Whoops

Carryfast:

Buckstones:

Carryfast:

Gembo:
If those coils are 10 ton each, how much force do they have at 40 MPH? A lot more than 10 ton.

^ This is the question in contention.The scientists seem to think that it’s all about the relative speed between the load and truck.According to them so long as that relative speed is kept to zero then all they need to do is apply an opposite 1g pull,or even less with a bit more friction against the load bed,to cancel out all of the kinetic energy contained within the 40 mph load which sounds like bollox to me.That’s even without factoring in any potential shear loadings against the straps by the load. :open_mouth: IE their whole case is that a 10t + lump of metal travelling at 40 mph contains no kinetic energy at all,so long as it’s supposedly stopped from moving relative to the load bed,by supposedly applying an opposing pull equal to its static gravitational weight. :confused: Which seems to defy all the laws of kinetics to me.

The laws of kinetics simply say Energy = Mass x (Velocity squared) divided by 2, Momentum (not Corbyn’s mates)
= Mass x Velocity.
These apply to the entire vehicle and its load, the issue of restraining the load (as for passengers with seatbelts) is purely what force is necessary to keep the load in place under a deceleration of the whole vehicle.

I never said the load had no kinetic energy, I used Newton’s Law of Force = Mass x Acceleration, from which one g of deceleration or acceleration will cause a mass to exert a force equal to its weight.
That is why you weigh what you do, and astronauts weigh nothing.

Great so we put a 10t + steel coil in the cargo hold of the space station with no straps on it at all then we reduce orbital velocity for a lower orbit and the thing will just stay there and not punch its way through the side because it weighs nothing ?.

Or we fire an 88 mm anti tank shell at the armour of a Sherman it will just stop and fall to the floor and do no damage when it hits the side because at that point it’s lost its velocity and is travelling at the same speed as the tank.As opposed to the laws of kinetics in which both the mass and the velocity subject the formula are exchanged for energy which just smashes and/or melts the armour.

On that note if you agree the 10t + load retains its kinetic energy,that being 40 mph’s worth in the case of the truck load and anyone’s guess in the case of the space station.How can you then only rate the load security based on 1 g of deceleration only supposedly exerting a force on anything in its way that’s only equal to its static weight.

When surely at least some of the formula should be based on the laws of kinetics and inertia. :confused:

On that note I’d guess that a race car 4 way safety belt system and anchorages or for that matter even a standard car system has a lot more strength than just that needed to suspend the static weight of a person above the floor ?.

OK. Taking the earth as our datum. A lump will have a kinetic energy relative to the square of it’s speed. Our ten ton lump has 4 times the energy at 40 kph as it does at 20 kph. Our truck as a whole has 4 times the energy at 40 kph as at 20 kph. All good. So our brakes need to dissipate 4 times the energy from 40 as from 20. (Thats why sports motorcycles have huge brakes, not because they’re heavy but because they’re very fast). That is all true.
But it is a but of a red herring.
We can see that our truck will not decelerate st more than 1g. (We’re ignoring aerodynamics as we’re not going too quick and have no spoilers). We are being slowed by brakes but our deceleration is decided by the amount of grip we have. With normal tyres on a dry road we’ll have a coefficient of friction of 1.0 at max.
So friction (the grip we have on the road) is the limiting factor. Out truck, call it 20 tons is pushing down (weighs) 20 tons and so (mu=1.0) receives a ‘pull’ from the road of 20 tons. In order for the 10 ton load to slow down at 1g too, that must receive a 10 ton pull. Assuming no headboard etc, that pull comes from the friction between load and bed. If we have mu of less than 1 we need straps or chains to ‘push’ that lump harder onto the bed.
The speed of the vehicle doesn’t affect the amount of FORCE to stop it, for any given rate of speed change. However you are correct the speed does affect the amount of ENERGY we need to get rid of. Stopping from 40 will get our brakes 4 times as hot, as from a stop from 20.
I’m confident that’s all correct. And I’m equally confident that I’ll never be asked to write a textbook.

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Quick addition.
Re race cars- because of their wings and ‘ground effect’ (the suction holdig them to the track*). They get ‘pushed’ onto the track at more than their mass. They may have a mass of 1 ton but push down with 3 tons. Also racing tyres are very soft so can get a mu of more than 1. (We also know they’re expensive and wear out quickly).
The result a lot of push down plus sticky tyres means we get more than 1g acceleration from F1 cars. The racing harnesses etc show this. Also racing harnesses need to hold a driver in a crash, where there will be acceleration of much more than 1g. In a crash the acceleration is due not to friction, but to hitting an object.

