Surely this can't be correct - or is it?
Comments
-
I never said that there is no difference between mass and weight, but weight is the force exerted by a given mass and the force is what the engine has to provide to move the complete outfit.
You have chosen to ignore the last statement made in my previous response, which I repeat here:
The car's mass includes the mass of its payload and the noseweight is part of the payload.
Towing implies a horizontal force that is required to move a given mass of a trailer, but part of that mass not being towed at all but being carried. The car would only be towing the total mass of the trailer if the jockey wheel were left down and there were a loose joint between the towbar and the trailer's coupling which doesn't transmit any vertical forces to the car's axles, but in a case like that the car's axle loads would be less by the amount of the missing noseweight and the towbar would be subjected to an equivalent higher horizontal force of tension between car and trailer.
0 -
Lutz, apologies about my post, got a bit too carried away, would love to argue this over a beer sometime if we're on the same site?
0 -
wow, posted at the same time!
To the car's engine it doesn't matter where the distribution of the weights of the trailer are does it? It only 'knows' that there is (say) 1000kg of trailer to move and keep moving. That's what I'm arguing.
Ok imagine the caravan is exactly set up so the whole weight is on the wheels. There is no weight on the car. The car has to overcome the frictional forces of 0.7 x 1000g, (and taking g as 10) = 7000N. I am assuming one big tyre rather than two and 'mew' =0.7
Then it has to overcome the car's frictional forces, say 0.7 x 1500g = 10500N. So total frictional forces are 17500N. If moving at a constant velocity the car's engine has to produce this.
Now if the nose weight is say 100kg, that means the weight to the rear wheels on the car will go up by a part of that and the front wheel will go down, don't know those figures but the car is now heavier by 100kg so the frictional forces now are 0.7 x 1600 x 10 = 11200N while the frictional force on the caravan is is 0.7x900x10 = 6300N, making a total frictional force of 17500N, Exactly as before !
So what I'm saying is that for the car, and by that I mean the car's engine it is still 'towing' the same force and hence the same weight.
What do you think?
0 -
Of course, as far as the engine is concerned, it matters not whether part of the trailer's mass is being towed or carried. It has to do the same amount of work to shift the total mass of the trailer, and for that matter, the complete outfit, but the horizontal tensile load on the towbar between car and trailer will be different depending on whether the full weight of the trailer is being towed or part of it carried.
Frictional forces that have to be overcome are those in the powertrain itself and running gear, not in the tyres. Friction between tyres and the road surface are really only an issue when braking. Far greater than friction in the system is aerodynamic drag (especially with something as big and boxy as a caravan) and, of course, the effort needed to get the outfit up any gradient.
The frictional forces that you mention 17500N are also way too high to be realistic. That's the equivalent of almost 1.8 tonnes. Add to that aerodynamic drag and the force required to pull the outfit up, say, a 12% incline and you would need something like a heavy commercial vehicle to tow even a modest caravan if your figures were right.
0 -
to a humble mathematician it's the towing force from the engine that we consider. Why do you say that the friction of tyres on road is only an issue when braking? as friction opposes motion won't the frictional force be the same either way? I used a simple model to show the towing force isn't affect by where the weight of the trailer is vertically. So what would a realistic frictional force be and how would you have calculated it?
0 -
I really have no idea what the frictional forces in the system are, but they would be a lot less than what you suggest. Your calculation is based on the force required to accelerate the outfit on level ground at a rate which is just below the point of wheelspin of the driven wheels (hence the reference to a µ of 0.7). Caravans aren't usually that quick off the mark so the actual force will be a lot less until the speed has increased to an extent that aerodynamic drag is significant.
The 1.8 tonnes required that you have calculated would mean that it would be impossible to push a car and caravan outfit of the size quoted by hand on level ground. I can assure you that it is quite possible for two relatively fit adults to do so although they wouldn't manage 1.8 tonnes.
0 -
I'm here early this morning, I've got my seat, a drink & some food. I'm looking forward to the next installment of the 'Corners & Lutz show' bring it on I say
3 -
I'm not into the learning bit SL I just want to see the outcome-who's right, who's wrong. Although I expect it to be a draw as these things tend to be. Each will concede certain points so we'll not actually know who the 'go to' person for trusted info is
I forgot the popcorn SL, did you bring any?0 -
And all I did in the first place was ask if the salesman was talking a load of bulls**t !
It has turned out to be quite interesting though. There is a lot more physics involved in this towing lark than probably 99.9% of us realise and certainly don't understand. Perhaps we will learn something from Lutz and Corners. Thank you, gentlemen, for keeping us entertained and educated
0 -
Of course, as far as the engine is concerned, it matters not whether part of the trailer's mass is being towed or carried. It has to do the same amount of work to shift the total mass of the trailer
First I'm glad you have finally conceded my point that the (extra) force required by the car, and when I say car, I mean the car's engine, will be the same no matter where the weight of the caravan is supported. This fits in with Newton's laws. This is what I found at university and this is what I said on page 2:
It doesn't matter where the weight is distributed as the weight acts vertically downwards, while the towing/engine forces are at right angles to this.
