Target position and Visual Angle

Sat Aug 1 00:44:37 UTC 2009

Could be wrong here, but ...
I don't think the pixel size is correct. A pixel 
is a unit with area .. a mm or cm is a unit of 
distance. Pixels are square (when used by modern 
compter displays), so their diagonal is 
1.41*side. It seems to me that this calculation 
returns the number of pixel sides that would fill 
the diagonal distance, not necessarily the number of pixels.

So you could possibly work out how many pixels in 
the diagonal by dividing 800 by 1.41, but I think 
this is wrong also - If you drew a 640*480 grid 
and then drew the diagonal, would it follow the 
diagonals of all of the pixels it passed through? I'm not sure that it would.

When it comes to computations of distance in 
terms of number of pixels, I think it is simple 
to define vertical, or horizontal planes, in this 
way, but very difficult for any other angles.

Oh. Visual angle is the angle an object subtends 
in the visual field, not where it is in the visual field.

http://en.wikipedia.org/wiki/Visual_angle

.. if you wanted a circle display that subtended 
3.2 degrees visual angle, then ,...

tan(visual angle) = diameter / viewing distance

.. say viewing distance of 1.2 metres
tan(3.2) = diameter / 1.2 metres
= diameter of 6.7cm

.. say viewing distance of 1.5 metres
= diameter of 8.3cm

.. say viewing distance of 0.9 metres
= diameter of 5.0cm

At 09:05 PM 31/07/2009, you wrote:

