Glueandcoffee
Junior Member
- Joined
- Feb 8, 2021
- Messages
- 77
- Reaction score
- 22
- Location
- Cork
To install this app on your iPhone:
Note: This feature may not be available in some browsers.
ThanksHi,
Nice project! In addition to what Alexander says you have to consider cavitation numbers, except for ventilation. In Tom Speers curves below, you might get a grip on what you need to go for(50kmh=27kn). About lifting forces you need to use CL3, that might be half or less of the CL2 numbers you have used. Let´s say you follow Alexanders advice to go for CL2=0.4 and hypothetical calcuations give CL3=0.2, you are down to one third of the force your previous calculations show. As you have a slanted foil the forces can be calculated for both horizontal and vertical lift. Curious, what profile are you using? These are my thoughts, but Alexander is the expert here.
View attachment 165949
In all honesty I don't have a clue what CL2 and CL3 are. I know what a coefficient lift is and I know what an aspect ratio is. Since I don't know what CL2 and CL3 are, I can't do the calculation. Even if I knew what CL2 was and did the calculation I wouldn't know what to do with CL3 and therefore what I'm expecting is beyond me. I'm also equally none the wiser to cavitation number. I know what cavitation is and it happens when the pressure in the liquid drops below the vapour pressure of said liquid(I think). The rest is like the stars... way over my head.Glueandcoffe, an OVER-simplified way of calculating CL3 is CL3=0.9*CL2*(AR/(AR+2)), where AR=span^2/projected area. Note this is very coarse, but it will anyway give you an idea of what to expect.
Definitely not as advanced as your Sailrocket design, though.
View attachment 165999
I must have been out the day they explained that in school. Haha.By "cl" I mean the lift per unit area divided by the free-stream dynamic pressure at one section somewhere along the foil's span. By "CL" with capital letters I mean the total hydrodynamic lift of the foil divided by the foil's projected under-water area and free-stream dynamic pressure. I think those definitions are quite common.
After the initial testing you will have a better idea about what you need.
Since the sail-force is not limited by the weight of the boat here, a model Sailrocket should be capable to sail as fast as the full-scale version, so you shall not be happy with just 30 or 40 knots!
Really interesting with the base-ventilated foils. Have to digest your info for a while.
simple electronic system with a GPS sensor, MPC and linear actuators controlling the foils to get optimal depth at any speed. If it doesn´t work as expected, I will anyhow have potentiometer-controlled incidence adjustment with the GPS removed.
Interesting idea with linking a GPS to an actuator for incidence control. My humble opinion would be to just increase the root angle of attack high up on the foil and have the tip(V intersection) angle of attack closer to 0 on the lower part of the foil. Might not be possible in your case but it seems simpler.
When I read your comment about how the sail-force and the force from the foil aThanks
Could you further explain cavitation numbers and the graph below. What do you mean by CL2 and CL3.
I do have a slanted foil but I'm not breaking the forces up into horizontal and vertical components as the lower section of the foil is parallel with the kant of the wing sail and the forces from each are resolved through the center of effort for each of them. Ie wing force = foil force . The upper part of the L foil will just lift the rear float clear of the water as it accelerates. the faster it goes the higher the ride height and the less area of the upper foil is in the water. Am I understanding your point correctly.
I'm also curious as to what profile I'm using. The first foil I made was cut and sanded from a piece of thick pvc pipe which was then heated and bent into shape. I had no design parameters in mind from the get go and it was very much done by eye. The second foil is not much better as I simply used the first one as a mold for the carbon fiber. Below the radius of the foil, the root chord is 80mm , thickness 12 mm , position of max thickness is about 25 mm from leading edge and I'm unsure how much camber there is but not much. I'll see what kind of naca profile I get with these dimensions. Keep in mind the profile is far from uniform along the span.
Do you really think it is simple(or cheap) to make a twisted foil? Mayfly used that configuration with success in the 70´s, with twisted foils milled from solid aluminium
.
Your opinion might hold for a vessel like yours, that will be sailed without heel. But a "conventional" foiler with two surface piercing mainfoils will and must have the same CL wether the leeward foil is deeper or shallower at a certain speed, depending on heel.
But back to your rear foil, use the lift equation and the numbers you used earlier, but substitute your CL2=0.6 with the value you get from the CL3 equation to get a more realistic force figure. But first of all use a CL2 of something like 0.3 as a NACA2315 with CL2=0.6 will cavitate in the ballpark of just under 25knots. So instead of 0.6x500x196x0.0225=1323 we get 0,9*0,3*((0,36^2/0,0225)/(0,36^2/(0,0225+2))*500*196*0,0225=441N with the rear foil completely immersed in water. 0,36m is the 14" span you mention. Curious, is the foil ment to be fully immersed whatever speed?
Might have misunderstood the concept, though.
Nothing in this world is ever cheap. Something I heard someone say a few years ago: if it flies, floats or f*cks... rent it.Do you really think it is simple(or cheap) to make a twisted foil? Mayfly used that configuration with success in the 70´s, with twisted foil milled from solid aluminium
![]()
But back to your rear foil, use the lift equation and the numbers you used earlier, but substitute your CL2=0.6 with the value you get from the CL3 equation to get a more realistic force figure. But first of all use a CL2 of something like 0.3 as a NACA2315 with CL2=0.6 will cavitate in the ballpark of just under 25knots. So instead of 0.6x500x196x0.0225=1323 we get 0,9*0,3*((0,36^2/0,0225)/(0,36^2/(0,0225+2))*500*196*0,0225=441N with the rear foil completely immersed in water. 0,36m is the 14" span you mention. Curious, is the foil ment to be fully immersed whatever speed?
Might have misunderstood the concept, though.