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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">Ciao Alessandra,</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0"><span class="Apple-converted-space ContentPasted0"> </span>I give the info. Al wires for the three targets and I have a comment for Fig.29.</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">I am attaching a pdf file for HDIce technical note 23 (actual 23 is different from this,</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">as you can see on HDIce wiki page).</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">(a)<span class="Apple-converted-space ContentPasted0">
</span>The diameters of Al wires and number of wires are as below;</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0"><span class="Apple-converted-space ContentPasted0"> </span>The lengths Al wires of three target cells are</span><span style="font-variant-ligatures: no-common-ligatures; color: rgb(180, 36, 25);" class="ContentPasted0">
6 cm</span><span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">, but the amounts of Al wires per cm are<span class="Apple-converted-space ContentPasted0"> </span></span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">proportional to the total areas of the wires:</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">(1) 21a: 51 um 769 wirea (0.002" = 2.54*0.002 =</span><span style="font-variant-ligatures: no-common-ligatures; color: rgb(180, 36, 25);" class="ContentPasted0"> 0.0051 cm</span><span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">
= 51 um)</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">(2) 19b: 38 um 965 wires</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">(3) 22b: 51 um 732 wires</span></p>
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<span style="font-variant-ligatures:no-common-ligatures"><span class="Apple-converted-space ContentPasted0"> </span></span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0"><span class="Apple-converted-space ContentPasted0"> </span>Ratio to total area of 19b to 21a:</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0"><span class="Apple-converted-space ContentPasted0">
</span>pi*38/2*38/2*965/(pi*51/2*51/2*769) = 0.697</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0"><span class="Apple-converted-space ContentPasted0"> </span>Ratio to total area of 22b to 21a:</span></p>
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<span style="font-variant-ligatures:no-common-ligatures"><span class="Apple-converted-space ContentPasted0"> </span></span></p>
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</span>pi*51/2*51/2*732/(pi*51/2*51/2*769) = 0.952</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0"><span class="Apple-converted-space ContentPasted0"> </span>These numbers are close to those in the forth row of the<span class="Apple-converted-space ContentPasted0">
</span>table at page 10.</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">The difference may come from the final number of wires.</span></p>
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<span style="font-variant-ligatures:no-common-ligatures"><span class="Apple-converted-space ContentPasted0"> </span></span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0"><span class="Apple-converted-space ContentPasted0"> </span>If use the narrow z vertex cut like yours (-9.5 < z <
</span><span style="font-variant-ligatures: no-common-ligatures; color: rgb(180, 36, 25);" class="ContentPasted0">-5.8 cm</span><span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">), the ratio of Al wires</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">contribution of 19b to 21a is just 0.7.</span></p>
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<span style="font-variant-ligatures:no-common-ligatures"><span class="Apple-converted-space ContentPasted0"> </span></span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">(b) If you would like to do discussions using Fig.</span><span style="font-variant-ligatures: no-common-ligatures; color: rgb(180, 36, 25);" class="ContentPasted0"> 29 in</span><span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">
your note, you may see better</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">way of separating contributions of Al wires from Kel-F windows in the empty targets<span class="Apple-converted-space ContentPasted0"> </span></span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">shown at page 11 and</span><span style="font-variant-ligatures: no-common-ligatures; color: rgb(180, 36, 25);" class="ContentPasted0"> 12 in</span><span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">
the attachment.</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">Instead of one gaussian, three gaussians may well reproduce the Kel-F distributions.</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">At the vertex z >
</span><span style="font-variant-ligatures: no-common-ligatures; color: rgb(180, 36, 25);" class="ContentPasted0">-4.8 cm</span><span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">, Kel-F is supposed to have more contriubutions than
Al wires,</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0">if assume both have same width of resolutions.<span class="Apple-converted-space ContentPasted0">
</span>This is my opinion.</span></p>
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<span style="font-variant-ligatures:no-common-ligatures" class="ContentPasted0"><span class="Apple-converted-space ContentPasted0">
</span>Ciao, Tsuneo Kageya.</span></p>
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