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Car Rolling (ODE Solution) (SI Units)
Created by Valery Ochkov
Area 1: Disclaimer
Area 2: User Notices
Input data
Car force N (cid:29)(cid:32) 0 kgf
car
Mass of the car mass(cid:29)(cid:32) 1200 kg
2
Area of the cut of the car A (cid:29)(cid:32) 2.5 m
s
Coefficient of resistance of air c (cid:29)(cid:32) 3.5
w
kg
Air density (cid:1)(cid:29)(cid:32) 1.2
3
m
Coefficient of friction of rolling f (cid:29)(cid:32) 0.65
k
Initial velocity of the car v (cid:29)(cid:32) 54kph
0
End time t (cid:29)(cid:32) 7s
end
Area 3: Converting to base SI units (dimensionless)
Given a balance of 3 Forces: Inertia force + Rolling resistance + Air resistance=0 (without units)
x'(t) 1 2
massx''(t) (cid:14) massgfk (cid:14) (cid:1)cwAsx'(t) = Ncar
x'(t) 2
... and initial conditions x(0s) = 0m x'(0s) = v
0
(cid:11) (cid:12) (cid:11) (cid:12)
x(cid:29)(cid:32) Odesolve t(cid:15)(cid:3)t Method - Fixed x t (cid:32) 126m
end end
m)
(
r 100
a
c
e
h
of t 50
oll
R
0 2 4 6
Time (s)
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Car Rolling (ODE Solution) (US Units)
Created by Valery Ochkov
Area 1: Disclaimer
Area 2: User Notices
Input data
Car force N (cid:29)(cid:32) 0 lbf
car
Mass of the car mass(cid:29)(cid:32) 2600 lb
2
Area of the cut of the car A (cid:29)(cid:32) 26.91 ft
s
Coefficient of resistance of air c (cid:29)(cid:32) 0.35
w
lb
Air density (cid:1)(cid:29)(cid:32) 0.0749
3
ft
Coefficient of friction of rolling f (cid:29)(cid:32) 0.25
k
Initial velocity of the car v (cid:29)(cid:32) 35 mph
0
End time t (cid:29)(cid:32) 7s
end
Area 3: Converting to base SI units (dimensionless)
Given a balance of 3 Forces: Inertia force + Rolling resistance + Air resistance=0 (without units)
x'(t) 1 2
massx''(t) (cid:14) massgfk (cid:14) (cid:1)cwAsx'(t) = Ncar
x'(t) 2
... and initial conditions x(0s) = 0ft x'(0) = v
0
(cid:11) (cid:12) (cid:11) (cid:12)
x(cid:29)(cid:32) Odesolve t(cid:15)(cid:3)t Method - Fixed x t (cid:32) 75ft
end end
(ft) 60
r
a
c
he 40
of t
oll 20
R
0 2 4 6
Time (s)
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Finding a Root of Equation x=f(x) by the Simple Iterations Method
Created by Valery Ochkov
Area 1: Disclaimer
Area 2: User Notices
Function of one argument - the right part of the equation x= f(x), where you want to find a root
(the value of the x at which the equation x= f(x) becomes an identity)
2.1
f(x)(cid:29)(cid:32) (cid:16)0.2(cid:152)x (cid:14) x(cid:14) 2
Range of Plot x (cid:29)(cid:32) 2.7 x (cid:29)(cid:32) 3.2
1 2
Guess Value x(cid:29)(cid:32) 2.8
You can change the value of the variable iteration and see on the plot the approximation
to the root of the equation x = f(x)
iteration (cid:29)(cid:32)
Area 3: Solution
(cid:11) (cid:12)
Iterations xi = f xi(cid:16)1
xx1
3.1
3
yy1
2.9
2.8
2.7
2.7 2.8 2.9 3 3.1
x
f(x)
Points of Iterations
Tolerance
T (cid:11) (cid:16)3 (cid:16)3 (cid:16)3(cid:12)
(x(cid:16) f(x)) (cid:32) (cid:16)0.262 0.097 (cid:16)0.04 0.016 (cid:16)6.474(cid:117) 10 2.605(cid:117) 10 (cid:16)1.05(cid:117) 10
Iterations to the root
T
x (cid:32) (2.8 3.062 2.965 3.005 2.989 2.995 2.993)
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Newton Method for Zero Finding of the Function y(x)
Problem. To find in the section [a, b] unique zero (root) of function y (x)
Created by Valery Ochkov
Area 1: Disclaimer
Area 2: User Notices
3 2
Function y(x)(cid:29)(cid:32) x (cid:16) 5x (cid:16) 4.5
Range of Plot x1(cid:29)(cid:32) (cid:16)2.5 x2(cid:29)(cid:32) 5.5
y1(cid:29)(cid:32) (cid:16)35 y2(cid:29)(cid:32) 15
First x x(cid:29)(cid:32) (cid:16)2 iteration (cid:29)(cid:32)
Area 3: Method
Area 4: Solution
x x
1
10
(cid:16)2 0 2 4
(cid:16)10
(cid:16)20
(cid:16)30
(cid:167) "n =" 0 1 2 3 4 5 (cid:183)
(cid:168) (cid:184)
(cid:168) "x =" (cid:16)2 (cid:16)0.98437 (cid:16)0.17667 2.32909 (cid:16)0.37706 0.87726 (cid:184)
Iterations(cid:32)
(cid:168) (cid:184)
"y(x) =" (cid:16)32.5 (cid:16)10.29882 (cid:16)4.66158 (cid:16)18.9888 (cid:16)5.26446 (cid:16)7.67281
(cid:168) (cid:184)
(cid:169)"error =" 0 (cid:16)1.01563 (cid:16)0.8077 (cid:16)2.50576 2.70615 (cid:16)1.25432(cid:185)
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Secant Method of Zero (Root) Finding of the Function y(x)
Created by Valery Ochkov
Area 1: Disclaimer
Area 2: User Notices
Function of one argument, where you want to find a zero (the value of the argument
in which the function ravnv zero)
3 2
y(x)(cid:29)(cid:32) x (cid:16) 5x (cid:16) 4.5
Range of Plot x1(cid:29)(cid:32) (cid:16)2.5 x2(cid:29)(cid:32) 5.5
y1(cid:29)(cid:32) (cid:16)35 y2(cid:29)(cid:32) 15
First x x (cid:29)(cid:32) 2.8
1
You can change the value of the variable iteration and see on the plot the approximation
to the zero function y(x)
iteration (cid:29)(cid:32)
Area 3: Method
Area 3: Solution
10
(cid:16)2 0 2 4
(cid:16)10
(cid:16)20
(cid:16)30
(cid:11) (cid:12)
x (cid:32) 2.8 x (cid:32) 2.801 x (cid:32) (cid:16)2.058 y x (cid:32) (cid:16)34.398
1 2 3 3
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Steiner Point
Problem. You need to design takings from the electric cable to two points (to two houses) so
that the total length of takings was minimal. You can change the coordinates of two points
of the branches (from 0 to 100), and see the circuit branches, as well as the coordinates
of the possible tee.