*I’m sure you’re familiar with the Venturi effect from carburettors.

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Franglais:
OK. Taking the earth as our datum. A lump will have a kinetic energy relative to the square of it’s speed. Our ten ton lump has 4 times the energy at 40 kph as it does at 20 kph. Our truck as a whole has 4 times the energy at 40 kph as at 20 kph. All good. So our brakes need to dissipate 4 times the energy from 40 as from 20. (Thats why sports motorcycles have huge brakes, not because they’re heavy but because they’re very fast). That is all true.
But it is a but of a red herring.
We can see that our truck will not decelerate st more than 1g. (We’re ignoring aerodynamics as we’re not going too quick and have no spoilers). We are being slowed by brakes but our deceleration is decided by the amount of grip we have. With normal tyres on a dry road we’ll have a coefficient of friction of 1.0 at max.
So friction (the grip we have on the road) is the limiting factor. Out truck, call it 20 tons is pushing down (weighs) 20 tons and so (mu=1.0) receives a ‘pull’ from the road of 20 tons. In order for the 10 ton load to slow down at 1g too, that must receive a 10 ton pull. Assuming no headboard etc, that pull comes from the friction between load and bed. If we have mu of less than 1 we need straps or chains to ‘push’ that lump harder onto the bed.
The speed of the vehicle doesn’t affect the amount of FORCE to stop it, for any given rate of speed change. However you are correct the speed does affect the amount of ENERGY we need to get rid of. Stopping from 40 will get our brakes 4 times as hot, as from a stop from 20.
I’m confident that’s all correct. And I’m equally confident that I’ll never be asked to write a textbook.

Firstly the kinetic energy released in bringing a 10t + load down from 40 mph to 0 will be the same regardless of how short or long the distance/time we do it over ?.On that basis I make the strain which the lashings and anchorages need to transmit from the load,into the load bed,through the chassis and then into the brakes is the equivalent of applying the sustained tension of a winch,subject to a force of 1445000 newton metres of torque,applied to the combined lashings and anchorages.IE a lot more than 10t ?.Which explains why the load deck/body needs all those metal fixings and fastenings to transmit that force to the chassis and subsequently to the brakes. :wink:

Although admittedly I might have got those figures wildly wrong being that this is all at if not past the outside fringes of my understanding of maths and physics.But then again what if Juddian,Gembo and myself are actually right. :open_mouth: :laughing:

Carryfast’s shell hitting the tank is a good point. The tank will rock back because of the impact, but it won’t get pushed over by it. It will however get it’s armour melted as he says.
The tank is not rolled over or pushed back by many metres because the Momentum is the product of the weight of the shell times it’s speed. It is going very fast but is quite light compared to the tank.
As he said before the Kinetic Energy is dependent on the square of the speed. Stopping a shell (even without explosives in it) involves dissipating huge amounts of energy. The armour melts because of the energy in the projectile.
High velocity rounds are much more damaging than larger but slower rounds.
Here’s another one. Remember that lump of rock that fell on Russia? It wasn’t more than a couple of tons was it? But because of the sheer speed it hsd massive amounts of energy.
This is why we always have film stars saving the planet fron asteroids. Not because their momentum is so great, but because their speed means that a two ton lump of inert rock travelling at very high speed will have more energy than an A bomb!

Another litlle point- the angle of a truck’s load bed to the horizontal should be such that it is a slope from the rear of the vehicle up towards the front. This means during braking the reaction to yje decelerating forces acting on the load give an increased normal reaction on to the bed.
Or the load will, because of the angle of the trailer push down harder onto the bed of the vehicle during braking, again giving a beneficial increase in friction.

And why cars carried carried on the peak of transporters are very prone to jumping off.!
As the truck stops the car (loaded perfectly horribly for dynamics) will tend to ‘lift off’ the bed and there will be no friction holding it in place. Only the restraining straps will be holding it.
I hope Juddian will confirm that cars loaded on an angle such thst the forward (relative to the truck) part is higher than the rearward, travel better than those where the car has it’s rear higher than the front?

