Portcullis Weight

[GodBeastX]

How much would a Portcullis weigh on a keep entrance? I'm looking for a general weight for them or how to calculate.

I'm doing a siege event in a Banestorm-like world so I wanted to get an idea if a Ogre SM-2 with 33 ST could lift a portcullis.

[Phantasm]

I think at one point we determined that a wrought iron portcullis weighed about a ton. Using "two-handed lift (BL x 8)" as the base, you'd need about ST 35 or so for "bench-press" purposes. Your ST 33 ogre might be able to do it on a successful Lifting skill roll, but it won't be easy for him, nor will he be able to maintain it for long.

[mlangsdorf]

How big of a portcullis? Iron weighs roughly 7 tons per cubic yard (46656 cubic inches). If you assume 1" square bars, a 15' by 15' portcullis (adequate for a gate that you can easily ride a horse or two through), and 6" of separation between bars, then each bar is 180 cu inches and there's about 900 of them. The total mess would weigh 24 tons. That's probably too much.

A smaller portcullis might be 7' by 3' - enough for a doorway. The 7 vertical bars are 84 cu in each, and the 15 horizontal bars are 36 cu in. That works out to a much more reasonable 340 lbs or so. Two handed lift is BLx8, so you'd need BL45 (ST 15) to lift it.

Also, some portcullises were made of wood, or iron reinforced wood, which would cut the weight by a factor of 6-8. If the horse gate is made from iron reinforced wood, it might weigh around 4 tons, which is still too much for the barbarian to lift alone (BL 1000 is ST70!) but might be possible with support from the martial artist using Power Blow and everyone burning fatigue like crazing to maximize their lift rolls. Especially if they only need to deadlift it a foot so that people can roll/crawl under it.

[Varyon]

DF2 has some "good enough for fantasy" values on page 18. You can find rules for lifting on page 8 (you can lift a portcullis weighing up to 8*BL). Note these portcullises have some pretty wide openings - bending/breaking a single bar is enough for a skinny person to squeeze through, two is enough for normal people to squeeze through.

[Dammann]

It seems like a portcullis might have counterweights like a casement window so that it is more useable for castle residents. Security is balanced against convenience, so making it easier to open makes sense if it's operated often.

[Varyon]

Originally Posted by Dammann

It seems like a portcullis might have counterweights like a casement window so that it is more useable for castle residents. Security is balanced against convenience, so making it easier to open makes sense if it's operated often.

During a siege, any counterweight is probably going to be detached in some way from the portcullis. An easily-reversible method like tying off the ropes or settling the counterweight on something would work just fine, and in a pinch you could just cut the ropes and let the counterweight fall. Having a special team get in and free the counterweight (and subsequently protect it from being cut) could make the ogre's job a lot easier, of course.

[Dammann]

Yeah, something like a carabiner and eyelet is what I'm picturing. That's clever since it gives you some parallel action for a different character.

[Anthony]

Originally Posted by Varyon

During a siege, any counterweight is probably going to be detached in some way from the portcullis.

Actually, you probably just put a heavy shaft underneath the counterweight; a counterweight that's not free to drop won't help lift.

[Curmudgeon]

A photo of Bodiam castle in England shows 13 visible vertical beams (the two beams at the groove edges are only half visible but counted as full) on the portcullis and the vertical and horizontal beams are clearly half-lapped with the horizontal beams solid side showing to the rear (interior of the castle and the vertical beams solid towards the front (exterior). It is also clear that the beams are squared rather than rounded logs. The castle gate is in the form of a double door but doesn't seem particularly squat. The height isn't really given but I'm assuming 16' overall. Illustrations of how a portcullis works suggest that the vertical grooves extend inside the castle wall by a foot and a half to two feet each or three to for feet overall and about the same at the top. I'm assuming that the points of the portcullis rest on the floor rather than a recess. The square spaces formed by the beam grillwork looks to be about one and a half times the thickness of the beams.

The primary question is whether the beams are 6" x 6" or 4" x 4". (Actually they could be anywhere in between those two values, but I'll assume the two extremes are all that apply.)

To have 13 visible vertical 6" x 6" beams, the gate would have to be 11 x 15" (6" beam + 9" space) +2 x 3" (half width edge beams) or 13' wide which would make the gateway look almost square, which it doesn't.

