Target position and Visual Angle

[Michiel Spape]

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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