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Franglais:
OK. Taking the earth as our datum. A lump will have a kinetic energy relative to the square of it’s speed. Our ten ton lump has 4 times the energy at 40 kph as it does at 20 kph. Our truck as a whole has 4 times the energy at 40 kph as at 20 kph. All good. So our brakes need to dissipate 4 times the energy from 40 as from 20. (Thats why sports motorcycles have huge brakes, not because they’re heavy but because they’re very fast). That is all true.
But it is a but of a red herring.
We can see that our truck will not decelerate st more than 1g. (We’re ignoring aerodynamics as we’re not going too quick and have no spoilers). We are being slowed by brakes but our deceleration is decided by the amount of grip we have. With normal tyres on a dry road we’ll have a coefficient of friction of 1.0 at max.
So friction (the grip we have on the road) is the limiting factor. Out truck, call it 20 tons is pushing down (weighs) 20 tons and so (mu=1.0) receives a ‘pull’ from the road of 20 tons. In order for the 10 ton load to slow down at 1g too, that must receive a 10 ton pull. Assuming no headboard etc, that pull comes from the friction between load and bed. If we have mu of less than 1 we need straps or chains to ‘push’ that lump harder onto the bed.
The speed of the vehicle doesn’t affect the amount of FORCE to stop it, for any given rate of speed change. However you are correct the speed does affect the amount of ENERGY we need to get rid of. Stopping from 40 will get our brakes 4 times as hot, as from a stop from 20.
I’m confident that’s all correct. And I’m equally confident that I’ll never be asked to write a textbook.

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Couple of bits. Why can’t a mass change speed at a rate more that g (10ms^2)? There’s a good chance I’m forgetting something, but that doesn’t ring a bell at all.
Speed is a huge factor in the amount of effort needed to stop a mass, Force =mass x acceleration, and we’re talking about changing a masses speed from whatever to 0 over a given time. The higher the starting speed, the higher the negative acceleration (assuming it’s over he same timeacale).

Captain Caveman 76:

Franglais:
OK. Taking the earth as our datum. A lump will have a kinetic energy relative to the square of it’s speed. Our ten ton lump has 4 times the energy at 40 kph as it does at 20 kph. Our truck as a whole has 4 times the energy at 40 kph as at 20 kph. All good. So our brakes need to dissipate 4 times the energy from 40 as from 20. (Thats why sports motorcycles have huge brakes, not because they’re heavy but because they’re very fast). That is all true.
But it is a but of a red herring.
We can see that our truck will not decelerate st more than 1g. (We’re ignoring aerodynamics as we’re not going too quick and have no spoilers). We are being slowed by brakes but our deceleration is decided by the amount of grip we have. With normal tyres on a dry road we’ll have a coefficient of friction of 1.0 at max.
So friction (the grip we have on the road) is the limiting factor. Out truck, call it 20 tons is pushing down (weighs) 20 tons and so (mu=1.0) receives a ‘pull’ from the road of 20 tons. In order for the 10 ton load to slow down at 1g too, that must receive a 10 ton pull. Assuming no headboard etc, that pull comes from the friction between load and bed. If we have mu of less than 1 we need straps or chains to ‘push’ that lump harder onto the bed.
The speed of the vehicle doesn’t affect the amount of FORCE to stop it, for any given rate of speed change. However you are correct the speed does affect the amount of ENERGY we need to get rid of. Stopping from 40 will get our brakes 4 times as hot, as from a stop from 20.
I’m confident that’s all correct. And I’m equally confident that I’ll never be asked to write a textbook.

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Couple of bits. Why can’t a mass change speed at a rate more that g (10ms^2)? There’s a good chance I’m forgetting something, but that doesn’t ring a bell at all.
Speed is a huge factor in the amount of effort needed to stop a mass, Force =mass x acceleration, and we’re talking about changing a masses speed from whatever to 0 over a given time. The higher the starting speed, the higher the negative acceleration (assuming it’s over he same timeacale).

Because we’re discussing trucks, not in crashes, but on the road, the acceleration of the truck and therefore of the load is limited by the tyres grip on the road. Assuming Mu max of 1. The mass of the truck is equal to it’s normal reaction.