>Hi Ashraf,
>         Sorry for responding so late - the 
> message below should be seen as answer here and 
> off-list. The numbers are either centimetres 
> (i.e. 1 inch is about 2.5 cm), or - more 
> traditionally angles. The paper you seem to be 
> reading will mention it, and if you just give a 
> reference, or paste the relevant passage here, 
> we would possibly be able to help you. At the 
> moment, your English makes it very difficult to 
> understand what you are saying. As a fellow 
> non-natively English speaker, I can empathise 
> with the difficulty you might be experiencing, 
> but as you will most likely be wanting to 
> publish in English, I believe you should try a little harder.
>
>         Anyway, if the numbers are cm:
>1. Write down the size of your monitor, 
>typically given in the diagonal size, in inches. 
>Mine used to be 19 inch, for example, which is about 47.5 cm.
>2. Write down the resolution used in your 
>experiment. This you can find under 
>edit>experiment>properties>devices>display>properties 
>(or something like that). It is 640x480 by 
>default, which is X (number of pixels) by Y 
>(number of pixels). I'll stick to this resolution for the current example.
>3. Calculate the diagonal number of pixels by 
>using Pythagoras' wisdom: A^2 + B^2 = C^2 --> 
>640^2 + 480^2 = C^2 --> 409600 + 230400 = 640000 --> SQRT(640000) = 800.
>4. Divide number of pixels (from 3) by number of 
>centimetres (from 1) to get the number of pixels 
>per centimetres: 800 / 47.5 = about 16.84 
>pixel/cm (also useful is number of cm per pixel): about 0.05 cm/pixel.
>
>
>Okay, so now we can use the fact that 1 cm 
>equals about 16.84 pixels on the monitor to 
>calculate the number of pixels used to create a 
>0.61 by 0.41 letter: about 10 by 7.
>
>It's a bit small, though, so I am actually 
>thinking that the authors use visual angle 
>rather than cm - 0.41 cm is pretty small for any 
>stimulus. Still, to understand visual angles 
>requires the information above. Also, you will 
>need to make a good guess (er.. I mean measure) 
>as to how far the monitor is placed from the 
>participant - typically about 40 to 60 cm. 
>Visual angle refers to the angle of the stimulus as relative to the eye, i.e:
>
>      s
>     /s
>Eye< s
>     \s
>      s
>
>..in which s is stimulus. Sorry if this becomes scrambled.
>
>Given that your screen is 40 cm away and 47.5 cm 
>in diagonal width, you can calculate that the 
>entire screen has a visual angle of the 
>arctangent of Y / X, i.e. ATAN(47.5/40), which 
>is about 38.66 degrees. As you may remember, 
>43.53 degrees should be equal to both 47.5 cm 
>(1) but also 800 pixels (3), you will realise 1 
>degree should be about 800 / 38.66 = 20.69 
>pixels. Therefore, if you want your stimulus to 
>be 0.61 degrees, it should be about 13 pixels.
>
>Okay, I'm not the best at trigonometry, so maybe 
>I made a few mistakes in the above simple 
>calculus. There are, of course, age old tools 
>which will give you the information without 
>headache: just use your measuring tape and 
>trigonometry triangle. Sit where you think a 
>participant sits, put the triangle in your eye 
>(note: I'm not liable for any damage) use a marker, note the angle, good.
>
>Best,
>Mich
>
>
>
>Michiel Spapé
>Research Fellow
>Perception & Action group
>University of Nottingham
>School of Psychology
>
>-----Original Message-----
>From: e-prime at googlegroups.com 
>[mailto:e-prime at googlegroups.com] On Behalf Of ashraf
>Sent: 30 July 2009 22:14
>To: E-Prime
>Subject: Re: Target position
>
>
>Thank you very much , you said "do not forget to tell e-prime the
>dimensions of your screen first" . execuse me i do not understand ,...
>Tell me  how could i make circle letters subtended, 0.61 by 0.41
>exactly
>and the Flanker letter  out of  the Circle subtended 0.81 by 0.51.
>
>Target position differs according to six possible position ,how
>   Flanker letter position differs according to two possible position :
>right/Lift,how
>
>On Jul 30, 11:40 am, liwenna <liwe... at gmail.com> wrote:
> > Hello Ashraf,
> >
> > E-prime does not offer the possibility to control target positions
> > relative to each other not in terms of 'place one target a centimeter
> > left to the other, and not in terms of place 6 targets in a circle).
> > The only way to position targets is by positioning each target
> > separately. In the slide object you can place multiple targets and
> > give each target an x (horizontal) and an y (vertical) position, this
> > can be done in eather pixels from the top left corner or percentages
> > of the total screen size.
> >
> > For your setup you should make an slide object with 7 textboxes: 6 in
> > the circle and 1 to the left or right. For the circle letters figure
> > out the correct x and y positions either by simply trial and error and
> > controlling with a set triangle. Yet I also think it should be
> > possible to simply calculate the desired x and y positons if you know
> > what the dimensions of your screen are, how big your textboxes are and
> > how big the circle should be. (do not forget to tell e-prime the
> > dimensions of your screen first however... find display properties
> > under the square with the e-prime E at the top of your experiment
> > tree). For the target letter-textbox the x position (left or right)
> > should be drawn from a list that holds the value for x (in a variable
> > called targetposition for instance) and is set to the value for left
> > (ie 25%) or rigth (75%) both in half of the trials. In the properties
> > of the targettextbox set the y value to center (assuming that it
> > shoudl appear in the vertical middle of the screen) and set the x
> > value to [xtargetposition] to make it refer t1o the variable with 25%
> > or 75% in it.  The content of the textboxes (i.e. the letters that
> > make up the target and distractors) should also be drawn from a list.
> > Make a list with 7 variables: distractor1 distractor2 etc and target:
> > and place your letters into this list. In the properties of the 7
> > textboxes do not fill in a text but fill in [distractor1],
> > [distractor2], etc. For each of the trials the textboxes will now take
> > their content from (the same level of ) the list.
> >
> > Alternatively, you could make the distractor circle arrays in a
> > separate program (perhaps paint, it is more easy I think with
> > photoshop, Gimp or another programs that offers working in multiple
> > layers). Then you can use an image like this 
> one:http://www.mathsisfun.com/geometry/images/degrees-360.gif, and place
> > it in a layer. In a new layer you can then place your letters and
> > delete the circle layer and save the image to use in in e-prime. (as
> > an imageobject in your slide). You would need to make quite a bunch of
> > these images however in order to not have the same circle of
> > distractoritems repeat too often.
> >
> > I hope that this info will help you start your experiment.
> >
> > Good luck and best regards,
> >
> > liwenna
> >
> > On Jul 30, 12:45 am, ashraf ashraf <ash2003r... at yahoo.com> wrote:
> >
> >
> >
> > > . I want to  make of six letters in E-prime 
> , and I want to present target letter 
> appears  in one out of six possible positions 
> in a circle and a distractor letter  presented 
> to the left or right of the circle,
> > > how can i maniplute Target position and distractor position .
> > >
> > > I read in some papers properties of stimuli 
> as " The task display consisted of
> > > a circle (1.61 radius) of six letters 
> centered at fixation, plus aperipheral 
> distractor letter, presented to the left or 
> right of the circle, 1.41 away from the nearest circle letter. Each of the
> > > circle letters subtended 0.61 by 0.41, and 
> the distractor letter subtended 0.81 by 0.51. "
> > > what do   these numbers mean and how could I control  it with E-prime
> > >  - Hide quoted text -
> >
> > - Show quoted text -
>
>
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>
>

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