曝光臺(tái) 注意防騙 網(wǎng)曝天貓店富美金盛家居專營(yíng)店坑蒙拐騙欺詐消費(fèi)者

In general, it is a good idea to start with a very simplified model to obtain
a ‘first idea’ on the problem and then iterate on more complex
representations to arrive to an accurate description.
Golf ball trajectory: mathematical formulation
!
U
A. Physical description
B. Mathematical formulation
C. Solution of the mathematical model
D. Interpretation and validation
✓ Parameters:
✓ Variables:
✓ Simplified equations of motion
Mass M, radius a, drag coefficient c = !"a2
2 CD
Position x = (x, y), velocity v = (x˙ , y˙)
! ¨x = 0
¨y = −g ! x(t = 0) = 0, x˙ (t = 0) = U cos !
y(t = 0) = 0, y˙(t = 0) = U sin !
Four steps to model an engineering problem
A. Physical description
B. Mathematical formulation
C. Solution of the mathematical model
- analytical solution if the problem is simple
enough (very rare!)
- numerically in most cases.
D. Interpretation and validation
!
U
Golf ball trajectory: solution
A. Physical description
B. Mathematical formulation
C. Solution of the mathematical model
D. Interpretation and validation
Analytical solution
x(t) = Ut cos !, y(t) = Ut sin ! −
1
2gt2
According to this model, the optimal angle is α=π/2.
! ¨x = 0
y¨ = −g ! x(t = 0) = 0, x˙ (t = 0) = U cos !
y(t = 0) = 0, y˙(t = 0) = U sin !
!
U
D
Maximum distance D (find tf such that y(tf)=0; then x(tf)=D)
tf =
2U sin !
g
, D= U2 sin 2!
g
Four steps to model an engineering problem
A. Physical description
B. Mathematical formulation
C. Solution of the mathematical model
D. Interpretation and validation
- check if the result of the model makes sense physically
- identify (and quantify) the possible sources of errors
This step is essential: no matter how elegant the model or the solution
are, they are worthless if they are not a good description of the initial
problem.
Golf ball trajectory: validation
!
U
A. Physical description
B. Mathematical formulation
C. Solution of the mathematical model
D. Interpretation and validation
x(t) = Ut cos !, y(t) = Ut sin ! −
1
2gt2
tf =
2U2 sin !
g
, D= U2 sin 2!
g
According to this model, the total energy of the ball is conserved (kinetic +
potential): when the ball hits the ground it has the same velocity as initially!!!
Something is wrong in our model: aerodynamic drag can not be neglected at such
high velocities: go back to step B and add the drag!
Golf ball trajectory: solution
!
U
A. Physical description
B. Mathematical formulation
C. Solution of the mathematical model
D. Interpretation and validation
There is no analytical solution, we need to solve this equation
numerically (see section on ODE later in the class)
! Md2x
dt2 = −c"x˙ 2 +y˙2x˙
Md2y
dt2 = −Mg − c"x˙ 2 +y˙2y˙ # x(t = 0) = 0, x˙ (t = 0) = U cos !
y(t = 0) = 0, y˙(t = 0) = U sin !
Results with friction
Without any air friction, the ball would travel 490 m before hitting the ground!
Including a quadratic drag to represent the effect from the air, this distance is
reduced to ~ 223 m
m = 46 g, a= 2.1 cm, c= 0.25
U = 70 ms−1, ! = 45",
g = 9.8 ms−2, " = 1kgm−3
−500 50 100 150 200 250
0
50
100
150
x
y
with friction
no friction
Optimal angle
The optimal angle for maximum distance has changed: the golfer should give
the ball an initial angle lower than 45o (about 38o in this case)
00 20 40 60 80
0.2
0.4
0.6
0.8
1
1.2
! (degrees)
D/Dmax
with friction
no friction
Possible improvements on this model
If a more precise answer is needed or to validate our simplifications, it is
important to understand the approximations that we made here:
➡ we neglected rotation: an additional lift is present due to Magnus’
effect.
➡ we neglected the structured surface of the ball
➡ we neglected the wind
➡ we neglected the variations of the drag coefficient in time (the ball
slows down)
➡ we neglected the variations of g with height
➡ we neglected Coriolis effects....
!
U
A. Physical description
B. Mathematical formulation
C. Solution of the mathematical model
D. Interpretation and validation
It is important to keep in mind the different sources of errors in the
　
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