Lecture 2 - Loops

7/29/2019 Lecture 2 - Loops

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Loop Design1ME 297 Roller Coaster Dynamics

Purdue UniversityFall 2010

Layout of a simple roller coaster track with lift hill/first drop and loop

1

Primary reference: Pendrill, A.-M., 2005, Rollercoaster loop shapes,Physics

Education, 40(6), pp. 517-521.


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3/15

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ReviewIn the last class, we discussed three different ways of describing the kinematics for the

planar motion of particles: Cartesian, path and polar. The results are summarized below:

a = !!x i + !!y j ; Cartesian

= !v et +v2

!en ; path

= !!r " r !#2( ) er + r!!#+ 2 !r !#( ) e# ; polar

acceleration vectorvelocity vector

v = !x i + !y j ; Cartesian

= v et ; path

= !r er + r!! e

!; polar

x

yi

j

P

path of P

Cartesiandescription

et

P

path of P

en

s

C

Pathdescription

r

er

P

path of P

e

Ox

y

Polardescription


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Path geometry

The path of a rollercoaster cart is probably most easily visualized in terms of its x,y( )

coordinates since having detailed information on the path y = y x( ) allows us to

accurately draw the shape of the path.

However, the path description is the most logical choice in terms of a performance-baseddesign. The fundamental parameters for the path description are introduced below and

related back to the more physically realizable Cartesian coordinates x,y( ) (see figure

below):

inclination angle, ! : the angle between the x-axis and the tangent to the path:

tan!=dy

dx

distance traveled along the track, s : ds = dx( )2 + dy( )2 . Alternately,

dx

ds= cos! (1)

dy

ds= sin! (2)

radius of curvature of the path, !: The radius of curvature of the path is the radius

of a circle tangent to the path and having the same curvature as the path. Since thearc length of a circle is found by ds = !d", we have

d!

ds

=

1

"(3)

C

!y or, h( )

xx0

y0

!

s = 0

path : y = y x( )

dx

dyds

!


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Path designThe design of a coaster track is typically done in terms of prescribing (or deriving) the

radius of curvature of the track! as a function of the distance traveled s along the path.

In other words, we will start out of a desired function ! s( ) and will end up with an

equation of the path in Cartesian coordinates x,y( ) . To this end, we do the following:

1. Using the known ! s( ) , determine the inclination angle by integrating equation

(3):

! s( ) = !0 +ds

" s( )0

s

# (3a)

where !0 is the inclination angle at the start of the segment of track.

2. Once we know ! s( ) from equation (3a) above, we can determine the equations

for the x,y( ) components of the track by integrating equations (1) and (2):

x s( ) = x0 + cos! s( )ds0

s

"(1a)

y s( ) = y0 + sin! s( )ds0

s

" (2a)

where x,y( ) = x0 ,y0( ) are the coordinates at the start of the segment of track.

From this we see that if one is given the radius of curvature of a path ! s( ) , along with

initial conditions x0 ,y0( ) and !0 , one can construct the equations describing theCartesian components of the path.

Note that the integrands of equations (1a), (2a) and (3a) are typically complicated enough

that we will often need to resort of numerical means for performing these integrations.


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TRACK DESIGN #1:circular path

Say that the cart starts with a speed ofv0 at s =!0 = 0 where the path has a desired

constant radius of curvature != r . As the cart moves beyond the starting point, the speed

is given governed by the conservation of energy equation:

1

2mv0

2=1

2mv

2+ mgh ! v

2= v0

2" 2gh

As the cart moves along the path, its centripetal component of acceleration is given by:

an = v

2

r= v0

2! 2

ghr

The path of cart can be derived from the fundamental equations (3a), (1a) and (2a) as:

! s( ) = !0 +ds

" s( )0

s

# =ds

r0

s

#

x s( ) = x0 + cos! s( )ds0

s

" = cos! s( )ds0

s

"

y s( )=

y0+

sin!

s( )ds0

s

"=

sin!

s( )ds0

s

"

y or, h( )

xs = 0

!0 = 0

r

x0,y0( )


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Numerical results forv0 = 30m / sec

an0 =

2g

an0 = 3g

an0 =

4g

an0 = 4.5g


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TRACK DESIGN #2:path with constant centripetal acceleration

Say that the cart starts with a speed ofv0 at s = 0 where the path has a desired radius of

curvature !0 and inclination angle !0 . As the cart moves beyond the starting point, the

speed is given governed by the conservation of energy equation:

1

2mv0

2=1

2mv

2+ mgh ! v

2= v0

2" 2gh

As the cart moves along the path, its centripetal component of acceleration is given by:

an =v2

!=

v02" 2gh

!

