Adding Air Drag to Motion Equations: Numerical Solutions using Euler Method, Lecture notes of Engineering

The concept of adding air drag to motion equations, which cannot be solved analytically. Students will learn how to use euler's method to find approximate solutions. The equations for a falling ball with air drag, the review of the euler process, and the trajectory of a projectile with air drag. The document also discusses the convergence of euler's method.

Typology: Lecture notes

2017/2018

Uploaded on 04/06/2018

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ME123 Computer Applications I
Day 34
โ€ข(Concept Question)
โ€ขAdding Air Drag
โ€ขReview Euler Process
โ€ขTrajectory with Drag
โ€ขEuler Convergence
โ€ข(Exercises)
ME123 Computer Applications I
Adding Air Drag
โ€ข Yesterday we used Eulerโ€™s method to solve
equations for which we knew the exact
solution
โ€ขWe only did that for practice
โ€ขToday we will solve equations which have no
exact solution
pf3
pf4
pf5

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Download Adding Air Drag to Motion Equations: Numerical Solutions using Euler Method and more Lecture notes Engineering in PDF only on Docsity!

ME123 Computer Applications I

Day 34

  • (Concept Question)
  • Adding Air Drag
  • Review Euler Process
  • Trajectory with Drag
  • Euler Convergence
  • (Exercises)

Adding Air Drag

  • Yesterday we used Eulerโ€™s method to solve equations for which we knew the exact solution
  • We only did that for practice
  • Today we will solve equations which have no exact solution

ME123 Computer Applications I

Adding Air Drag

Suppose we have a ball falling: ๐‘š๐‘” ๐‘š ๐‘‘๐‘‰ ๐‘‘๐‘ก = โˆ’๐‘š๐‘” + ๐‘˜๐‘‰^2 ๐‘‘๐‘‰ ๐‘‘๐‘ก = โˆ’๐‘” + ๐‘˜ ๐‘š ๐‘‰^2 This one is much harder to solve analytically. ๐‘˜๐‘‰^2

Adding Air Drag

๐‘š๐‘” ๐‘‘๐‘‰ ๐‘‘๐‘ก = โˆ’๐‘” + ๐‘˜ ๐‘š ๐‘‰^2 ๐‘‰ 0 = 0 You will work with these equations in the exercises. ๐‘‘๐‘ฆ ๐‘‘๐‘ก = ๐‘‰ ๐‘ฆ 0 = 0 ๐‘˜๐‘‰^2 This is the only new term!

ME123 Computer Applications I

Review Euler Process

  1. Isolate ๐‘ฅ๐‘–+ 1 ๐‘ฅ๐‘–+ 1 โˆ’ ๐‘ฅ๐‘– โˆ†๐‘ก

Review Euler Process

  1. March in time starting from initial condition ๐‘ฅ 1 = ๐‘ฅ( 0 ) ๐‘ฅ 2 = ๐‘ฅ 1 + (โˆ†๐‘ก) ๐‘“(๐‘ฅ 1 , ๐‘ก 1 ) ๐‘ฅ 3 = ๐‘ฅ 2 + โˆ†๐‘ก ๐‘“ ๐‘ฅ 2 , ๐‘ก 2 ๐‘ฅ๐‘–+ 1 = ๐‘ฅ๐‘– + (โˆ†๐‘ก) ๐‘“(๐‘ฅ๐‘–, ๐‘ก๐‘– )

ME123 Computer Applications I

Trajectory with Drag

Launch a projectile with air drag: This one has NO exact solution. It MUST be solved numerically. ๐‘š๐‘” ๐‘˜๐‘‰^2 The air drag always acts to oppose the motion of the projectile. Its magnitude depends on the square of the magnitude of the velocity.

Trajectory with Drag

Launch a projectile with air drag: ๐‘š๐‘” ๐‘š ๐‘‘๐‘‰๐‘ฅ ๐‘‘๐‘ก = โˆ’๐‘˜ ๐‘‰๐‘ฅ ๐‘‰ ๐‘‰^2 ๐‘‰๐‘ฅ 0 = ๐‘‰๐‘™๐‘Ž๐‘ข๐‘›๐‘โ„Ž cos ๐œƒ ๐‘‘๐‘ฅ ๐‘‘๐‘ก = ๐‘‰๐‘ฅ ๐‘ฅ 0 = 0 ๐‘˜๐‘‰^2 x y ๐‘š ๐‘‘๐‘‰๐‘ฆ ๐‘‘๐‘ก = โˆ’๐‘˜ ๐‘‰๐‘ฆ ๐‘‰ ๐‘‰^2 โˆ’ ๐‘š๐‘” ๐‘‰๐‘ฆ 0 = ๐‘‰๐‘™๐‘Ž๐‘ข๐‘›๐‘โ„Ž sin ๐œƒ ๐‘‘๐‘ฆ ๐‘‘๐‘ก = ๐‘‰๐‘ฆ ๐‘ฆ 0 = 0 You will work with these equations in the Exercises.