Sliding the 5th wheel doesn't change the distance from the pin to the center of the trailer, or from the pin to the trailer axles. The pin is a FIXED PART OF THE TRAILER!!! The ONLY thing sliding the 5th wheel does is move the weight on the TRACTOR, affecting how much weight is being distributed to the steer and drive axles. The trailer axle weights WILL NOT be affected by sliding the 5th wheel.
What do you think, @Chewy352? Can you tell the gent on how to load a trailer? BTW, @Pedigreed Bulldog is correct about the 5th wheel slide debate. Can anyone explain why, in cornbread language for all to understand, in order to end once and for all, the 5th wheel debate about sliding the 5th wheel get remove weight off the trailer axles?
(this should be fun) @passingthru69, you had a sliding 5th wheel on your jeep and a slide on your tractor (on the 13 axle setup). If you were over on your trailer weight by 15000 lbs, how many clicks on each 5th wheel would it take to get the trailer weight legal?
The 5th wheel debate will never end. I wish I had kept my weight tickets. When I was over weight with my lumber load and was playing with my 5th wheel I was able to move maybe 50 lbs off my trailer axle onto my drives. I will cede that this amount is so small that it doesn't really matter. But it did move weight. When I had my flat I always ran with the 5th wheel all the way back for 2 reasons. 1 prostars are already heavy on the steers and 2 the slide pins were not exposed on my 5th wheel to give them a tap and I was not able to slide it under a load. So I always ran ready for a front overhang. As far as where to center the load. Measure from the kingpin to the center of the spread. Divide that by 2 and then measure back from the kingpin that amount. In general that is where you want to center your load. I say in general because I have a question for professor six. How do my axle weights affect this? My empty drive weight is 12460 my spreads are 7960. So I can put 21540 on my drives and 26040 on my spreads (to equal 34k, I know I can put 40k but we're talking about centering).
So when loading coils you need to know a few things. 1. Empty weight with full tanks 2. Centerline of the trailer Most trailers the centerline is the side light. If the trailer has a coil package the center beam of the coil package is the centerline. Some trailers have a sticker that states "place single coil here" it is usually slightly behind the centerline. This is because of the arch of the trailer or the design of the coil package. Generally speaking 1ft front or rear equals 1,000 lbs shift of weight. Most loaders will help you out if you ask. In the bigger mills you are on your own look at you load ticket it gives the weight and number of coils and how they have to be loaded sucide or shotgun. Once you get everything set you rarely have to scale yourself. Most major shippers scale you in and out for gross weight. 5th wheel setting varies on type of truck and wheelbase of tractor. Most coil haulers it's all the way back or centered between the tandems.The industry standard is the ability to haul 48,000. That's not saying all coils will be that heavy. Most shippers want you to have the ability to haul that amount.
There's a 4500lb difference between the drives and the trailer. You know how to find the center to load. If you load dead center, and your drives are 34000, what will be the weight on the trailer axles? If the load is 20 ft long, loaded dead center, how far back would you have to move the load so that the trailer axles have exactly the same weight as the drives? Someone is thinking, "Whats with all the math? We can get 40k on the spread." Ignore this post and stick with what you know. "But Six, why would you ever need to get the trailer and the drives exactly the same?" In order to axle the MAXIMUM weight on a 5 axle rig with a non divisible load, the drives and the trailer are going to have to be dead even. If you don't know how, you're going to run back and forth, back and forth through the Cat scale.
With a 48k lb load my spreads would be 34420 If the load is 48k lbs I would need to move it just over a foot back from center. To be at 34k on the drives.
If my loaded drives are 34000 lb then subtract my empty weight of 12460 lb and that means I have 21540 lb of cargo on my drives. Subtract that from a 48000 lb load and I have 26460 lb of cargo on my spreads. Add my empty weight of 7960 lb and my spreads weigh 34420 lb. For the second one If the load is 20 feet long and equal weight through out and weighs 48000 lbs then you divide 48000 by 20. That gives you 2400 lb per foot. If I put the load dead center ill be at 36460 lb on my drives. That's 48000/2=24000 24000+12460=36460 36460-34000=2460 So that's 2460 lbs that need to be moved back towards the spread which is just a hair over a foot. Does that make sense professor? By the way professor is not meant badly. The way you teach and ask questions just makes me feel like i'm back in collage again. I think you'd be a good teacher.
