dtoc
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Much discussion has been put up in boards on why the canting keel systems are failing, including support structure, ram pistons and hydraulics. I thought it might be interesting to look at the forces involved and the equipment used to exert them.
The canting keel works in a lever mechanism, where the bulb sits on one end of the lever and the hydraulic rams push/pull on the other. In order for the keel to stay in place relative to the structure, the torques created by the rams and the bulb must be in balance.
In order to do these calculations the masses and geometries must be known. For this I used information posted on the web, specifically the ram info from Cariboni for their special Volvo 70 Titanium ram and the general rule dimensions for a Volvo 70 along with some estimates based on the CAD drawings shown.
The draft is assumed to be 14.5 ft, or 4.42m. I used a bulb mass of 6000 kg, but also calculated based on 4500 kg. While the draft is 14.5 ft, the distance from the pivot point to the cg of the keel will be less than that. This is due to the pivot point being below water level, even though it is inside the hull, and that the cg is not at the max distance from the pivot point. Initially I used 14 ft (4.27 m), but I will show where this is in the calcs in case you want to split the fin and bulb weights and calculate a detailed cg and effective mass. However, the exact mass of the bulb is a highly guarded secret for the teams.
The other end of the lever is easier to calculate. The max stroke of the custom ram is 386 mm. For it to swing the keel through 80 degrees (+/- 40) and have this range of motion the geometry can be solved for with the dimensions given. The 1 unknown is the trunnion pivot point of the ram. Its distance from the end of the cylinder, height above the keel pivot point and the lateral distance from the pivot point. Assuming a pivot point of 1/3 of the distance up the cylinder, an estimate from the cad drawings, the other 2 numbers can be solved for. There is a relationship between the pivot point and the length of the lever. I solved this to maximize the torque applied. The other assumption is that the 2 rams connect to different arms, at an angle away from 0 when the cant is 0 degrees. This is to maximize the pull force when at max cant. From the drawings I assumed +/- 20 degrees. The optimum value can be solved for, but I haven’t. These calculations gave me a lever length of 340 mm with the pivot point 20 mm higher. The combined force perpendicular to the respective lever connection points is 1.675 times the push force.
Steady state with 0 degrees heel and 40 degrees cant
6000*4.27*SIN40=Fpush*1.675*0.340
Fpush=(6000*4.27SIN40)/(1.675*0.340)
Fpush=28930 kg
Steady State with 20 degrees heel and 40 degrees cant
Fpush = 38976 kg
The ratio of these forces to the rated loads for the ram can be viewed as a factor of safety or as the number of g’s before that limit is reached.
Case 1: 2.2 to working load 6.6 to ultimate
Case 2: 1.6 to working load 4.9 to ultimate
My question to the naval architects is what is range of g forces a boat like this would encounter when landing? Or to anyone who was keeping track of the onboard data from the VOR website? I remember seeing 4 g’s.
This does not seem like much of a factor of safety, which may be why these titanium rams have been failing.
The next subject to discuss is the hydraulic pressure transients created by the shock loads and what can be done to control / utilize them. Which is the reason I did all of these calcs. As has been previously mentioned pressure relief valves are not effective.
The canting keel works in a lever mechanism, where the bulb sits on one end of the lever and the hydraulic rams push/pull on the other. In order for the keel to stay in place relative to the structure, the torques created by the rams and the bulb must be in balance.
In order to do these calculations the masses and geometries must be known. For this I used information posted on the web, specifically the ram info from Cariboni for their special Volvo 70 Titanium ram and the general rule dimensions for a Volvo 70 along with some estimates based on the CAD drawings shown.
The draft is assumed to be 14.5 ft, or 4.42m. I used a bulb mass of 6000 kg, but also calculated based on 4500 kg. While the draft is 14.5 ft, the distance from the pivot point to the cg of the keel will be less than that. This is due to the pivot point being below water level, even though it is inside the hull, and that the cg is not at the max distance from the pivot point. Initially I used 14 ft (4.27 m), but I will show where this is in the calcs in case you want to split the fin and bulb weights and calculate a detailed cg and effective mass. However, the exact mass of the bulb is a highly guarded secret for the teams.
The other end of the lever is easier to calculate. The max stroke of the custom ram is 386 mm. For it to swing the keel through 80 degrees (+/- 40) and have this range of motion the geometry can be solved for with the dimensions given. The 1 unknown is the trunnion pivot point of the ram. Its distance from the end of the cylinder, height above the keel pivot point and the lateral distance from the pivot point. Assuming a pivot point of 1/3 of the distance up the cylinder, an estimate from the cad drawings, the other 2 numbers can be solved for. There is a relationship between the pivot point and the length of the lever. I solved this to maximize the torque applied. The other assumption is that the 2 rams connect to different arms, at an angle away from 0 when the cant is 0 degrees. This is to maximize the pull force when at max cant. From the drawings I assumed +/- 20 degrees. The optimum value can be solved for, but I haven’t. These calculations gave me a lever length of 340 mm with the pivot point 20 mm higher. The combined force perpendicular to the respective lever connection points is 1.675 times the push force.
Steady state with 0 degrees heel and 40 degrees cant
6000*4.27*SIN40=Fpush*1.675*0.340
Fpush=(6000*4.27SIN40)/(1.675*0.340)
Fpush=28930 kg
Steady State with 20 degrees heel and 40 degrees cant
Fpush = 38976 kg
The ratio of these forces to the rated loads for the ram can be viewed as a factor of safety or as the number of g’s before that limit is reached.
Case 1: 2.2 to working load 6.6 to ultimate
Case 2: 1.6 to working load 4.9 to ultimate
My question to the naval architects is what is range of g forces a boat like this would encounter when landing? Or to anyone who was keeping track of the onboard data from the VOR website? I remember seeing 4 g’s.
This does not seem like much of a factor of safety, which may be why these titanium rams have been failing.
The next subject to discuss is the hydraulic pressure transients created by the shock loads and what can be done to control / utilize them. Which is the reason I did all of these calcs. As has been previously mentioned pressure relief valves are not effective.
