The east-west runway is just a little under 700 feet long. The Taylorcraft was down and stopped in about 400 feet. It used just a little more getting out. The approaches are clear to the east, and only a fence and cornfield is on the west.
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( 2.9 / 103 )Cut and fit the lower aileron diagonals today.
1 hr
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( 2.6 / 73 )In my earlier post, I derived the performance envelope for the aircraft for which we now need to find the airloads. Let's go in alphabetical order and start with estimating the airloads in Condition A. We will determine the loads on the front and rear spars by summing moments about the rear and front spars, respectively. This is done on a spanwise per-unit-length basis using the local coefficient of lift in each of about ten wing segments.
Condition A represents Va, the design maneuvering speed. From the previous posts, the load factor, n, is +3.8, and the airspeed is 85 kts. The aircraft weight is 1200 lbs. One new piece of information we need today is that the wing weight is estimated at 220 lbs. This acts at the wing CG which I've estimated at 26.6% of the chord length back from the leading edge. This weight includes 90 lbs of fuel (7.25 gallons per side), the tanks and fittings and half of the strut weight in addition to the basic airframe and covering.
To get the loads on the front spar, we add all the twisting forces acting around the rear spar, and divide by the distance between the front and rear spars to find the force required by the front spar to keep things in equilibrium. There are four forces/moments at work in our simplified (CAM 04 Appx IV) analysis:
1) Force of lift acting at 25% of the chord
2) Force of drag acting at 25% of the chord
3) Pitching moment of the airfoil
4) Weight of the wing acting at the wing CG
;)
The lift force is defined as follows:
...where Cl_local is the local coefficient of lift, A is the wing planform area, and q is the dynamic pressure:
...which works out to 24 psf for this speed at sea level. FAR 23, Appendix A in paragraph A23.7(e)(1) states that we need to increase the positive loads by 5%. We also need to account for the angle of attack, as the diagram above illustrates. Multiplying the resulting lift force by the moment arm (distance from the rear spar to the aerodynamic center) gives us the moment about the rear spar due to lift. Using the chord length instead of the area in the equation above, and dividing the result by the moment arm of the front spar gives the force per unit length in the front spar. The equation looks like this, where rac and rfrontspar are the moment arms for the aerodynamic center and the front spar, respectively, about the rear spar:
Drag is made up two components, form drag and induced drag. The equation for the combined drag coefficient for our example looks like this:
AR is the aspect ratio of the wing, Cd0 is the zero-lift drag coefficient from the airfoil data (0.011), and Cl is the wing coefficient of lift. Coefficient of lift can be calculated using the following equation:
S is the gross wing area (135 sf), Wgross is the max gross weight of the aircraft (1200 lbf). This Cl computes to 1.4 which should, and does, come out pretty close to the max wing coefficient of lift of 1.38. Substituting that result into our drag coefficient equation gives a Cd of 0.105.
The equation to find the required reaction in the front spar due to the contributing component of the drag force is as follows:
To be continued...
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( 3 / 48 )I attached five of the noseribs today. I also glued in a few more of the aileron diagonals. All the top diagonals are finished. When the bottom diagonals are all in, it will be time to cut the ailerons free.
2 hours
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( 3.4 / 63 )I cut out the eleven nose ribs today. They were roughed out from 1/4" birch plywood with a bandsaw then matched to my noserib template using a laminate trimmer router bit that has the bearing. These were all clamped together and stack sanded. While they were clamped, I also cut the notch in the nose for the leading edge moulding.
2 hours
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( 2.5 / 53 )The maximum coefficient of lift used in the last post was determined from computer model data. The airfoil on the RW11 appears quite similar to the NACA 4415, but for the thicker section around the rear spar. I've run the airfoil through both XFoil and JavaFoil and produced two polars. This is a very cambered airfoil capable of producing a lot of lift.
The polars listed above are 2D estimates assuming the wing is infinitely long with no spanwise flow, and give a Cl Max from 1.6 to 1.8 at a Reynolds number of 3x10^6. Wind tunnel data from NACA report 824 indicates the Cl Max for the 4415 to be somewhat lower than that, about 1.3 for similar Re. It has never been clear to me whether I should treat the numbers from the NACA report as 2-D or 3-D. That the data is from an actual airfoil and the values are considerably lower than computer model estimates suggest to me that it is 3D. However, the report does say the data is from the Langley 2-D tunnel.