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Franglais:

Captain Caveman 76:

Franglais:
OK. Taking the earth as our datum. A lump will have a kinetic energy relative to the square of it’s speed. Our ten ton lump has 4 times the energy at 40 kph as it does at 20 kph. Our truck as a whole has 4 times the energy at 40 kph as at 20 kph. All good. So our brakes need to dissipate 4 times the energy from 40 as from 20. (Thats why sports motorcycles have huge brakes, not because they’re heavy but because they’re very fast). That is all true.
But it is a but of a red herring.
We can see that our truck will not decelerate st more than 1g. (We’re ignoring aerodynamics as we’re not going too quick and have no spoilers). We are being slowed by brakes but our deceleration is decided by the amount of grip we have. With normal tyres on a dry road we’ll have a coefficient of friction of 1.0 at max.
So friction (the grip we have on the road) is the limiting factor. Out truck, call it 20 tons is pushing down (weighs) 20 tons and so (mu=1.0) receives a ‘pull’ from the road of 20 tons. In order for the 10 ton load to slow down at 1g too, that must receive a 10 ton pull. Assuming no headboard etc, that pull comes from the friction between load and bed. If we have mu of less than 1 we need straps or chains to ‘push’ that lump harder onto the bed.
The speed of the vehicle doesn’t affect the amount of FORCE to stop it, for any given rate of speed change. However you are correct the speed does affect the amount of ENERGY we need to get rid of. Stopping from 40 will get our brakes 4 times as hot, as from a stop from 20.
I’m confident that’s all correct. And I’m equally confident that I’ll never be asked to write a textbook.

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Couple of bits. Why can’t a mass change speed at a rate more that g (10ms^2)? There’s a good chance I’m forgetting something, but that doesn’t ring a bell at all.
Speed is a huge factor in the amount of effort needed to stop a mass, Force =mass x acceleration, and we’re talking about changing a masses speed from whatever to 0 over a given time. The higher the starting speed, the higher the negative acceleration (assuming it’s over he same timeacale).

Because we’re discussing trucks, not in crashes, but on the road, the acceleration of the truck and therefore of the load is limited by the tyres grip on the road. Assuming Mu max of 1. The mass of the truck is equal to it’s normal reaction.

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From memory, rubber tends to have a friction coefficient of higher than 1, especially modern compounds of rubber used in tyres.

Unless you’re using it as a theoretical maximum to make the scenario easier to work with. Doh! You’ll have to excuse me, I’ve been up a long time and I haven’t read all of Carryfast’s work yet.

Soft F1 rubber maybe. But not by much I think.
Check out the acceleration times of dragsters. Comoste that to 1g of 9.86 metres per sec per sec? And the best figures are at the end of a hot dry day when soft tyres are running on layers of sticky rubber lrgt from previous runners.

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Franglais:
Soft F1 rubber maybe. But not by much I think.
Check out the acceleration times of dragsters. Comoste that to 1g of 9.86 metres per sec per sec? And the best figures are at the end of a hot dry day when soft tyres are running on layers of sticky rubber lrgt from previous runners.

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I’m always a bit wary of Internet “facts” but I did do a quick Google and found this.

sccoia.org/articles/top-fuel … ast-facts/

I’ve been out of physics for some time now and the stuff on that Web page is completely outside of my sphere of experience, but it’s fascinating nonetheless. I do hope their drivers have pressurised race suits, more than 4 or 5 g can cause the body problems.

Franglais:
And why cars carried carried on the peak of transporters are very prone to jumping off.!
As the truck stops the car (loaded perfectly horribly for dynamics) will tend to ‘lift off’ the bed and there will be no friction holding it in place. Only the restraining straps will be holding it.
I hope Juddian will confirm that cars loaded on an angle such thst the forward (relative to the truck) part is higher than the rearward, travel better than those where the car has it’s rear higher than the front?

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Its something most drivers of such are always wary of and why i always used 4 straps on such vehicles, when we still tied down underbody (now its chocks/wheelstraps), i would always have three straps pulling uphill/backwards and only one needed to stop it going uphill, that last strap only really being in place to keep the tensions right, it was never going to go uphill angled forwards anyway, also during a multi drop that car might well be lowered to level so that one forward pulling tensioning strap becomes vital.

Obviously the one flipped up on the peak is the most dodgy, as thats the one that as you rightly allude to is going to do its best to escape in the event of a hard stop, it shouldn’t be able to jump in theory on modern designed transporters cos there are good stops which the wheels ■■■■ up to, these could be higher still but modern designs of cars got so low that the height of the stops can only be what it is.

