Sixteen Experimental Investigations from the Harvard Psychological Laboratory by Hugo Münsterberg (good novels to read in english TXT) 📖

answer, “We are dealing with the phenomena of after-images.” The

bands persist as after-images while new ones are being generated. The

very oldest, however, disappear pari-passu with the generation of

the new. We have already seen (p. 169) how well these authors have

shown this, in proving that the number of bands seen, multiplied by

the rate of rotation of the disc, is a constant bearing some relation

to the duration of a retinal image of similar brightness to the bands.

It is to be noted now, however, that as soon as the rod has produced a

band and passed on, the after-image of that band on the retina is

exposed to the same stimulation from the rotating disc as before, that

is, is exposed to the fused color; and this would tend to obliterate

the after-images. Thus the oldest bands would have to disappear more

quickly than an unmolested after-image of the same original

brightness. We ought, then, to see somewhat fewer bands than the

formula of Jastrow and Moorehouse would indicate. In other words, we

should find on applying the formula that the ‘duration of the

after-image’ must be decreased by a small amount before the numerical

relations would hold. Since Jastrow and Moorehouse did not determine

the relation of the after-image by an independent measurement, their

work neither confirms nor refutes this conjecture.

 

What they failed to emphasize is that the real origin of the bands is

not the intermittent appearances of the rod opposite the lighter

sector, as they seem to believe, but the successive eclipse by the rod

of each sector in turn.

 

If, in Fig. 2, we have a disc (composed of a green and a red sector)

and a pendulum, moving to the right, and if P represents the

pendulum at the instant when the green sector AOB is beginning to

pass behind it, it follows that some other position farther to the

right, as P’, will represent the pendulum just as the last part of

the sector is passing out from behind it. Some part at least of the

sector has been hidden during the entire interval in which the

pendulum was passing from P to P’. Clearly the arc BA’ measures

the band BOA’, in which the green stimulation from the sector AOB

is thus at least partially suppressed, that is, on which a relatively

red band is being produced. If the illusion really depends on the

successive eclipse of the sectors by the pendulum, as has been

described, it will be possible to express BA’, that is, the width of

a band, in terms of the widths and rates of movement of the two

sectors and of the pendulum. This expression will be an equation, and

from this it will be possible to derive the phenomena which the bands

of the illusion actually present as the speeds of disc and rod, and

the widths of sectors and rod, are varied.

 

[Illustration: Fig 2.]

 

Now in Fig. 2 let the

width of the band (i.e., the arc BA’) = Z

speed of pendulum = r degrees per second;

speed of disc = r’ degrees per second;

width of sector AOB (i.e., the arc AB) = s degrees of arc;

width of pendulum (i.e., the arc BC) = p degrees of arc;

time in which the pendulum moves from P to P’ = t seconds.

 

Now

arc CA’

t = ––-;

r

 

but, since in the same time the green sector AOB moves from B to B’,

we know also that

arc BB’

t = ––-;

r’

then

arc CA’ arc BB’

––- = ––-,

r r’

 

or, omitting the word “arc” and clearing of fractions,

 

r’(CA’) = r(BB’).

But now

CA’ = BA’ - BC,

while

BA’ = Z and BC = p;

therefore

CA’ = Z-p.

Similarly

BB’ = BA’ + A’B’ = Z + s.

 

Substituting for CA’ and BB’ their values, we get

 

r’(Z-p) = r(Z+s),

or

Z(r’ - r) = rs + pr’,

or

Z = rs + pr’ / r’ - r.

 

It is to be remembered that s is the width of the sector which

undergoes eclipse, and that it is the color of that same sector which

is subtracted from the band Z in question. Therefore, whether Z

represents a green or a red band, s of the formula must refer to the

oppositely colored sector, i.e., the one which is at that time

being hidden.

 

We have now to take cognizance of an item thus far neglected. When the

green sector has reached the position A’B’, that is, is just

emerging wholly from behind the pendulum, the front of the red sector

must already be in eclipse. The generation of a green band (red sector

in eclipse) will have commenced somewhat before the generation of the

red band (green sector in eclipse) has ended. For a moment the

pendulum will lie over parts of both sectors, and while the red band

ends at point A’, the green band will have already commenced at a

point somewhat to the left (and, indeed, to the left by a trifle more

than the width of the pendulum). In other words, the two bands

overlap.

 

This area of overlapping may itself be accounted a band, since here

the pendulum hides partly red and partly green, and obviously the

result for sensation will not be the same as for those areas where red

or green alone is hidden. We may call the overlapped area a

‘transition-band,’ and we must then ask if it corresponds to the

‘transition-bands’ spoken of in the observations.

