# Calculating induced drag on Tapered Wing

Tags:
1. Nov 2, 2014

### ChimM

1. The problem statement, all variables and given/known data
When Cl = 0.9, find the difference in induced drag for a rectangular wing of 36-ft. span and 6-ft. chord and a tapered wing of the same area, the tapered wing having a 6-ft. chord at the root, the leading and trailing edges being tangent to a circle of 2-ft. radius at the tip, and the fuselage being 3 ft. wide. The airspeed is 100 miles
per hour.

---From that question, I have been able to calculate the induced drag for the rectangular wing. How can I solve the average chord of the tapered wing by this,"the tapered wing having a 6-ft. chord at the root, the leading and trailing edges being tangent to a circle of 2-ft. radius at the tip, and the fuselage being 3 ft. wide".

2. Nov 2, 2014

### SteamKing

Staff Emeritus
Well, making a sketch of the wing layout would be a big help here.

You are given the wing span, the width of the fuselage (the thingy in the middle to which the wings are attached), and the width of the root chord.

The layout of the wing tips is a little trickier, but basically, on your sketch, lay out a circle of radius = 2 feet such that the center of the circle lies at the mid-chord line of the wing plan. The leading and trailing edges of the wing will connect the ends of the root chord and be tangent to this circle.

3. Nov 2, 2014

### ChimM

I'm still confused. Am I doing it right? Manually, how can compute for it. Attached here is the illustration of the given data.

Hoping for your kind clarifications.. Thank you..

#### Attached Files:

• ###### tip chord.JPG
File size:
22.3 KB
Views:
257
Last edited: Nov 2, 2014
4. Nov 3, 2014

### SteamKing

Staff Emeritus
It's clear from your diagram that you are having trouble understanding how an airplane looks in plan view.

The fuselage is the body of the airplane, which is occupied by passengers and/or cargo. It's the structure to which the wings are attached. In other words, the wings stick out from the sides of the fuselage.

The wingspan is the total distance measured from wing tip to wing tip, and this distance runs perpendicular to the centerline of the fuselage.

This diagram might help envision the wing layout rather than using wordy explanations:

Instead of having the wing tips squared off as shown in the diagram above, lay out a circle of radius = 2 feet such that the center of the circle lies at the mid-chord line of the wing plan. The leading and trailing edges of the wing will connect the ends of the root chord and be tangent to this circle. You could probably use a lot of math to calculate these intersections, but making a simple sketch to scale on a piece of paper would be quicker and more accurate.

5. Nov 3, 2014

### ChimM

Thanks a lot! I got the answer! :)

6. Nov 3, 2014

### ChimM

Hi, i'm here again. I have been able to solve the problem by using AutoCAD. I also searched for computations and calculations of the chord created by the tangential lines, unfortunately, i haven't seen any. Please help me to solve the chord at the tip :(

7. Nov 3, 2014

### rcgldr

You didn't mention what formula(s) you're using to calculate induced drag. The formula is probably simplified to ignore issues like tip and trailing edge vortices.

8. Nov 3, 2014

### ChimM

Di = [(Coefficient of lift)^2 (density at SSLC/2) (span) (velocity in ft/s)^2] / (pi x AR)

Wherein AR = b/c
For tapered wing, (Root chord + tip chord) / 2

I'm having a hard time to calculate for the tip chord.[/SUB][/SUB]

9. Nov 4, 2014

### rcgldr

Hint - one way to calculate average chord length is to divide the area of a section by it's span.

10. Nov 4, 2014

### ChimM

But, on the tapered, there's no given data for span :(

11. Nov 4, 2014

### rcgldr

The problem states that the tapered wing has the same area as the rectangular wing, so the total area is known. You also know that the root is 6 feet, the tip is 4 feet, with a 2 foot radius semi-circle attached to each tip. You can calculate the area of a semi circle based on it's 2 foot radius. Then subtract that from the total area (per wing), to get the area for the trapezoid portion of the wing, then use the formula for trapezoid area to get the span of the trapezoid portion of each wing. Then the total span for each wing equals the trapezoid portion plus the 2 foot radius of the semi-circle at the wing tip.

Last edited: Nov 4, 2014
12. Nov 4, 2014

### ChimM

Wow! Thanks for that! I got it. I was confused for the usage of the given data about the fuselage. Anyway, I used 3.82 instead of 4 for the tip of the chord. Because of the tangential line. Is that okay? But the area of the circle will also change. Do I need to stick with 4ft tip?

Here is my solution:

Wing area = 216
Root chord = 6
Tip chord = 3.82

Area of the sector = 7.4814

Area of trapezoid = 216 - 7.4814 = 208.5186
Altitude of the trapezoid as the wing span = 42.3625 + 2 ft radius = 44.3625

THANKSSSS!! :)

Last edited: Nov 4, 2014
13. Nov 4, 2014

### rcgldr

The problem states it's a 2 foot radius circle. It doesn't state how the wing tapers, so you could assume it tapers equally from the leading and trailing edges, so the chord at the base of the semi-circle would be 4 feet. Normally, a "wing span" includes the diameter of the fuselage, but the wing area does not.

14. Nov 4, 2014

### ChimM

Okay. Thanks for the information :)

Hmm, one more thing.

Aspect ratio = wing span / chord or Aspect ratio = wing span ^2 / wing area

Which do I need to use in terms of exact computation? Because using both computation has a minimal difference on Drag. But for the computation of Coefficient of induced Drag, difference in decimal value matters.

Coefficient of induced Drag = (coefficient of lift)^2 / (pi x AR)

15. Nov 4, 2014

### rcgldr

These are the same:

AR = wing_span^2 / wing_area = wing_span^2 / (wing_span x average_chord) = wing_span / average_chord

It's not clear how the problem wants you to take the fuselage into account.

16. Nov 5, 2014

### ChimM

It seems that the given data about fuselage was at the problem for confusion. Thanks a lot sir! :)