lehmann

I looked at the drawings for the carronades provided on the CD from the USS Constitution Museum. They provide two designs, one from the 1927-31 rebuild and another that is "1812 Era" (1985 drawing). The first drawing shows two wooden "t-nuts" that fit in the slot in the skid and hook under the skid. Each t-nut is attached to the slide with four 1.25" bolts. There is 1/2" clearance between the bottom of the skid and the hook edge on the t-nut. The 1812 design just shows a 3.75" fighting bolt that runs in the slot and goes through a large (8" diameter) ring that runs under the base. There is no other block in the slot to stop the slide from rotating on the skid shown on the drawings. I did some engineering calculations to estimate the forces on the various components of the 1812 design. (I'll post them later). Assuming the thrust force from the barrel is 1000 lbs, then: Compression force on the elevation screw: 350 lbs Tension (lifting) force on the fighting bolt: 260 lbs Force of the back edge of the slide on the skid: 540 lbs There's a lot of friction between 1) the end of the slide and between the skid and 2) the fighting bolt ring that slows the recoil: assuming a friction coefficient of 0.5 (wood on wood and metal on wood), then the friction force is (0.5)(540+260) = 400 lbs. Most, if not all, of this force will be transferred to the pivot pin next to the gun port. (I wonder if this means more energy goes to the ball since the barrel recoil velocity must be smaller. Also, since the recoil is absorbed over a period of time, it should mean less impact on the hull when the breech cable tightens up.) As Jud showed in his drawing, the elevation screw is threaded into the cascabel, so when the screw is not 90 deg to the top of the sled then there is a bending torque on the screw. However, since the bottom of the screw is just sitting on the surface, there is some friction involved, which reduces the bending forces by about 60%. Going back to the original question of why there there is only one pin in the slot. If the fighting bolt (pin) is not exactly under the bore then there will be torque that will twist the barrel sideways (looking from above). This will happen also if the blast force is not symmetric to the bore or the friction between the underside of the slide and the top of the skid were not the same on either side of the slot (actually, this would be outright unstable), so I would think it would be a good idea to have a second slide in the slot. There are, however, several reasons why the slide will recoil straight back: that the moment of inertia of the barrel is so large that it would take a lot of torque to make it turn as it recoils. there is a lot of friction between the breech end of the slide and the skid, which will limit rotation of the slide. the force of the explosion acts on the breech, so as long as the fighting bolt (pin) is out-board of the breech, then this force will push the barrel straight back. These factors also apply to army artillery mounted on two wheels - they recoil straight back. I have to think this out some more, but I think it's important that the center of mass of the barrel is inboard of the fighting bolt. Overall, it looks like getting the geometry right does most of work, and friction helps to take care of the imperfections. Bruce

Other factors: - they didn't have calculators (maybe slide rules!) and few could even do arithmetic. - have you ever tried to measure the diameter of a rope, even with calipers? Circumference is far more reliable and repeatable. Bruce

You may want to recheck the reference to 9". I suspect it refers to circumference, not diameter (because it's easier to measure). Also, from what I've read, when converting from circumference to diameter, the practice was to just divide by 3, not 3.1416. Bruce

I think the main reason for the tear-out (fuzzies) is that you're using a rip tooth for cross-cutting. The rip saw tooth is basically a square chisel that tears, more than cuts, grains. A cross-cut tooth needs to have a angle on the tooth face and ideally on the top as well. This design creates a high point on one corner of the tooth that cuts/scores the wood fibres. I haven't seen any small circular saw cross-cut saws being offered for sale anywhere (is Jim Byrnes listening?), but you can re-sharpen a rip saw with a small triangular file. It's not difficult. See http://www.vintagesaws.com/library/primer/sharp.html for hand saws, but the same idea holds for filing circular saws. The second possible reason for tear-out that the feed is too fast. This obviously creates a larger bite for each tooth and larger forces on the wood grains. At the end of the cut, these grains are unsupported, so they bend rather than get cut. This the reason that zero-clearance saw throats and miter guides are used - to support the fibres at the end of the cut. To make use of this, I would have fed the piece from the other side so that the long (outboard) edge would be against a sacrificial extension on the miter. As you've done it, each step of the profile is unsupported at the end of the cut. The third reason is related to the size of the tooth relative to the depth of cut. Each tooth has to have enough volume in the triangular section (called the gullet) to hold and carry all the sawdust that is produced at the tooth tip. As the feed or depth of cut increases there is more sawdust to be stored in the gullet. If the gullet gets too much sawdust then it will be under pressure so that at the end of the cut the sawdust pressure will blow the final fibres out before they are cut. Generally, smaller teeth (more teeth in a saw) have smaller gullets, which is one of the reasons fine-tooth saws can't be fed fast. As a last point, I would have made the profile to run across the grain, then ripped with the grain. Phillip Reed shows this in his book "Period Ship Modelmaking". As you've done it, the carriage cheeks are very fragile and the surface is rough, which means a lot of sanding or a lot of paint. Bruce

