[sourdough]

Originally Posted by markcee

The following applies at least to a Ford.....Higher trim, cab config, 2WD vs. 4WD, bed length, engine option and also a potential GVWR 'paper derate' (like I have), eat payload.

The payload on my modestly equipped F350 XLT is 3,271 pounds. This is somewhat due to the paper de-rating of my vehicle's GVWR capacity (for licensing purposes) to 10K. De-rating is not a result of shortcuts in components (frames, brakes, suspension etc.) but rather is done done due to potential registration $$ savings. This is actually listed as option 68D on the Ford build site.

To be clear, available payload is easily calculated. It is your truck's physical weight, subtracted from it's door sticker GVWR. When the truck comes off the assembly line, they weigh it (and for Ford they also assume a full tank of fuel) and subtract it from GVWR. What's left goes on the sticker as 'cargo carrying capacity' (payload). In my case, since my sticker GVWR was de-rated to 10K, I lost some legally usable payload

So, how much GVWR should your truck have without any type of paper de-rate? In looking at the build guide for the 2019 Ford, I think I was only reduced by 200 lbs from what MY max should be. If I'm reading the guide correctly, a 160" wb, 4x2 with a 6.2L should be rated at 10.2K GVWR. Conversely, a longbed 4x4 with diesel can be obtained with an 11.4K GVWR.

The moral to the story: if looking at a truck on the lot, open the driver door AND LOOK at the yellow/white sticker to see what you have to work with. If ordering, be familiar with the build guide for each vehicle and it's potential delimiters with regard to GVWR.

When a member is trying his best to figure out weights and keep his family safe, your above post trying to muddy the waters with "I wish I had" just ends up confusing folks which isn't helpful IMO. All you had to do is jump to your last paragraph and repeat what everyone else has tried to point out.