Calculating Coefficient of Lift: Angle of Attack of 7°

  • Thread starter Thread starter yecko
  • Start date Start date
  • Tags Tags
    Coefficient Lift
Click For Summary
SUMMARY

The discussion centers on calculating the coefficient of lift (Cl) for an angle of attack of 7°. The user initially calculated Cl incorrectly due to a misunderstanding of units, mistaking pounds-force (lb(f)) for pounds-mass (lb(m)). Correcting for this, the user found that dividing by 32.2 yields a coefficient of lift of approximately 0.23, aligning with standard atmospheric density of 0.000889378 slugs/cubic foot. This highlights the importance of unit conversion in aerodynamic calculations.

PREREQUISITES
  • Understanding of aerodynamic principles, specifically lift calculations.
  • Familiarity with unit conversions between slugs and pounds.
  • Knowledge of standard atmospheric conditions and their impact on lift.
  • Basic proficiency in using equations related to lift and angle of attack.
NEXT STEPS
  • Research the relationship between angle of attack and coefficient of lift in aerodynamics.
  • Study unit conversion techniques, particularly between slugs and pounds in fluid dynamics.
  • Explore the impact of atmospheric density on lift calculations.
  • Learn about the significance of using correct units in engineering calculations.
USEFUL FOR

Aerospace engineers, students studying fluid dynamics, and anyone involved in aerodynamic analysis or aircraft design will benefit from this discussion.

yecko
Gold Member
Messages
275
Reaction score
15
1. Homework Statement
https://i.imgur.com/tG9unNs.png
tG9unNs.png


2. Homework Equations
in picture below

3. The Attempt at a Solution
https://i.imgur.com/egZC3Ay.jpg
egZC3Ay.jpg

I calculated the angle of attack is "7", yet the answer is "1", which implies the coefficient of lift is about 0.2 from the answer. I am not sure where of my answer have got mistake...
can anyone help?
thank you very much!
 
Physics news on Phys.org
I think there is a problem of slugs versus lbs. Standard atmosphere has density 0.000889378 slugs/cubic foot, not lbs/cubic foot. Using your numbers, I get Cl = 7.3, which is way too high. But dividing by 32.2 gives Cl = 0.23
 
  • Like
Likes   Reactions: yecko
Thank you very much!
FactChecker said:
Standard atmosphere has density 0.000889378 slugs/cubic foot, not lbs/cubic foot.

It is because https://i.imgur.com/a43giFt.png is what I have got in my assignment and I misunderstood lb(f) as lb(m) represented by lb in the question.
a43giFt.png

So, I do not need to multiply 32.2 on the right side.

FactChecker said:
But dividing by 32.2 gives Cl = 0.23

Thank you so much for reminding me!
 
  • Like
Likes   Reactions: FactChecker

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
9K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
10K