By Stan Hall - Soaring, October 1978, Pages 41- 45
HP-14 Installation (Updated 1/19/02)
The Gazelle/Gorilla Syndrome
One of the complaints frequently voiced by pilots of flapped sailplanes concerns the high handle forces required to lower the flaps. Typically, the pilot tiptoes into the landing pattern with small light touches on the stick and rudder pedals, only to be faced at the last minute with the Herculean task of muscling back on a heavily loaded flap handle. The sailplane transforms immediately from gazelle to gorilla, with the result that the keen edge of pilot judgment required for precision landing is blunted by the distraction of all that heavy pulling. His right hand is caressing the sailplane with the fingertips and wrist while the whole left side of his body is wrenching on that blasted flap handle. Annoying, to say the least, and potentially dangerous.
There are ways to get those flap handle forces down. One way to do this is to design in as much flap handle travel as practicable in the cockpit. The commonly-used single-stroke lever, which provides not more than 10 to 12 inches travel at the top, is not the answer. Better to have four to five times that much.
For those readers who have forgotten their high school physics, the reason for having lots of available flap handle travel is shown in the principle of the chain hoist: You can lift a lot of weight with a chain hoist with very little tugging on the chain - but you have to pull a lot of chain through the device to do so. A small force applied over a long distance will move a large one over a short distance. And that's the situation with flaps. They provide high loads that need be moved through short distances, and the way to keep the force down in the cockpit is to have that force operate over a long distance.
The more capable designers have provided the equivalent of large flap handle travels by the use of gears, ratchets, cranks, and multiple flap handles. An example of the multiple handle technique is described by Paul Bikle in the October 1970 issue of Soaring. Here, he describes the flaps on his T-6 as driven by a long push rod (a tube, actually) in the cockpit. One handle is secured to the forward end of the tube. In the flaps-up position, this handle is so far forward the' pilot can't reach it. However, a second handle is secured to the same tube, back where the pilot can reach it. To lower the flaps he first pulls on the nearer handle and locks it in the aft position. This brings the first handle within range and he reaches forward and grabs another handful. The old Chain Hoist Trick.
According to his article, Bikle gets 22 inches flap handle travel this way. But even so, he reports that at 80knots,. 100 pounds pull on the handle are required to fully depress the flaps to 68' - which force, he suggests (with a twinkle in his eye?) is within the average pilot's capability. At 55 knots he reports 55 pounds - which for a skinny old graybeard like me is still much, much too high. You too, maybe.
A simple and effective way to depress the flaps, and still have forces as low as you like in the handle, is to do the job hydraulically which is what this article is all about.
The column is presented in two parts. Part I goes through some of the simple arithmetic of design. As in all forms of design, a little prior arithmetic always minimizes expensive and frustrating time-consuming trial-and-error. If you want to determine the basic design parameters of your own system, you'll find I've done most of the arithmetic for you and all you'll need do is fill out the form in Table 1.
Next month, Part 2 will examine a neat little system built by Sherb Klein of Redwood City, California, for his HP-14. The design is,, with a few mods, the one originally worked up with good results by Dick Schreder. Although I (and a great number of other glider guiders) consider Sherb a master craftsman, I think you'll agree after reading about his system that you can build one like it yourself - without the prerequisite of Sherb's talents.
The Basic System
As shown in Figure 1, there isn't, hardware-wise. really much to the system. It comprises a double-acting hand pump, a hydraulic actuator to drive the flaps, a relief valve and a reservoir, plus piping. Except for the -pump, you can build everything you need yourself. There are a couple of lathe-jobs, though. If you're really clever and a confirmed workaholic, you can even build the pump. You get two additional Attaboys for this - but no how-to-do-it information from Homebuilders’ Hall.
Operationally, to lower the flaps you merely pump back and forth on the handle, and the flaps will stay put when you stop. To raise the flaps (all at once or incrementally), you push another handle which opens the relief valve. This permits the flaps to push the hydraulic oil back to the reservoir in the process of coming up. Springs in the flap circuit help the flaps come snugly against the up-stops as the-air-load on the flaps decays.
The relief valve also provides protection against damaging the system in the event of overspeeding. The valve and springs are adjusted so that when the flap loads reach a predetermined level, the valve will vent oil out of the actuator, back to the reservoir, and the flaps will come up until the load is again at a safe level.
