Re: back of the envelope

["Mark M. Ito" <marki@jlab.org>]

Eugene,

I re-derived the expression you gave for the deviation given your stated 
assumptions and I agree with it in the approximation where the steps are 
small. I also looked at the data from the plots I posted; for the track 
I showed, the nominal radius of curvature is about 42 cm. After moving 
about 5.2 cm, the deviation I see in the data is 32 microns. Plugging 
into the formula you gave also gives 32 microns, which I will call 
excellent agreement.

In my last message, the "1/2 cm per % of B field deviation..." statement 
implies a linear dependence of the deviation with track length which is 
not correct. The dependence is quadratic; thanks for pointing that out. 
The plots and the deviation as shown in the post are correct (as 
suggested by the aforementioned agreement); it is just that there is no 
linear scaling coefficient.

  -- Mark

Eugene Chudakov wrote:
> Mark,
>
> At small steps, the deviation is proportional to the square of the 
> step size.
>
> Let us assume that the trajectory has been measured up to
> a certain point and a certain radius of curvature is reconstructed 
> (say, R).
> The next step (which length is d radially, d<<R ) is done in an area 
> where the field does actually deviate from our map by (dB/B). Then, the
> trajectory would deviate from the predicted one by (approx):
> d**2/2/R*(dB/B).
> For d=50mm, dB/B=0.01, R=1000mm the deviation is 0.012mm, which is
> negligible (unless I made a mistake with these simple calculations).
>
> It is interesting, what is your deviation at a 5cm step in radius.
>
> Eugene
>
> On Fri, 27 Feb 2009, Mark M. Ito wrote:
>
>> Eugene,
>>
>> All of the tracks start with zero separation and at this level of 
>> approximation, the separation is 1/2 cm per % of deviation of B field 
>> per 60 cm of track length or roughly 80 microns per % of B field 
>> deviation per cm of track length (again for this particular 
>> configuration). So if we take a B field deviation of 10^-4 uniformly 
>> and a track length of 50 cm, we get a position error of 40 microns. 
>> This is small when added in quadrature with a 150 to 200 micron 
>> position resolution and will be "smaller" still when multiple 
>> scattering is considered. Again a uniform deviation is not what we 
>> are worried about, but this is just to set the scale (as you 
>> originally and aptly suggested). As you know, the problem is that the 
>> non-uniform distortions in field are very difficult to calibrate away.
>>
>> -- Mark
>>
>> P. S. Went back to cc'ing the list again.
>>
>> Eugene Chudakov wrote:
>>> The separation grows with the radius.
>>> At the last 5 cm step the separation increased somehow.
>>> What is this increase? I must be small (0.5mm?), but it is
>>> hard to say from the plots.
>>>
>>> Thanks,
>>> Eugene
>>>
>>>
>>> On Fri, 27 Feb 2009, Mark M. Ito wrote:
>>>
>>>> Not sure I understand the question. The separation of 1 cm is 
>>>> mention is just obtained by eye looking at the plot.
>>>>
>>>> Eugene Chudakov wrote:
>>>>> Hi Mark,
>>>>>
>>>>> as far as I see these are X-Y plots (not r-phi).
>>>>> I guess, the deviation between two tracking steps matters. The 
>>>>> full deviation of the trajectory will be absorbed in the momentum
>>>>> measurement. The bottom plot shows the range in X (close to R for 
>>>>> this area) of 5cm, which is about the distance between CDC
>>>>> super-layers and is about 1 step in tracking. What is the 
>>>>> increment between
>>>>> the curves' separations at this step?
>>>>>
>>>>> Thanks,
>>>>> Eugene
>>>>>
>>>>>
>>>>> On Fri, 27 Feb 2009, Mark M. Ito wrote:
>>>>>
>>>>>> Folks,
>>>>>>
>>>>>> At last Monday?s tracking meeting Eugene suggested that someone 
>>>>>> make an estimate of the position error caused by a mistake in the 
>>>>>> knowledge of the magnetic field. Here is an example: 
>>>>>> http://tinyurl.com/btp8jx
>>>>>>
>>>>>> -- Mark
>>>>>>
>>>>
>>>>
>>
>>