Please Need Help with Finger Skirt Design

[kemo1988]

When considering lift from an integrated fan you should use the manufacturers software with the fan static pressure set to the required fan duty static pressure (or cushion pressure +10% nominal) to calculate the whole fan volume at that pressure. the lift will be the 18.5% (or lift area %) of this figure (Also note the power absorption). HOWEVER........ you must then calculate the thrust for this fan at the SAME pitch but at 50-100 Pascals (nominal), (again noting the power absorption) deduct the lift area, add the power figures and check that the overall power required is available to you.

you came to the point which exactly what i wanted to ask, and you made me more confused..

now the fan i selected according to my understanding is 1000/8-8/ppg @2000rpm

first the lift: as you said the fan now is set to 600 Pa @ 2000 rpm and it give 20 m3 and take 39.2 HP, this is the whole volume right, so 20*0.185 = 3.7 m3 for lift.

now to calculate the thrust : @ 100 Pa 2000 rpm the fan gives 23.8 m3 and take 34 HP,..... then what

[kemo1988]

When in flight the gap under the front skirt will be about 25mm and under the rear 0mm (given correct balance) to give sufficient drag for steerage. if more air is fed to the cushion the gap will increase, less air and it will reduce giving more drag and possibly (if too little air) a plough-in.

Paul Fitz

I dont actually understand how the front skirt will have 25mm hovergap and the rear 0mm, what i know is to make the craft steer you need to make for the craft at the front an axis to rotate around, and to do that you need more drag at the front than at the rear.

[Paul Fitz]

then what

Using Multi-wing data - set 1000/8-8/5Z using the AMCA B test method, Using your initial figures of 20m3/sec and 40HP, I get a working point between 40 deg and 45 degrees. For clarity I moved the working point to the 40 degree blade angle and get:-

1000/8-8/5Z/40 deg
19m3/sec @603Pa 40.1 HP 60% eff ---- 19x0.185= 3.51m3/sec Lift --- 40.1x0.185 = 7.4 HP

reset to 100Pa but stay on the same 40 degree curve.

23,2m3/sec @ 99.9Pa 34.6HP 55% eff ---- 23,2x0.815= 18.9m3/sec Thrust ---- 34.6x0.815= 35.6 HP

therefore total power absorption = 35.6 + 7.4 = 35.6HP

(0.815 being the % thrust area of the fan)

Figures will vary slightly between calculators and the results are somewhat crude as they do not take into account turbulence around the splitter plate. The distance between the splitter and the rear of the fan can also affect the result, as air 'bleeds' around the leading edge of the splitter. It is however the easiest method of obtaining a reasonable estimate of the overall power absorption for a given splitter position.

Paul Fitz

[Paul Fitz]

I dont actually understand how the front skirt will have 25mm hovergap and the rear 0mm, what i know is to make the craft steer you need to make for the craft at the front an axis to rotate around, and to do that you need more drag at the front than at the rear.

You are correct. However if the craft is completely clear of the surface with virtually no skirt or wave drag at speed it will be difficult to steer, as every little movement of the rudder would impart a turning force. Some drag is necessary. When the craft is trimmed it is normal to have more weight to the rear to counter the effect of thrust from the fan trying to push the nose down. This makes the craft run tail down giving the required drag. If a greater volume of air is fed to the cushion, the hovergap will increase. With practice the driver will automatically sense the craft trim and adjust it by moving his weight. In a turn the driver can move forward and sideways toward the turn to create drag on one front corner, allowing the drag to reduce at the rear and the rudder to assist in pushing the rear of the craft around the maximum point of skirt drag.

When you look at most craft travelling on water, you will see that the skirt starts to drag causing spray from the contact point about 2/3 down the length of the craft. Some of this spray is caused by direct skirt contact and some by air being forced under the rear side segments as the craft moves forward because the rear segments form a tighter seal to the water. This air forces water out from under the skirt contact point adding to the spray.

[kemo1988]

Thrust ---- 34.6x0.815= 35.6 HP

therefore total power absorption = 35.6 + 7.4 = 35.6HP

Paul Fitz

Thanks Paul, there was a couple of thoughts in my mind none of them was multiplying the power to the percentages. I think the thrust power is 34.6 * 0.815 = 28.2 HP and Total 7.4 + 28.2 = 35.6 HP.

I have the latest version of optimiser, i dont know where that difference came from, anyway, Now i have recalculated these numbers to just get the feeling:

Fan: 1000/40/PPG/5Z @ 2000 RPM, Pressure = 600 Pa, lift % = 18.5, thrust % = 81.5

For Lift: 20m3 @ 600 Pa for whole fan-------20 * 0.185 = 3.7m3 for lift------Power consumed for lift = 39.2 * 0.185 = 7.3 HP

For thrust: 23.8 @ 108 Pa for whole fan ------- 23.8*0.815 = 19.4m3 for thrust ----- Power consumed for theust = 34.4 * 0.815 = 28 HP

Now for air velocity, area lift = 0.145 m2, area thrust = 0.64 m2

For lift: 3.7/0.145 = 25.5 m/s giving a dynamic pressure of 390 Pa + 600 (static) = 990 Pa (Total).

For thrust: 19.4/0.64 = 30.3 m/s giving a dynamic Pressure of 550 Pa + 100 (static) = 650 Pa (Total).

For static thrust: 19.4 * 30.3 * 1.2 = 705 N = 158.8 Ibf

As you stated in the PDF the air velocity is lower for the lift area, but how the total pressure differs it is supposed to be constant along the fan.

For the drag force can i use the total drag curve in the PDF or it will differ to get the actual craft speed, and i didnt find the "Drag calculation" PDF if you have it please attach it.

Kareem

[Paul Fitz]

As you stated in the PDF the air velocity is lower for the lift area, but how the total pressure differs it is supposed to be constant along the fan.

For the drag force can i use the total drag curve in the PDF or it will differ to get the actual craft speed, and i didnt find the "Drag calculation" PDF if you have it please attach it.

Kareem

Total pressure (Pt) at the fan is usually considered to be constant across the disc. In fact this will not be true for a typical hovercraft fan due to the marked difference in performance at different parts of the blade.The total pressure within any part of a system will be dependent on the value of system losses up to that point (see downloads page - "Principles of hovercraft design"- section - pressures in a ducted fan system).

An integrated fan is an unusual and special case. The thrust part of it has a very low resistance to flow, but the lift section has a relatively high resistance to flow. So theoretically the area immediately behind the fan should have a constant Pt across the disc, but just within the entry to the area under the splitter Pt can be lost because of the different size and conditions of the duct at that point. Above the splitter Pt will be almost constant between the fan and the discharge, (a very slight loss due to wall friction losses). Each part of the fan should be considered as a different "system", as would two seperate fans.

I am not aware of a "Drag calculation PDF". There used to be an Excel Drag Calculator on the Old THCC website, which may still be available from Juergen Schoepf.

If any of our American cousins are listening in and still have contact with Juergen, perhaps they can help obtain this for re-publication on the HoA and HCGB websites.