In this connection, you must reread Bikle's article in the October 1970 Soaring. The article states that he was diving the T-6 at 102 knots (straight down or better) when the flaps broke. "Fortunately, " he writes, "both right and left flaps faded at the same time and in the same manner so that recovery was possible without undue rolling"! (the exclamation mark is mine). . Paul has to be the coolest cat in town. Aside from that, had the T-6 been equipped with a hydraulic system such as we are discussing, the flaps would have relieved themselves at the appropriate airspeed, thus reducing the possibility of the pilot finding himself doing the same thing in the cockpit. No report from Paul on this.
Basic Design Procedure
As indicated earlier, the initial phase of design calls for doing a bit of simple arithmetic, which can be expedited through the use of Table 1. Follow me through on this table. and we'll both "design " the system Sherb Klein used on his HP-14. After you see how simple the process is, you can fill in your own numbers for that new system you are already contemplating.
Steps I through 4: We start by filling in those numbers we know right off, like the maximum flap deflection (90' in Sherb's case) and the airspeed at which the flaps will be deployed (55 knots). We also fill in the total flap area (14.82 sq.ft. for both flaps) and the average flap chord (6 in.).
Step 5: Write in the "K" factor taken from Table 2. This value is a function of the airspeed at which we propose to lower the flaps and the maximum flap angle - in Sherb’s case 55 knots and 90' gives a K of 6.9. For those readers who may be interested in what this K-factor means and where it comes from, these are treated at the end of Part 1.
Put Table 1 away for a while and tape a piece of drawing paper to the top of your kitchen table. Lay out the flap torque-tube drive horn and the actuator center lines in accurate scale (full size, if possible), showing both the flaps-up and flaps-down positions as indicated in Figure 2. Make the horn radius any dimension that seems appropriate to the surrounding structure, considering clearances, etc. If the radius turns out to be incorrect, you'll know it soon enough, and you can go back and change it.
The relationship between horn and actuator should be such that, at the flaps-down (maximum-torque) position, a line drawn through the center of the horn makes a 90'angle with the actuator center line. The reason for this, of course, is that you want the longest lever arm you can get where the flap torque is the highest. This assures the lowest maximum load in the actuator and in turn the lowest maximum force in the cockpit control handle.
Step 6: From the layout, measure off the horn radius and enter it as shown in the table. Sherb's horn radius is 3.25 inches.
Step 7: Compute the actuator load as shown in the table. Observe that you will be using the K factor here as determined in Step 5. By this technique the maximum load in Sherb's actuator computes to 189 pounds.
Step 8: Estimate the inside diameter of the flap actuator cylinder. Initially, we'll have to do a little creative guessing on this. If necessary, we'll come back later and make corrections.
One of the more practical considerations which determine cylinder I.D. is the size of the 0-ring seal on the piston, and you'll need to use a standard size; one you can buy from your local hardware store, auto parts, or aircraft supply house. You can refer to Table 3 for some standard sizes.
With his experienced eyeball, Sherb picked I-% inches. Although that size 0-ring is .005-inch larger in O.D. than the I.D. of the standard 1.5 x .065 wall SAE 1020 steel tube he used as the actuator cylinder, the 0-ringgroove in the piston provided the right "squeeze" for the 0-ring.
Steps 9 through 12.- Compute the effective piston area in the actuator. It is most desirable that the hydraulic fluid be introduced on that side of the actuator piston which places the piston rod. in tension as the flaps are lowered. If you place the rod in compression, it will need be larger in diameter. Homebuilders' Hall for January 1974 addresses tubes in compression. Reread it.
Since the fluid is on the rod-side of the piston, the rod -cross-section area has to be subtracted from the piston area to yield effective area. In Sherb's actuator the effective piston area is 1.398 sq. in.
Step 13: Go back to the layout on your kitchen table and measure the actuator piston stroke from flaps up to flaps down. Enter the dimension as shown in the table, where Sherb shows 3.75 inches.
Step 14: Knowing the effective piston area and the piston stroke, multiplying one by the other gives the amount of fluid volume required to lower the flaps through their full range. Enter that value in the table, where Sherb shows 5.24 cu.in.
Step 15: Measure the pump fluid output per stroke or obtain it from the manufacturer or seller.