Cars flipped backwards (sloping downhill towards the rear of the vehicle) obviously are going to be pushed into the deck harder under heavy braking, but those at the rear end are going to see a much more bouncy ride which you see if you look at the centre car you can see from the back on a typical 11 plus transporter as it goes over road undulations, and again good strapping is a must…just to mention wheelstraps momentarily, if any wheelstraps are going to slip off as they can and do due to suspension movement of lorry exacerbated then by the transported car’s own suspenders, then the vehicles at the rear behind the trailer axles are the likely suspects.
I never did like the wheelstrap method and the blanket move to such was one of the many reasons initially for me looking to get out of the game, not only is it not as safe IMHO, almost never did underbody straps come off cos the strapped vehicle could not move to shake its straps off, it also means each deck has to be higher from the cars below to allow for suspension bounce (almost eradicated when underbody strapped) hence the overall CoG is higher than needed, its also very hard on your knees and joints when you’re scrabbling down on your knees so many times day in day out on a stamped deck :imp:

Sorry i’ve waffled a bit there.

Interesting thread this.

Waffling on? That was me waiting for the tacho to go over 11hrs this morning. Fresh coffee, world service radio and fingers busy on the phone screen. At drop now so more time to kill.

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I’ve not carried steel coils for years, last time was the late 90’s but we didn’t have wells in the trailers so we carried 2x 10t coils on end! I never liked it and when I said so was told to use as many ratchet straps as I could but always a minimum of 6 per coil. I drove according to the load and never had a problem unlike one lad that took the wrong exit of a roundabout near hagley and ended up going downhill into a housing estate. When he realised his mistake he hit the brakes too hard and from what we were told one of the coils broke loose, fell off the trailer, rolled down the hill and crushed a car! Lucky escape for him not getting crushed in the cab!

I’ve just read through this thread and have to say I agree with something Kate said about shadow training. Learn it practically from a guy that’s done the job for years and seen all the ■■■■ ups or from a guy in a classroom that’s not actually done the job in the real world just read about it… [emoji57]

Franglais:
Carryfast’s shell hitting the tank is a good point. The tank will rock back because of the impact, but it won’t get pushed over by it. It will however get it’s armour melted as he says.
The tank is not rolled over or pushed back by many metres because the Momentum is the product of the weight of the shell times it’s speed. It is going very fast but is quite light compared to the tank.
As he said before the Kinetic Energy is dependent on the square of the speed. Stopping a shell (even without explosives in it) involves dissipating huge amounts of energy. The armour melts because of the energy in the projectile.
High velocity rounds are much more damaging than larger but slower rounds.
Here’s another one. Remember that lump of rock that fell on Russia? It wasn’t more than a couple of tons was it? But because of the sheer speed it hsd massive amounts of energy.
This is why we always have film stars saving the planet fron asteroids. Not because their momentum is so great, but because their speed means that a two ton lump of inert rock travelling at very high speed will have more energy than an A bomb!

So on the basis that we’re agreed at the point when an AP shell actually smashes/melts its way through the armour of a tank there is no velocity difference relative to the tank and is at that point just transferring its kinetic energy into the armour ?.

Which therefore leaves the the question of the conundrum should the relevant calculation,for the loads applied to the lashings and anchorages of a steel coil for example,be based on transferring all/most of the kinetic energy contained in it,according to its actual speed,relative to the outside world,to the brakes where its dissipated as heat and/or transferring resulting inertial loads when changing direction ?.

Or is it just a case of cancelling out the potential ( g force ),based on the speed differential between the load and truck,by an equal opposing pull ?.

Bearing in mind the totally different formulas and the resulting final sum figures respectively. :open_mouth: :confused:

Quick answer yes, all the Kinetic Energy is converted into heat. We all know a heavily loaded vehicle’s brakes get hotter than an empty one? Thst is correct. The calculation concerning the acceleration and friction etc. don’t concern themselves with this though.
Can’t say more gotta load.

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Carryfast:

Franglais:
Carryfast’s shell hitting the tank is a good point. The tank will rock back because of the impact, but it won’t get pushed over by it. It will however get it’s armour melted as he says.
The tank is not rolled over or pushed back by many metres because the Momentum is the product of the weight of the shell times it’s speed. It is going very fast but is quite light compared to the tank.
As he said before the Kinetic Energy is dependent on the square of the speed. Stopping a shell (even without explosives in it) involves dissipating huge amounts of energy. The armour melts because of the energy in the projectile.
High velocity rounds are much more damaging than larger but slower rounds.
Here’s another one. Remember that lump of rock that fell on Russia? It wasn’t more than a couple of tons was it? But because of the sheer speed it hsd massive amounts of energy.
This is why we always have film stars saving the planet fron asteroids. Not because their momentum is so great, but because their speed means that a two ton lump of inert rock travelling at very high speed will have more energy than an A bomb!