Google has a 360 interactive "scan" of all the decks. The's a link to another from the museum website https://www.google.com/maps/views/view/streetview/us-highlights/uss-constitution-deck/Jjoh75kl7doAAAQZN_qRbA?gl=us&heading=317&pitch=74&fovy=75

Salute Gaetan, I've seen the post. My interpretation is that It covers the number of threads of a given diameter that make a strand of a given diameter. I covered the next steps: the diameter of three strand rope for a given strand size and the diameter of a cable made up of three ropes. For instance, I can make a rope using three strands that each have have 25 threads. If you get a chance to look at the Chapman reference, he talks about a progression of sizes, starting (I think, haven't got it all straight yet) from threads and yarns, which are spun, then moving up to ropes, which are twisted (as on a ropewalk), then larger ropes, which are twisted from several ropes. As modelers, we'll have to decide, depending on the scale we're working and the amount of patience, what will be the fundamental fibre to start making our ropes. I see on another post that someone what trying to start the process from 100 or 120 cotton thread. Another was pulling apart threads and using the smaller "yarns" that make up thread. Bruce PS. I've just found another old book on rope making: "ModernFlax Hemp and Jute Spinning and Twisting" by H.R. Carter 1909. https://ia902307.us.archive.org/3/items/modernflaxhempju00cartrich/modernflaxhempju00cartrich.pdf I hope it will clarify the definitions of yarn, thread, lessom, etc.

I found "Treatise on Rope Making - Description of the Manufacture, Rules, Tables of Weights, etc." by Robert Chapman (Master Ropemaker of HM Dockyard, Deptford), 1869. The topics start with harvesting hemp, twisting fibres, spinning, tarring, amount of yarn.threads for various sizes and types of cable, sizes of hearts, weights of cables by length, It even covers manufacturing costs, by operation, including management (officers) Although published in 1869, the English seems older (it is a revised edition), and there is a lot of jargon that will take a while to decipher. The method of showing calculations is not easy to follow. Not an easy read, and not a place to find quick answers. Download of pdf, eBook and other formats from the Smithsonian. http://library.si.edu/digital-library/book/treatiseonropema00chap This is probably a candidate for posting in the References list.

Frank Thanks for the posting and the chart. A quick check shows that my formula works for the 3 strand ropes. In you post you say Cebelia, but the chart lists Cordonnet. Could you clarify? Is one a better quality for rigging, or? Thanks Bruce

I was beginning to think about getting/making the correct size ropes and cables for my project. It occurred to me that if I want a certain size finished cable then how do I select the diameter of the strands used in the cable?. There doesn't seem to be any data or formula available for this. I guess most ropewalkers have developed their own by trial and error. I started making a drawing of a 3 strand rope to see the relationship if the strand were assumed to stay circular. I then took the 3 strand rope and used that to make a larger 3 strand cable (now made of 9 strands). Have a look at the drawing in the attached pdf. I then tried to figure out how the helix angle (lay angle) of the rope affects this. As a quick simplification I assumed that this reduced the diameter by the cosine of the angle. Based on my model, a rope made up of three strands of diameter d and a helix angle of A will have an outside diameter of 2.155 x d x Cos(A). Or, for a 9 strand rope, [2.155 x Cos(A)]2 = 4.64 x Cos2(A). Theory is good, but it needs a reality check, so I measured some ropes and string around the shop. The rope measurements and the calculated results are tabulated in the attachment. It turns out the formula works quite well. The numbers I have are for fairly large size ropes, not model size. If some one could measure some of their home made rope, or try to use the formula to make a given rope size, that would be a good test. Also, if anyone had access to actual cable (say a 6" hauser), that would be very interesting. I can update the table or refine the formula as needed. If this works, I could try stroud layed 4 strand rope. Bruce RopeSizeData.pdf