The centripetal component of acceleration at the start (! = !0 , != !0 and h = h0 = 0 ) is

therefore:

an0

=

v02

!0

In order for the particle to maintain a constant centripetal component of accelerationthroughout the path, we need to have:

an0 = an =v02! 2gh

"#

"=v02! 2gh

an0

=

v02

an0! 2

gh

an0= "0 ! 2

h

an0 / g( )

This result says that in order for the cart to maintain a constant normal g-level, an0

/ g ,

the radius of curvature must DECREASE at a linearrate with the change in elevation h .

y or, h( )

xs = 0

!0

!0

path


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an0

= 2gan0 = 3g

an0 = 4g

an0=

4.5g


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TRACK DESIGN #3:path with constant normal force (constant apparent weight)

Say that the cart starts with a speed ofv0 at s = 0 where the path has a desired radius of

curvature !0 and inclination angle !0 . As the cart moves beyond the starting point, the

speed is given governed by the conservation of energy equation:

1

2mv0

2=1

2mv

2+ mgh ! v

2= v0

2" 2gh

If we sum forces on the cart in the direction normal to the path, we get:

Fn! = N" mgcos#= mv2

$%

N= m gcos!+v2

"

#

$%

&

'(= m gcos!+

v02 ) 2gh

"

#

$%

&

'(

The normal force acting on the cart at the start (! = !0 , != !0 and h = h0 = 0 ) is

therefore:

N0 = m gcos!0 +v02

"0

#

$%

&

'(

To maintain a constant normal force on the cart as it moves through its path, we setN = N0 , or:

m gcos!+v02 " 2gh

#

$

%&

'

()= m gcos!0 +

v02

#0

$

%&

'

()*

#=v02 " 2gh

g cos!0 " cos!( ) +v02

#0

=

#0 an0 / g( )" 2hcos!0 " cos!+ an0 / g( )

=

#0 " 2h / an0 / g( )1+ cos!0 " cos!( ) / an0 / g( )

y or, h( )

xs = 0

!0

!0

path

y or, h( )

x

!

path

mg

N

!


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an0 =

2g

an0 =

3g

an0

= 4g

an0 = 4.5g


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TRACK DESIGN #4:path with linearly-decreasing radius of curvature (clothoid)

Say that the cart starts with a speed ofv0 at s = 0 where the path has a desired inclination

angle !0 . As the cart moves beyond the starting point, the radius of curvature of the path

decreases with s as:

! s( ) =1

bs

where b is a design parameter for the loop. Note that ! 0( ) = " .

Equation (3a) gives:

! s( ) = !0 + bs ds0

s

" = !0 +1

2bs

2

The x,y( ) coordinates of the path for this design can then be found from equations (1a)

and (2a) as:

x s( ) = cos! s( )ds0

s

" = cos !0 + bs2

/ 2( )ds0

s

"

y s( ) = sin! s( )ds0

s

" = sin !0 + bs2

/ 2( )ds0

s

"

Unfortunately, this integrals cannot be expressed in terms of elementary functions2.

Therefore, numerical integration is needed to determine the x,y( ) coordinates of the path

for a given parameter value b . However, we do know that, by definition, this path has a

monotonically-decreasing radius of curvature as one moves along the path. In simpleterms, these equations represent a spiral, commonly referred to as a clothoid. Afigure representing the general shape of this spiral is shown below.

2These integrals are known as the Fresnel integrals. Numerical values for these

integrals can be found in tabulated form in many handbooks and Matlab routines exist for

their evaluation.

y or, h( )

xs = 0

!0

!0

path


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For our application to the design of coaster loops, we will take the segment from != 0 to

!= " and join it with its mirror image, as shown below to produce a symmetrical

clothoid loop.

! = "

! = 0

! = "

! = 0 != 2"

symmetrical

clothoid loop


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b = 0.004

b = 0.003

b = 0.002

b = 0.001


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Summary: numerical results for v0 = 30m / sec and an0 = 4g

circular

const. normal acc.

const. normal force

clothoid

(b = 0.003)

circular

const. normal acc.

const. normal force

clothoid

(b = 0.003)

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Lecture 2 - Loops