Sherb purchased his pump from Schreder. It puts out 0.55 cu.in. per stroke. Bob Klemmedson and Mac Snyder (RH-7 and HP-18, respectively) bought their pumps from Palley Supply Co., 11630 E. Burk Ave., Whittier, Calif. Sherb advises that at last report the smallest pump available through Palley puts out 1. 5 cu. in. per stroke, or about 3 times what Schreder's does. If you use Sherb's 1.370 I.D. cylinder (why not?) and Palley's pump, you'll need increase the flap drive-horn radius by the ratio of 1.5 to 0.55 (8.66 inches) to achieve the same mechanical advantage. More about mechanical advantage later.
Increasing the horn radius will, of course, change the actuator stroke, so you 11 have to re-do some of the steps in the table.
Step 16: Determine the number of flap handle (pump) strokes required to deliver the actuator volume developed in step 14. This is simply total actuator volume divided by pump output per stroke. Sherb shows 9.53.
Step 17: Measure the travel at the flap handle per stroke, hard aft to hard forward. Sherb shows 6.25 inches.
Step 18: Determine the total flap handle travel required to depress the flaps. This is simply the number of strokes (step 16) times the per-stroke travel of the flap handle (step 17). Sherb shows 59.58 inches (versus Bikle’s 22 inches).
Step 19: Determine the mechanical advantage of the system from the flap handle to the flap horn. This is total flap-handle travel (step 18) divided by total actuator stroke (step 13). Sherb gets 15.89.
Step 20: Magic time. Compute the cockpit handle force required at the flaps-down position. This is actuator load (step 7) divided by the mechanical advantage (step 19). Sherb divides 59.58 inches by 3.75 and gets 11.89 pounds. Better than 55 pounds, right? Step2l: If you’re interested, you can compute what the flap handle forces would be at greater speeds by multiplying the force you just computed in the step preceding by the ratio of the speed that interests you to the speed you used in the table in the first place, squared. Don't forget the square. The table shows how this works out for Sherb's HP- 14.
Now, If You Don't Like the Forces You Came Up with . . . .
If, in computing the handle forces for your own sailplane, you decide they're still too high for your preference, there are some things you can do. You can increase the actuator cylinder diameter, which will increase the amount of fluid you will have to pump. This will require additional pump strokes - and more time to get the flaps down. Note that providing additional pump strokes increases the mechanical advantage. Alternatively, you can lengthen the flap drive-horn radius. This will have the same effect as increasing the cylinder diameter because you are now increasing the actuator stroke - which increases the fluid volume required - which requires more pump strokes, etc. etc.
If the handle forces turn out to be too low (heaven forbid!), you can reverse the procedures described above. Isn't all this fun?
About That K-Factor
The forces in the flap handle come, of course, from the loads in the actuator, and those loads come from the air loads on the flaps. Numerically, the flap loads equate to the Flap-Hinge Moment divided by the flap-horn lever arm (which in the flaps-down position is the horn radius, provided the actuator and horn centerlines are 90' to one another).
Flap loads are normally expressed in terms of Hinge Moment Coefficient, which permits the calculation of hinge moments for flaps differing in size and operating airspeed from those of the flaps actually tested.
The equation for flap-hinge moment (H) is H = CH q Sf cf, where CH is the hinge-moment coefficient, q is the dynamic pressure (a function of airspeed), Sf is the total flap area, and cr is the flap chord. If cf is expressed in feet, the hinge moment comes out in foot pounds. If expressed in inches, the hinge moment comes out in inch pounds.
The K factor in Table 2 is merely CH q. CH comes from the -measurements Bikle reported in the October 1970 Soaring - with one exception. The largest flap angle Bikle shows is 68. 1 extrapolated his data to 900 - not normally a "safe " technique, but one I feel is appropriate for the purpose intended.
As shown in Bikle's article, the T-6.average flap chord is 17.7 percent of the average wing chord. The ratio of flap chord to wing chord has an influence on the magnitude of the hinge moment coefficient, but not much in the range of 10 to 30 percent.
There are some other sailplane-specific factors, too, involved in determining flap-hinge moments, but they would likely bore you; factors like wing lift curve slope, type of flap (plain, split, slotted) etc. etc.
For the purpose, the K factors shown in the table are likely to be accurate enough. Anyway, prediction need not be all that precise. When you grab that handle, you don't do it with a calibrated, laboratory load-measuring device. You grab it with your hairy, imprecise fist.
Next Month - From the kitchen to the shop. Sherb Klein's system. The nitty gritty of making and assembling the parts, Drawings, tables, and all like that.