Find Airspeed for Min Horsepower w/ 1800 lb Clark Y Wing

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SUMMARY

The minimum airspeed for a Clark Y wing monoplane with a total weight of 1800 lbs is calculated to be 67.03 ft/s. This calculation utilizes the formula that relates weight to the maximum coefficient of lift, density at sea level, wing area, and minimum airspeed. The maximum coefficient of lift for the Clark Y wing is established at 1.56. Additionally, the discussion highlights the significance of the equivalent flat plate area, which is 3.8 sq ft, in determining the coefficient of drag and its impact on overall performance.

PREREQUISITES
  • Understanding of aerodynamic principles, specifically lift and drag
  • Familiarity with the Clark Y airfoil characteristics
  • Knowledge of basic equations of motion in aerodynamics
  • Ability to perform calculations involving coefficients of lift and drag
NEXT STEPS
  • Research the impact of equivalent flat plate area on drag calculations
  • Explore the relationship between angle of attack and lift for different airfoils
  • Learn about the performance characteristics of the Cessna 172/182 for comparative analysis
  • Investigate methods to optimize airspeed for minimum horsepower in various aircraft designs
USEFUL FOR

Aerospace engineers, aviation students, and anyone involved in aircraft performance analysis will benefit from this discussion.

ChimM
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Homework Statement


For monoplane of Clark Y wing, 36-feet by 6-feet, with 3.8 sq ft. equivalent flat plate area, what should be the airspeed for minimum Horsepower when fuel has been burned so that the total weight is 1800 lb?

Homework Equations


Wing Area = 36 x 6 = 216
Weight = Lift
Weight = Max. Coefficient of lift × (density at sea level/2) × wing area × (Minimum Airspeed)^2

○ Gasoline consumption
HPreq/V = W/375 × (((Coefficient of Drag) + (1.28Ae/ wing area))/Coeffcient of Lift)

The Attempt at a Solution


Given:
Clark Y wing
Wing span = 36; wing chord = 6
Wing area = 216
Ae = 3.8 sq. ft
W = 1800 lbs

Reqd: Airspeed

Solution:
From the formula
W = Max. Coefficient of Lift × (density at sslc/2) × (Wing area) × (Minimum Airspeed)^2

Since, the problem stated that the wing is Clark Wing, Max. Coefficient of Lift of the airfoil selection is 1.56.

By substituting the given data,

Minimum Airspeed = 67.03 ft/s

My question is, does my answer correct? What is the relevant of the equivalent flat plate area on the problem?

Thank you..
 
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Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
ChimM said:
3.8 sq ft. equivalent flat plate area ... What is the relevant of the equivalent flat plate area on the problem?
I assume that is how you're supposed to calculate the coefficient of drag. Not sure if that includes the drag related to the fuselage.

ChimM said:
Clark Wing ... Max. Coefficient of Lift of the airfoil selection is 1.56.
Max coefficient of lift occurs at a specific angle of attack, and the angle of attack required for a specific amount of lift decreases as the aircraft speed increases. It's more likely that what you're looking for is the speed that corresponds to the lowest amount of drag for the amount of lift required which equals the weight of the aircraft. Using a Cessna 172/182 as an example, looks like the longest range speed is around 65 to 70 knots, which is slower than the best glide ratio speed which is closer to 80 knots. Since longest range is affected by engine and propeller efficiency, I don't know if it is the same as the speed of lowest power output (thrust (equals drag) times air speed). Also the stated speed of 65 to 70 knots was posted by another person as opposed to the article that mentioned best glide ratio speed was around 80 knots.
 
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