Distance in feetbetween first andlast axles of ve-hicles or combi-nation of vehicles. | Maximum load inpounds on allaxles. | |
4......... | .40,000 | |
5.......... | 40,000 | |
6.......... | 40,000 | |
7.......... | 40,000 | |
8.......... | 40,000 | |
9.......... | 44,140 | |
10.......... | 44,980 | |
11.......... | 45,810 | |
12.......... | 46,640 | |
13.......... | 47,480 | |
14.......... | 48,310 | |
15......... | .49,150 | |
16.......... | 49,980 | |
17.......... | 50,810 | |
18.......... | 51,640 | |
19.......... | 52,480 | |
20.......... | 53,310 | |
21.......... | 54,140 | |
22.......... | 54,980 | |
23.......... | 55,810 | |
24.......... | 56,640 | |
25.......... | 57,470 | |
26.......... | 58,310 | |
27.......... | 59,140 | |
28.......... | 59,970 | |
29.......... | 60,810 | |
30.......... | 61,640 | |
31.......... | 62,470 | |
32.......... | 63,310 | |
33.......... | 64,140 | |
34.......... | 64,970 | |
35.......... | 65,800 | |
36.......... | 66,610 |
W = 500((LN ÷ (N-1)) + 12N + 36)
where W = the overall gross weight on any group of two or more consecutive axles to the nearest 500 pounds; L = the distance in feet between the extremes of any group of two or more consecutive axles; and N = the number of axles in the group under consideration. Such overall gross weight of any vehicle or combination of vehicles may not exceed 80,000 pounds, including all enforcement tolerances.
W = 500((LN ÷ (N-1)) + 12N + 36)
where W = overall gross weight of the vehicle to the nearest 500 pounds; L = distance in feet between the extreme of the external axles; and N = number of axles on the vehicle. However, such overall gross weight of any vehicle or combination of vehicles may not exceed 80,000 pounds including all enforcement tolerances.
Fla. Stat. § 316.535
Former s. 316.199.