Forum Discussion
dballentine
Jan 24, 2014Explorer
In a word, NO. As in "NO, the manager doesn't have a clue what he's talking about."
1534 is the amount by which you would exceed the legal/safe total vehicle weight, if you loaded 4250 on the front axle and 6084 on the rear axle.
The front axle rating plus the rear axle rating will always exceed the maximum vehicle weight (GVW). That's so the end user will have some amount of flexibility in terms of where (front or rear) the payload is located. One person might want to hang a snowplow off the front; another an RV off the back.
Using your truck as an example: GVW = 8800, UVW = 6000, Payload = GVW-UVW = 2800. As a guess*, let's say the UVW splits 3200 Front, 2800 Rear. You could put 1000 on the front and 1800 on the rear; 0 Front and 2800 Rear; or any combination in between. What you can't do (legally) is put more than 2800 payload in the pickup, counting passengers, cargo, hitch, and FW pin weight.
* The only way to know these weights exactly is via a trip to the nearest truck stop that has CAT scales. Cost = $10.
1534 is the amount by which you would exceed the legal/safe total vehicle weight, if you loaded 4250 on the front axle and 6084 on the rear axle.
The front axle rating plus the rear axle rating will always exceed the maximum vehicle weight (GVW). That's so the end user will have some amount of flexibility in terms of where (front or rear) the payload is located. One person might want to hang a snowplow off the front; another an RV off the back.
Using your truck as an example: GVW = 8800, UVW = 6000, Payload = GVW-UVW = 2800. As a guess*, let's say the UVW splits 3200 Front, 2800 Rear. You could put 1000 on the front and 1800 on the rear; 0 Front and 2800 Rear; or any combination in between. What you can't do (legally) is put more than 2800 payload in the pickup, counting passengers, cargo, hitch, and FW pin weight.
* The only way to know these weights exactly is via a trip to the nearest truck stop that has CAT scales. Cost = $10.
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