Thirteen visible vertical 4" beams would occupy a width of 11 x 10" (4" beam and 6" space) + 2 x 2" (half width edge beams) or 9' 6" which would look about right for a gateway approaching twice as tall as it is wide. The individual gate doors would then be about 4' 9" wide which also sounds right.

The wood used for portculli was usually oak and could be faced with iron.
There are no indications of solid iron portculli ever existing.

Portcullis height would have been adequate for a mounted man to ride through, so 14’ to 16’ height and again 1’ 6” to 2’ remaining unseen across top. I'm going with 14' of visible height and a real height of 16'.

English Brown oak has a density of 45 lb/cu. ft.
Iron has a density of 491 lb./cu. ft.

Assuming the vertical grooves have a two foot depth, there are three additional spars for each groove for a total 19 vertical spars each 16’ high.

The sixteen foot height allows for (16’ -4”)/10" spacing = 15’ 8”/10” +1 = 182/10 +1 = 18 +1 = 19 horizontal spars 13’ 6” long.

Half laps = 19 x 19 = 361 half laps of 4” x 4” x 4” = 361 x 64/1728 cu ft = 13 10/27 cu. ft. material lost to joinery

Portcullis volume is 19 x 16’ x 4” x 4” (vertical beams) + 19 x 13’ 6” x 4” x 4” (horizontal beams) = 28 6.5/9 cu. ft. -13 10/27 cu. ft. (joinery loss) = 15 9.5/27 cu. ft

15 9.5/27 (oak volume) x 45 lb./cu. ft. (oak density) = 690.8 lb. or about 691 lb. for a wooden portcullis of oak.

We need the surface area of the wooden lattice and the thickness of facing to determine the weight of the iron facing. I'm assuming that the front and back of the interior groove beams are also faced to keep the portcullis from being shaken within the groove.

Surface area = 19 x 4” x 16’ + 19 x 4” x 13.5’ (– 19 x 19 x 4” x 4”) = 248 13/18 sq. ft. for the front and the same for the back or 497 8/9 sq. ft. overall.

The interior of the lattices may or may not be faced, so we'll calculate it separately.

The interior of the lattice within the groove is not considered to be faced, so the lattice will use 10 (number of horizontal row spaces) x 17 (number of vertical rows) x 4 (top, right, left and bottom piece for each grill space) x 6” (height of each grill space) x 6” (width of each grill space) = 170 (spaces) x 4 pieces x 36 sq. in. x 1/144 sq. ft./sq. in. = 170 spaces x 1 sq. ft. = 170 sq. ft. to face the lattice interior.

Assuming metal is 3/8” thick on all facings total volume is:
497 8/9 sq. ft. x 3/8 in. x 1/12 ft./in. = 15 17/288 cu. ft. if iron is only put on the front and back of the portcullis,

or

(497 8/9 + 170 sq. ft.) x 3/8 in. x 1/12 ft./in. = 20 251/288 cu. ft. of iron if the lattice interior is faced as well.

15 17/288 cu. ft. x 491 lb./cu. ft. = 7,393 lb. iron if only the front and back are faced,

Or

20 251/288 cu. ft. x 491 lb./cu. ft. = 10,171 lb. iron if lattice interior is faced as well.

Total weight 691 lb. oak + 7,393 lb. iron = 8,084 lb.
Or
Total weight 691 lb. oak + 10,171 lb. iron = 10,862 lb.

[Varyon]

Originally Posted by Anthony

Actually, you probably just put a heavy shaft underneath the counterweight; a counterweight that's not free to drop won't help lift.

Indeed, this is one of the possibilities I mentioned (setting the counterweight on something). I was using an overly-broad definition of "detached" which meant "make it cease functioning as a counterweight."

Originally Posted by Curmudgeon

Total weight 691 lb. oak + 7,393 lb. iron = 8,084 lb.
Or
Total weight 691 lb. oak + 10,171 lb. iron = 10,862 lb.

That's fairly close to mlangsdorf's estimate of 4 tons, so sounds like that's probably one of the more realistic options. That means somewhere between effective ST 71 and 83 (BL 1010.5 or 1357.75 are the minimums). You'll need between 5 and 7 ST 33 ogres to pop that thing open. Assuming SM-2 is a typo and it's supposed to be SM+2, you're going to have difficulty fitting more than a couple ogres for lifting, so that thing probably isn't coming up without using the actual mechanism. If those ST 33 ogres are indeed SM-2, they'll fit without much issue and be ready to tear apart the guards on the other side. Strong little buggers.