument this is the centre of the rear axle in the moments calculation. The CofG and the mass is what dictates the individual axle loadings but again you need a pivot point for this calculation which is a simply supported beam, in my calculation the CogG would replace the 4 in 4000x4 the 4 is the centre of the chassis with a UD load.

Martin

Ps never heard of a synaptic connection, might google it later and learn something else.

 

[Jim]

I've amended the article to read

If you have a Tag Axle; W should be measured from the centre of your front wheel to the [centre of the rear-most wheel]* OR [A point in the middle of the two rear wheels]* *After reading this you can decide for yourself the best place to measure.

 

[DBK]

But a pivot has to be a point that can be measured to make the calculation, if we are looking at overhang loads for a single axle there is no argument this is the centre of the rear axle in the moments calculation. The CofG and the mass is what dictates the individual axle loadings but again you need a pivot point for this calculation which is a simply supported beam, in my calculation the CogG would replace the 4 in 4000x4 the 4 is the centre of the chassis with a UD load.

Martin

Ps never heard of a synaptic connection, might google it later and learn something else.

I'm reluctant to reply as Jim's post is a good way to close this but your assumption with a twin axle MH a load added to the rear pivots around the rear axle is misleading. Of course you can analyse the moments around any axle, assuming it pivots around that axle, you can do it around the the front axle if you want.

But in reality adding a load to the rear causes the front of the vehicle to rise and the rear to drop. If the MH really pivoted around the rear axle of this four wheel MH then the rear suspension wouldn't change, everything would just pivot around the axle, of course the rear axle load increases and the vehicles sits lower on the rear axle.

I think the confusion is between pivoting and rotation. The MH does really pivot around anything as it doesn't have a pivot on the c of g, but it does rotate approximately around the c of g. I say approximately because nonlinear springs or air suspension will mean it would rotate around a point either in front of or behind the c of g.

But Jim's assumption of the mid-point is fine given the whole point of a tag axle is to divide the load which would otherwise be on one axle onto two, and a well designed tag axle should split this evenly, until someone adds a scooter of course but in that event both tag axles see an increase in load with the rearmost getting a little bit more.

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[grumps147]

Interesting related article cocerning weighing enforcement Guidelines, and the usual + and - allowance for scientific innacuracy in weighing instrumentation.. https://www.gov.uk/government/publi...-of-practice-enforcement-weighing-of-vehicles

 

[funflair]

Hi DBK

I will certainly agree with your last paragraph, the rest I am sure we could debate till the cow's come home another time possibly, over a beer.

Martin

 

[jonandshell]

Newtons 3rd Law prevails!

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[TheCaller]

There would seem to be two separate questions.

The first, which Jim asked, was 'where is the pivot point for the purpose of the weight transfer calculation'.
The second question that then follows, is how is the additional weight we have now calculated, shared between the rear pair of axles.

The answers to both questions will be affected by the way that the two rear axles are connected to the chassis & to each other (if at all). With the simple systems, the two axles are independent of each other & simply each attached to the chassis in the same way as a single axle would be. In this case, Geo's answer would seem to apply & the centre line of the rearmost axle is the answer to question 1.

With more complex interlinked systems, designed to better share the load between the axles over undulating surfaces, the pivot point could be further forward, but I can't envisage an inter-connect system that would move it further forward than the mid-point between the two centre lines. That would apply for a full bogie system, where the two axles act on the chassis as if they were a single axle.

The more sophisticated the interlinking, the more evenly the additional load will be shared between the two axles. On an unlinked system, the rearmost axle will support the bulk of the extra weight, but how much will depend on the relative spring rates of the two axles, because in practice, the chassis will not pivot about the rear axle or axles, but a point somewhere well in front of the rear axles, as the vehicle sinks onto it's springs. Thus the springs of both rear axles will compress & take some of the additional load, the rear axle taking more than the front of the pair. It would be a complex calculation to work out the ratio & would require a detailed knowledge of the springs & tyres to do it.

At the other end of the scale, the sophisticated interlinked systems are designed to spread the load (although not necessarily equally), so that as long as you know what they are designed to do, the calculation would be relatively easy.

 

[2657]

There would seem to be two separate questions.

The first, which Jim asked, was 'where is the pivot point for the purpose of the weight transfer calculation'.
The second question that then follows, is how is the additional weight we have now calculated, shared between the rear pair of axles.

The answers to both questions will be affected by the way that the two rear axles are connected to the chassis & to each other (if at all). With the simple systems, the two axles are independent of each other & simply each attached to the chassis in the same way as a single axle would be. In this case, Geo's answer would seem to apply & the centre line of the rearmost axle is the answer to question 1.

With more complex interlinked systems, designed to better share the load between the axles over undulating surfaces, the pivot point could be further forward, but I can't envisage an inter-connect system that would move it further forward than the mid-point between the two centre lines. That would apply for a full bogie system, where the two axles act on the chassis as if they were a single axle.

The more sophisticated the interlinking, the more evenly the additional load will be shared between the two axles. On an unlinked system, the rearmost axle will support the bulk of the extra weight, but how much will depend on the relative spring rates of the two axles, because in practice, the chassis will not pivot about the rear axle or axles, but a point somewhere well in front of the rear axles, as the vehicle sinks onto it's springs. Thus the springs of both rear axles will compress & take some of the additional load, the rear axle taking more than the front of the pair. It would be a complex calculation to work out the ratio & would require a detailed knowledge of the springs & tyres to do it.

At the other end of the scale, the sophisticated interlinked systems are designed to spread the load (although not necessarily equally), so that as long as you know what they are designed to do, the calculation would be relatively easy.

That's what I thought

 

[Diasy]

Oh dear. This is all very technical and I really don't understand. We have a new autosleeper to which we have added a satallite dish and tow bar. We will put two bikes on the tow bar. Looking at the technical detail in the AS handbook I see that the MTPLM is 3500kg. The MOT is 3112 giving a payload of 388kg which I think is about 6 stone. I weigh 9 stone my husband about 16 stone so he is about 5 stone over the drivers allowance. It seems we must be over the limit before we even think about food,clothes, hoses,aquarolls etc,etc. Am I misunderstanding something. Will Simply getting a weighbridge weight give me the answer I need? Is a tow bar worse or better than a bike rack?

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[2657]

The only way to find out is to weigh the vehicle, each axle and a total weight.

The allowed weights will be on the manufacturers or converters plate, the manufacturers plate will be under the bonnet, the converters could be anywhere.

BTW 388kilo is about 60 stone not six!

 

[Diasy]

Oh that's a relief. We will be well within limits. But will defo go to weighbridge for peace of mind. This

 

[PORKSTER]

Just read this myself as I have a tag axle and found it interesting and baffling g at same time. I did however pick up on being nose heavy and it effecting weight. It may explain my weight differences in rear axles here.

https://www.motorhomefun.co.uk/forum/threads/axle-weights.113011/

I can update mine to 5000kg but guy never let me know what my axles would be once uprated. So I didn't go ahead but I really want to get a scooter for garage so will have to do some jiggling around. I am Lu my I have a free visa weighbridge very close by

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