# Help: making a compound shape with LITRANI

Hi, rooters. I’m working on LITRANI (Gentit’s) trying to combine two shapes (tsbriks) as one flourescent piece. But in doing so I reached two problems.

First, It seems to be that litrani put revetment between the shapes composing the piece: I compair the composition of two tsbriks (model A) and one tsbrik alone (model B). Model A and B had the same dimensions, the same light detectors and were exposed to the same beam of pions. However, the efficency (saw in dostats) was different (1.2% vs 0.8%) due to absortion in revetment How do I tell litrani to consider model A as just one piece?

Second, When I take off the revetment in the construction of the tsbriks (to obtain equivalence between the two models), detectors doesn’t work. Am I doing something wrong or it we have to define a revetment?

dimensions of tsbriks are ( 10, 10, 5 )
dimensions of detectors - fibers ( 0.1, 0.1, 0.1 ) -> 4 fibers on one side of the plastic

example of definition of one of the tsbriks
pieza1 = new TSBRIK( “PIEZA1”, “PIEZA1”, “PLASTICO”, “TEFLON”, hdx, hdy, hdz );

contact between pieces:
union = new TContact( “UNION”, “UNION”, “PIEZA1”, “PIEZA2”, identical, “none” );

example of detector:
fibra1 = new TSBRIK( “FIB1”, “FIB1”, “AIRE”, “TOTABS”, hdxf, hdyf, hdzf );

example of contact between fiber and plastic:
UnFibra1 = new TContact( “UNFIB1” , “UNFIB1”, “FIB1”, “PIEZA1”, contained, “none” );

Thanks for the attention,

Carlos,

These classes are not ROOT classes. You should communicate with FX.Gentit

Rene

To understand your problem, I must have a look at your example. I will do it as soon as I can. Notice the following point about revetment in Litrani: when a face of a shape is declared as having a revetment, but part of this face (or the whole face) is in contact with an other shape, Litrani assumes that there is no revetment for the part in contact.

I have read your macro and I have found it correct. Then I have tried it giving to “NumPiezas” values of 1,2,5 or 10 and I always found the same efficiency within the statistical error: 0.86 ± 0.01. [86% not 0.8%!]. So I do not see the problem? Could you explain what are the numbers you quote 1.2% vs 0.8% ?