To be useful, we need the lift coefficient to be corrected for 3D effects. For that, it is again easiest to consult Mr. Wilford's spreadsheets. S405-WindDesign.xls specifically. This uses the Schrenk method to approximate the lift coefficient distribution over the wing, and then the wing coefficient of lift. Using that spreadsheet and the lower computer model 2D Cl from XFoil of 1.6, I arrived at a corrected Cl in the clean condition of 1.38.
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( 2.8 / 63 )Thus far, the wings on my RW-11 have several notable changes from Roger's plans to support the higher gross weight that I am designing the airplane to handle. The first order of business is to determine what the member stresses will be based on the performance envelope and loading.
To do that we need to know what that performance envelope looks like. The easiest way to go about that is to use Neal Wilford's S306-Airloads-FabricWings.xls spreadsheet. The calculations are not in any way difficult, but Neal has gone through all the trouble of sorting through the various load cases in compliance with FAR 23 Appendix A and CAM 04. For the sake of education, we'll work through the calcs here.
Let's outline what operating conditions my RW-11 will be designed for:
Gross Weight (W): 1200 lbs
Maximum Operating Speed: 100 knots
Positive Load Factor (n1): 3.8 (Normal)
Negative Load Factor: -1.9 (Normal)
We also have some values based on the geometry of the wing:
Wing Area (S): 135 S.F.
Flaps up CL Max: 1.38
This gives us the information required to find points A, C, B, E, F, and G on the flight envelope diagram. With the gross weight, load factor, wing area, and CL Max we can calculate the minimum design maneuvering speed (A,G):
or 
...and the minimum design dive speed (D,E):
...based on the Simplified Design Load Criteria in Part 23, Appendix A. The minimum design cruise speed:
...is greater or equal to what I expect my RW-11 to actually be capable of, so n3 and n4 are 1.0 and -0.5, hence points C and F have 3.8 and -1.9 as a load factor, respectively. These are the same as the max positive and negative load limits specified above. Using the equations and data above, my V speeds look like this:
Minimum Design Flap Speed: 64 knots
Minimum Design Maneuvering Speed (A,G): 85 knots
Minimum Design Cruising Speed (C,F): 99 knots
Minimum Design Dive Speed (D,E): 139 knots
I've also created a PDF of my calculations.
Up next, we will see how those V speeds translate into flight loads.
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( 3 / 37 )The old method of logging progress on a web page was getting very tedious. It got to the point where I didn't even bother updating the site with my progress.
A little over a year ago I installed some blogging software (Simple PHP Blog) on my site, and am just now moving all the previous information into this new format. I'll be adding entries for the progress made over the last year as well.
I didn't do the greatest job taking pictures all the time, so over the last couple of days, I tried to catch up on that as well. Most of the images are posted with the related entries. However, here are a couple more pictures showing the current status.
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( 3.1 / 40 )I scarfed together 4 strips of 1/16" plywood to form the gusset strip covering the top of the rear spar. This gusset attaches the center and rear section of ribs to the spar. It isn't a structural necessity to scarf this part, it just makes a nicer joint.
I had a little extra epoxy, so I also whipped up a couple of gussets that hold the wing tip bow on.
2 hr
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( 2.9 / 29 )I made a jig to help shape the top of the wing tip bow to match the ribs. It is simply a straight 1"x6" that is long enough to bridge several ribs. The end is fashioned to attach a Dremel tool. I have a 1/4" collet in the Dremel with a 3/4" straight cut router bit. The bit is set flush with the bottom of the 1x6. By laying the jig perpendicularly across the inner ribs, I can follow the rib profile with the cutter.
I set the bit high at first and profiled the tip bow in multiple passes. A little follow-up with the sanding block, and the tip bow looks better than I had hoped. I profiled the bow from the front spar back to the trailing edge. I only had one spot where I wasn't careful enough with the profiling jig, and the router cut a little deep on the inside edge of the bow. It's a small nick that will eventually be beneath a gusset, so no worries.
I also made the final glue joint where the bow attaches to the front of the front spar. The bow had a fair amount of spring in it to force it into the proper shape, so I added a small filler block to reinforce the joint. Eventually the leading edge skin will act as a gusset at this joint, but it will probably take some abuse from sanding and shaping before the L.E. skin is attached.
2 hr
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( 2.7 / 36 )
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