Created by Valery Ochkov
Area 1: Disclaimer
Area 2: User Notices
coordinates 1-st point coordinates 2-d point
x1(cid:29)(cid:32) 0 y1 (cid:29)(cid:32) x2 (cid:29)(cid:32) y2 (cid:29)(cid:32)
Area3:Method
Area 4: Solution of the Task
110
100
90
80
70
60
50
40
30
20
10
electric cable
(cid:16)10 0 10 20 30 40 50 60 70 80 90 100 110
(cid:16)10
(cid:167)27.679(cid:183)
Steiner Point(cid:32) (cid:168) (cid:184) Length(cid:32) 157.942
(cid:169) 54.02 (cid:185)
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Geometry and Breaking Forces of the Catenary (SI Units)
Problem: To find the geometry and breaking forces of the catenary
Created by Valery Ochkov
Area 1: Disclaimer
Area 2: User Notices
Area 3: Method
(cid:167) (cid:167)x(cid:16) x (cid:183) (cid:183)
0
The catenary function: y(cid:11)x(cid:15)(cid:3)x (cid:15)(cid:3)h(cid:15)(cid:3)a(cid:12)(cid:29)(cid:32) h(cid:14) a(cid:152)(cid:168)cosh(cid:168) (cid:184) (cid:16) 1(cid:184)
0
(cid:169) (cid:169) a (cid:185) (cid:185)
h (cid:29)(cid:32) 15m
2
h (cid:29)(cid:32) 7m
1
L(cid:29)(cid:32) 30m
Minimal catenary length l(cid:29)(cid:32) L2 (cid:14) (cid:11)h (cid:16) h (cid:12)2 (cid:32) 31.05m
1 2
Catenary length S > l S(cid:29)(cid:32) 39m
Given
Guess values x (cid:29)(cid:32) 12m h(cid:29)(cid:32) 1m
0
(cid:11) (cid:11) (cid:12) (cid:12)
a(cid:29)(cid:32) 3m a(cid:29)(cid:32) root h (cid:16) y 0(cid:15)(cid:3)x (cid:15)(cid:3)h(cid:15)(cid:3)a (cid:15)(cid:3)a (cid:32) 12.89m
1 0
(cid:11) (cid:12)
Fixing at the left end h = y 0(cid:15)(cid:3)x (cid:15)(cid:3)h(cid:15)(cid:3)a
1 0
(cid:11) (cid:12)
Fixing on the right end h = y L(cid:15)(cid:3)x (cid:15)(cid:3)h(cid:15)(cid:3)a
2 0
L
(cid:180)
(cid:181) 2
Catenary length S = (cid:181) 1(cid:14) (cid:168)(cid:167)d y(cid:11)x(cid:15)(cid:3)x (cid:15)(cid:3)h(cid:15)(cid:3)a(cid:12)(cid:183)(cid:184) dx
0
(cid:181) (cid:169)dx (cid:185)
(cid:182)
0
(cid:167)x (cid:183)
0
(cid:168) (cid:184)
(cid:11) (cid:12)
(cid:168) h (cid:184) (cid:29)(cid:32) Find x0(cid:15)(cid:3)h(cid:15)(cid:3)a x0(cid:32) 12.46m h(cid:32) 50.14mm
(cid:168) (cid:184)
(cid:169) a (cid:185)
15
x
0
10
5
h
0 5 10 15 20 25 30
kg lb
Linear density of chain lpc(cid:29)(cid:32) 0.5 (cid:32) 0.336
m ft
Area 5: Angles and forces calculation
(cid:1)1(cid:32) (cid:16)50.445° <--Angles of inclination of the catenary at the it's ends--> (cid:1)2(cid:32) 63.316°
Weight of the catenary G(cid:32) 19.5kgf
F (cid:32) 9.568kgf F (cid:32) 13.568kgf
1 2
F (cid:32) 6.093kgf <--- Forces of tension on the catenary ends ---> F (cid:32) 6.093kgf
x1 x2
F (cid:32) 7.377kgf F (cid:32) 12.123kgf
y1 y2
Forces of breaking the chain
Forces of breaking the catenary, kgf
F
2
x
0
12
10
F
1
8
0 10 20 30
x, m
