McSiMBarrel Strips upgrade geometry

Steve McMahonRAL

The Aim : What am I trying to do?• Provide a tool for fast (minutes) visualization of the layout of

the upgrade strip barrel (barrel in the first instance, end-cap may come later) to answer some rudimentary questions concerning the layout.

• The things I would imagine that we will be able to provide are: module overlaps (active layers), clearances (of envelopes around modules), coverage (hole maps), momentum resolution, and their sensitivity to small changes in layout (radius, tilt angle, stereo angle) or removal of various section

• It is NOT a replacement for a CAD model or a full detector simulation.

How : The idea, the tools

• The code is written in ROOT in a general way.• It should be configurable in a few minutes (ie

create a new geometry) and run in seconds. It should be run by anybody on a laptop.

• It should be stand alone (no monster external libraries).

• Small amount of documentation.

Basic strip barrel geometry

r

X

Y

f1

Start with thebasic 2D “line” geometry

Y

XP0A

P0B

P1A

(0,+w/2)

(0,-w/2)

(0,0)f1

qC1

C0

P1B

r

r

Template Module at Origin : PTA(0,-w/2) PTB(0,+w/2)

Angle of module Centre i: fi = i . 2p/ Ni=0,N-1 (N=4m) m is integer

Template Module at Module Centre i R(fi+q). PTA(0,-w/2) R(fi+q). PTB(0,+w/2) Module Centre (red dot) at Module Centre i

x(Ci) = r cos(fi ) r(Ci) = r sin(fi )

PTA

PTB

cossinsincos

)(Rf1

Co-ordinates of module corners at every point on a circumference with N modules

++

+++

2

0.

)2cos()2sin(

)2sin()2cos(

)2sin(

)2cos(w

Ni

Ni

Ni

Ni

NirNir

yx

PBAi

qpqp

qpqp

p

p

Module Centre

Rotation of template moduledefined at the origin

Template module

X,Y coordinates of two-corners of the i th module of N

BA

Co-ordinates of module corners formodules 0 and 1

+

2

0

cossinsincos

00 wryx

PBA qq

qq

++

+++

2

0

)2cos()2sin(

)2sin()2cos(

)2sin(

)2cos(1 w

NN

NN

Nr

Nr

yx

PBA

qpqp

qpqp

p

p

Now restrict ourselves to P0B and P1A

+

++

++

+++

)2cos(2

)2sin(

)2sin(2

)2cos(

2

0

)2cos()2sin(

)2sin()2cos(

)2sin(

)2cos(1

qpp

qpp

qpqp

qpqp

p

p

Nw

Nr

Nw

Nr

w

NN

NN

Nr

Nr

yx

P A

+

+

q

q

qqqq

cos2

sin2

2

0

cossinsincos

00 w

wrwr

yx

P B

Y

XP0B

P1A

g0 g10

01

1

1101

1

11

tantan

tan

XY

XY

XY

i

gg

g

Visualisation in 2 dimensions of “line” detector

Angular Overlap

Y

XP0B

P1A

g0 g1

++

+

q

q

qpp

qpp

ggsin2

cos2tan

)2sin(2

)2cos(

)2cos(2

)2sin(tan 11

01 wr

w

Nw

Nr

Nw

Nr

Note overlap is negative until there is gap when it goes positive

Let’s take a simple, but useful, example

• Let’s take the strip barrel: – r=407mm, N=28 (4x7), w = 97.5mm– Let’s look at the overlap and its sensitivity to q (tilt)

++

+

q

q

qpp

qpp

ggsin2

cos2tan

)2sin(2

)2cos(

)2cos(2

)2sin(tan 11

01 wr

w

Nw

Nr

Nw

Nr

+++

++

+

qq

qqgg

qq

qpp

qpp

gg

sin75.48407cos75.48tan

)224.0sin(75.488.396)224.0cos(75.4856.90tan

sin75.48407cos75.48tan

)14

sin(2

)14

cos(407

)14

cos(75.48)14

sin(407tan

1101

1101 w

Fractional Overlap

++

+

1

sin2

cos2tan

)2sin(2

)2cos(

)2cos(2

)2sin(tan

21

20

121

2

1

1

0

01

0

1

0

01

q

q

qpp

qpp

ggg

gg

ggg

wr

wN

wN

r

Nw

Nr

F

Overlap in # strips and mm

)(28.1)(1280)(

121

2 0

1

0

01

minpitchFOmmOFOstripsOS

gg

ggg

Angular overlap, various radii

r=407mm

r=762mm

Tilt-Angle/degrees

Ove

rlap

/ deg

r=407mmOverlaps

Ove

rlap

/ fra

ction

Ove

rlap

/ mm

(80

mic

ron

pitc

h)O

verla

p in

strip

s

Tilt angle

Tilt angleTilt angle

Extend into 3 dimensions, add an extra piece of silicon and a stereo angle.

Add some nice visualization of two barrels at a single Z

Now make 2 concentric barrels

Tracking

• Now adding (overlaying on the detector) samples to tracks to calculate maps of “holes”.

• Traditional to quote results for samples of 1GeV tracks

• Some numbers– PT [GeV/c] = 0.3 x B[T] x R[m]– For ATLAS : B = 2T– For a 1 GeV PT track R[m] = 1/0.6 ~ 1.6m

Basic track geometry

1GeV track, radius Rcylinder, radius r

r R

R

(x1,y1)

R-x

y

x2+y2=r2

(R-x)2+y2=R2

x = r2/2Ry = ± r(1-r2/4R2)1/2

(x2,y2)

Y

X

Intersection of track with cylinder

Basic geometry : rotate track

1GeV track, radius Rcylinder, radius r

r

(x1,y1)

x

y

x2+y2=r2

R

X

Y

C’(x,y)

C’(x)=Rcos()C’(y) =Rsin()

R.cos()-X

Y-R.sin()

R2=[y-R.sin()]2 + [R.cos()-x]2

After the algebra has stopped• Set Z=r2/2R• X2-[2.Z.cos()].X + [Z2-r2sin2 ()]=0• Solve with normal quadratic form to get two solutions.• This gives the intersection of the two circles

• The intersections of the tracks with the silicon will be done with some rudimentary ray-tracing algorithms.

• Only really need to do this for ¼ in phi and ½ length in Z to get a real picture of the coverage.

Upgrade Strips: Currently using

Conclusion• A general non-specialist tool is being developed

to answer some rudimentary questions about layout and layout evolution or changes.

• If you want something added … just ask.• Anyone can contribute…• I have a summer student (Varun Varahamurt)

working with me on this from 4th June to 10th August.