Question about reduced target size

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When setting up a reduced target put at a shorter distance, that distance must be measured from the sensor lense, not from the shooter. It makes sense but I would like to understand how it deals with the 10m target (the nearest of all) as follows:

As a 10m pistol shooter, my front foot must be behind the 10m line. I measured that my eye is 10cm behind that foot (with the head turned in shooting position), my wrist is 70 cm ahead of my eye and the sensor lense is about 30cm further, which puts the sensor at 9,10m from the real target, not 10m. Since this is the usual situation at a 10m stand, obviously the Scatt system is calibrated in such a way that it gives exact results here.

Now let's consider I'm setting up a custom stand in some unknown place, like a field. I fix a target holder, make a ground mark, take my shooting position and measure 9.10m from the sensor to the target. Hence I must use the Scatt reduced target printing tool, type in 9.10m and get a target 9% smaller than standard.
Therefore I am about to shoot at a reduced target 9% smaller although the sensor is exactly at the same distance as with a standard 10m target.

How does the Scatt deal with this situation and give exact results there?
 
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AJV

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When setting up a reduced target put at a shorter distance, that distance must be measured from the sensor lense, not from the shooter. It makes sense but ...

Not for me.
Obviously the optic sensor needs to measure it's distance to the target so it can do the maths, but seems to me that the target distance must be measured to the shooter.
If you print a target with the distance till the sensor, you will get a image with a wrong proportion of the sights and the "black".
Also if you print a target for a distance of 10m you will see it's sized as a "real" target.

There are a lot of unexplained things regard the scatt, and very poor information. This forum seems to be the debut of something good.

PS I'm a new user of scatt pro.
PSS Another thing that needs a much better explanation is the calibration.
 
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seems to me that the target distance must be measured to the shooter
That's the point: Scatt instructions are to measure from the sensor.
needs to measure it's distance to the target so it can do the maths,
not really in theory: Since you select a target (therefore the distance) when starting a session, the software knows what the aspect ratio is and can do the maths from the digital image from the sensor only.
If you print a target with the distance till the sensor, you will get a image with a wrong proportion
Agreed. Although the discrepancy will be roughly proportional to the angle difference / distance ratio hence negligeable for longer distances (cosinus~=1). This is why I took the 10m air target example, for which the difference is not so negligeable.
wrong proportion of the sights
Agreed again, you will see the target smaller in your sights. But I'm also wondering if it will give good scores.
 
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I've also noticed that measuring from the sensor (such as using the the old IR USB sensor's measurement), the target ends up smaller than it should be, so I add about 1m (to simplify the math) to whatever the sensor says the distance is, and then the aperture settings match, and, since the IR sensor isn't looking at the target dot anyway, no problems. Then, if I'm using the MX-02, it seems to work just fine with the slightly too-large-for-the-closer-sensor dot. I'm shooting at about 9m, with a target printed for 9m, but the USB Pro reports a distance of about 8m. It might be more of an issue with a target much closer, however.
 
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As far as I'm aware, the size of the aiming mark used doesn't matter with Scatt USB. I presume that Scatt calculates the distance of the sensor from the target based on the angle subtended at the sensor by the known dimensions of the IR source array on the target, and then scales everything based on the known angular dimensions of the scoring rings on the selected target at its nominal distance (10m/50m/300m/etc.). Perhaps Peter can confirm this.
The key question is how does the camera-based Scatt (MX-nn) scale things? Does it do it based on the target to sensor distance dialled onto the sensor, or does it do it based on the angle subtended by the aiming mark that it is seeing compared with the angle subtended by the selected target at its nominal distance? If the former, it should not matter if the aiming mark used is not correctly scaled for the target to sensor distance (i.e. you can use a target that is scaled for the target to eye distance, so that it looks 'correct' when aiming). If the latter, it will be critical to print an aiming mark/target that is correctly scaled for the target to sensor distance (in which case it will not look 'right' when in the aim). In this latter case, there would be an advantage in moving the sensor as close to the eye as possible, so that the printed scale target/aiming mark looks as close as possible in angular size to the real target at its correct distance. Peter - please advise.
 
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The key question is how does the camera-based Scatt (MX-nn) scale things
Well simply because when starting a session you select a target. Therefore the software knows its size and distance, hence all the angular mesures. The sensor is an optical device, with known magnification and focal length, therefore pixels on the captor directly relate to widths at a given distance. With a closer target setup, angles do not change (Thales theorem). The only reason why the target must be smaller is that the shooter wants the same aspect ratio in his sights.
The dial on the sensor is only a focus setting so that the sensor doesn't see a too blurred image. You can see the focus effect on the calibration preview display.
 
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Well after some additional thinking this is a solved question with proved answer. See it here: "Tricking Scatt with Fake Targets"
Thanks for your comments.
Good test! This shows that MX-nn camera sensor Scatt is simply looking for the centre of the shape as its reference point and everything is calculated based on the angular displacement of the sensor from that centre, using the known angular dimensions of the scoring rings on the target selected.
As you say, this allows you to print a scaled target that matches the angular size of the real target, as seen by the eye, so the distance of the target from the eye is the distance required when printing the target.
 
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is simply looking for the centre of the shape
Well, maybe not. Some shapes have no obvious center as shown in my trick test. I have additional hypothesis on that to be tested.
 
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