So on the basis that we’re agreed at the point when an AP shell actually smashes/melts its way through the armour of a tank there is no velocity difference relative to the tank and is at that point just transferring its kinetic energy into the armour ?.

Which therefore leaves the the question of the conundrum should the relevant calculation,for the loads applied to the lashings and anchorages of a steel coil for example,be based on transferring all/most of the kinetic energy contained in it,according to its actual speed,relative to the outside world,to the brakes where its dissipated as heat and/or transferring resulting inertial loads when changing direction ?.

Or is it just a case of cancelling out the potential ( g force ),based on the speed differential between the load and truck,by an equal opposing pull ?.

Bearing in mind the totally different formulas and the resulting final sum figures respectively. :open_mouth: :confused:

Quite why we’re talking AP shells - an object moving, hitting a stationary object… so completely different to a load on a trailer, I don’t know. Until the shell hits the tank there is no problem, both systems are independent of each other. When the AP shell hits the tank there is a mismatch between the energy stored in each of them, and it will have to equalise. The shell is so light in comparison to the tank that very little energy can be balanced by transferring movement, so the only other way to balance the energy in the (now single) system is through deforming both the shell and the armour, but that will only use up a relatively small amount too, leaving heat as the final method of balancing the energy.

The entire time that an object is secured to the bed of trailer, it behaves as if it is an integral part of the trailer. This is why, so long as the object is secured in such a way that it doesn’t move, the force it experiences under panic braking is the identical to the rest of the truck. 1g, or in VOSA handbook terms 100% of the load’s weight. The object doesn’t have any kinetic energy of it’s own, in this scenario. The brakes will have to generate more heat because the whole think is heavier, but that’s the only change.

As soon as the object moves independently of the trailer bed, that is when you get serious problems. That is when you are trying to dissipate the kinetic energy of the load by itself (which results in things getting smashed up) because, as with the AP shell, the energy in the two systems must balance when they make contact. Unlike the shell, a 10ton top hat weighs more than the headboard and cab of the unit, so could balance the kinetic energy mismatch by moving and by deforming both of them.

AP wouldn’t do much to a tank, don’t you mean HESH

Franglais:
Quick answer yes, all the Kinetic Energy is converted into heat. We all know a heavily loaded vehicle’s brakes get hotter than an empty one? Thst is correct. The calculation concerning the acceleration and friction etc. don’t concern themselves with this though.

So we know that it’s the kinetic energy figure which applies to the energy dissipated in the form of heat by the brakes.So how doesn’t that same figure apply in the case of the lashings and anchorages which obviously have to transfer that energy into the load deck,which then transfers it into the chassis by the body to chassis attachments,which then transfers it to the brakes via the axle attachments.

Seems to me like it’s the idea of only needing to exert an opposing pull equal to the static weight of the load which is the weak link in that chain.All the others having arguably been calculated based on transferring and eventually dissipating the kinetic energy of the weight of the vehicle to/at the brakes ?.

IE straps to transfer/dissipate the kinetic energy of a 10t + coil seems like using a Sherman when you actually need a Challenger.Then telling the crew don’t worry the armour only needs to stop the round from moving relative to the speed of the tank,not from wrecking the armour by the transfer of its kinetic energy when it’s stopped it. :bulb: :wink:

Don’t really know, Grumpy Dad, I’m FAR too much of a wimp to have ever been in the forces, but whether AP means Armour Piercing or Anti Personnel, the rest of what I wrote should make sense! :laughing:

Gembo:

commonrail:
This is a Tata load
0

Personally- I wouldn’t be happy with that.
Worse case scenario, kid runs out in front of you, you stand on the brakes, that lot is off toward the unit.
In an ideal world, its going nowhere but in the event of a sudden stop, I wouldn’t trust those straps!
If those coils are 10 ton each, how much force do they have at 40 MPH? A lot more than 10 ton.

That is the Tata method.
Conform or no load.