Yes I am sure you're right about my simplistic model, it was merely meant to show the engine force is the same no matter where you support the vertical weights.
One thing I am really impressed with your post however, how did you get µ on here? I think I've answered my own question actually - copy and paste?
0 -
To Steve Rocky and John:
I didn't realise anyone was actually following this and I'm glad you enjoyed it.
I think the differences in thinking are due to the definition of towing? is it pulling the extra weight of the caravan (my view) where it doesn't matter about where vertical forces are, or is towing pulling the remaining weight of the van? Is the towing force the force provided by the engine (my view) or the tension in the tow bar?
Btw the imperial unit of mass was the slug
0 -
I think we are getting closer. Although the power required to shift the outfit is independent of whether the towing vehicle is carrying part of the trailer or not, the towing vehicle is actually towing only the axle load. (The loads imparted to the towbar, both vertically and horizontally, are definitely different depending on whether a noseweight is present or not and it's the horizontal component which is referred to as the towload).
µ is easy. It's a combination of the 3 keys CTRL+ALT+M
0 -
Yes, as I said it's all to do with the definition of towing. My view (perhaps due to my mathematical thinking (and remember an engineer is just a mathematician with half his brains knocked out
) my argument is that for the car, horse, even you, the force needed to pull the weight of the trailer is the same no matter where that weight is supported, (all on the trailer's wheels or shared between car and trailer) and so I think the whole weight of the trailer has to be pulled not just the part on the part left over on the trailer's wheels, and unless you need to calculate the tension it is best to think whole outfit as the tensions will then cancel out. That is not the same as the tension being zero btw although this can and does happen, only the resultant tension.As I said if you were pushing the van holding it up you're not really bothered with which weights are supported where you just want to push that van.
Why I say that (thinking the way I do as), is that the equations of motion are done using vectors with Newton's laws are simply (to me anyway) horizontal and vertical and then solving. Any forces at other angles are simply resolved.
0 -
CTRL+ALT+M mew doesn't work for me
0 -
" (and remember an engineer is just a mathematician with half his brains knocked out) "
But you still have to pay us to fix the things you can't and don't understand
0 -
Heck yes, if it continues much longer I'll make up some advertising posters-'The battle of the sages'
'Weight or Mass-they'll kick your a*s'
0 -
very true, that's why we have you
Just to let you know I am absolutely useless at DIY, fixing things, even flat pack furniture is difficult for me.
0 -
The frictional forces that you mention 17500N are also way too high to be realistic. That's the equivalent of almost 1.8 tonnes. Add to that aerodynamic drag and the force required to pull the outfit up, say, a 12% incline and you would need something like a heavy commercial vehicle to tow even a modest caravan if your figures were right
Sorry Lutz I'm going to have to challenge you on this, First I said that's the combined force from the engine. the car's frictional force was 10500N and the caravan's was 7000N. I'm think I'm about right. have a look at this video here . This chap has done the same calculation apart from using a figure of 0.6 for µ and calculates for a 1200Kg car a max frictional force of 3600N for each driven axle. assuming a 4wd that's 7200N.
I know you've said that I really have no idea what the frictional forces in the system are, but they would be a lot less than what you suggest.
but that's not good enough is it? You say my calculations are way out (apparently not or that chap and me are both wrong) but you can't say what they should be or even hint? Bit of a cop out? Over to you.
I realised this is the max static friction (but as the part of the tyre in contact with the road is static even when moving we can use static friction) that can be generated, but its give the idea of my view that the car is 'towing' the whole weight and where the vertical weights are don't matter.
0 -
ever tried pushing a slug?
0 -
Another retired engineer,
Not that great with maths & physics but know enough. In 2009 I did a complete nut & bolt restoration on my classic car, I have also, over the years built my own loft conversion and conservatory on my house.
I rather be a hands on practical person than an Academic any day
Anyway, all you need now is a smart phone with a decent calculator.
0 -
I rather be a hands on practical person than an Academic any day
I would say they are both equally important. Particularly in cutting edge technologies like space exploration. You might need the engineer to propel you, but if you want to end up at Saturn rather than Uranus 😂you need the academic to do all the calculations.
0 -
Love that old joke Steve, still makes me chuckle
. I've actually found that reading round my hobbies has actually enriched my enjoyment of them. When I got interested into photograph at 16 I read all could about it, not only about composition but all the technical stuff as well. Like I said each to their own, you enjoy your leisure time as you will.
And I would rather be an 'academic' (btw never been called one yet ) than a practical person as it's far less hard work all round, but each to their own.
0 -
very true, but then we can always find someone to do the hard work
0
!!