 

Now the formula obtained for Z includes two such transition-bands, one

generated in the vicinity of OB and one near OA’. To find the formula

for a band produced while the pendulum conceals solely one, the

oppositely colored sector (we may call this a ‘pure-color’ band and

let its width = W), we must find the formula for the width (w) of a

transition-band, multiply it by two, and subtract the product from the

value for Z already found.

 

The formula for an overlapping or transition-band can be readily found

by considering it to be a band formed by the passage behind P of a

sector whose width is zero. Thus if, in the expression for Z already

found, we substitute zero for s, we shall get w; that is,

 

o + pr’ pr’

w = ––- = ––

r’ - r r’ - r

Since

W = Z - 2w,

we have

rs + pr’ pr’

W = ––— = 2 ––,

r’ - r r’ - r

or

rs - pr’

W = ––— (1)

r’ - r

 

[Illustration: Fig 3.]

 

Fig. 3 shows how to derive W directly (as Z was derived) from the

geometrical relations of pendulum and sectors. Let r, r’, s, p, and

t, be as before, but now let

 

width of the band (i.e., the arc BA’) = W;

 

that is, the band, instead of extending as before from where P

begins to hide the green sector to where P ceases to hide the same,

is now to extend from the point at which P ceases to hide _any

part_ of the red sector to the point where it just commences again to

hide the same.

 

Then

W + p

t = ––- ,

r

and

W + s

t = ––- ,

r’

 

therefore

W + p W + s

––- = ––- ,

r r’

 

r’(W + p) = r(W + s) ,

 

W (r’ - r) = rs - pr’ ,

and, again,

rs - pr’

W = ––— .

r’ - r

 

Before asking if this pure-color band W can be identified with the

bands observed in the illusion, we have to remember that the value

which we have found for W is true only if disc and pendulum are

moving in the same direction; whereas the illusion-bands are observed

indifferently as disc and pendulum move in the same or in opposite

directions. Nor is any difference in their width easily observable in

the two cases, although it is to be borne in mind that there may be a

difference too small to be noticed unless some measuring device is

used.

 

From Fig. 4 we can find the width of a pure-color band (W) when

pendulum and disc move in opposite directions. The letters are used as

in the preceding case, and W will include no transition-band.

 

[Illustration: Fig. 4]

 

We have

 

W + p

t = –—,

r

and

s - W

t = –—,

r’

 

r’(W + p) = r(s - W) ,

 

W(r’ + r) = rs - pr’ ,

 

rs - pr’

W = ––— . (2)

r’ + r

 

Now when pendulum and disc move in the same direction,

 

rs - pr’

W = ––– , (1)

r’ - r

 

so that to include both cases we may say that

 

rs - pr’

W = ––— . (3)

r’ ± r

 

The width (W) of the transition-bands can be found, similarly, from

the geometrical relations between pendulum and disc, as shown in Figs.

5 and 6. In Fig. 5 rod and disc are moving in the same direction, and

 

w = BB’.

 

Now

W - p

t = ––- ,

r’

 

w

t = – ,

r’

 

r’(w-p) = rw ,

 

w(r’-r) = pr’ ,

 

pr’

w = ––- . (4)

r’-r

 

[Illustration: Fig. 5]

 

[Illustration: Fig. 6]

 

In Fig. 6 rod and disc are moving in opposite directions, and

 

w = BB’,

 

p - w

t = ––- ,

r

 

w

t = – ,

r’

 

r’(p - w) = rw ,

 

w(r’ + r) = pr’ ,

 

pr’

w = ––— .

r’ + r (5)

 

So that to include both cases (of movement in the same or in opposite

directions), we have that

 

pr’

w = ––— .

r’ ± r (6)

 

VI. APPLICATION OF THE FORMULAS TO THE BANDS OF THE ILLUSION.

 

Will these formulas, now, explain the phenomena which the bands of the

illusion actually present in respect to their width?

 

1. The first phenomenon noticed (p. 173, No. 1) is that “If the two

sectors of the disc are unequal in arc, the bands are unequal in

width; and the narrower bands correspond in color to the larger

sector. Equal sectors give equally broad bands.”

 

In formula 3, W represents the width of a band, and s the width of

the oppositely colored sector. Therefore, if a disc is composed, for

example, of a red and a green sector, then

 

rs(green) - pr’

W(red) = –––––– ,

r’ ± r

and

rs(red) - pr’

W(green) = –––––– ,

r’ ± r

 

therefore, by dividing,

 

W(red) rs(green) - pr’

––– = ––––––- .

W(green) rs(red) - pr’

 

From this last equation it is clear that unless s(green) = s(red),

W(red) cannot equal W(green). That is, if the two sectors are

unequal in width, the bands are also unequal. This was the first

feature of the illusion above noted.

 

Again, if one sector is larger, the oppositely colored bands will be

larger, that is, the light-colored bands will be narrower; or, in

other words, ‘the narrower bands correspond in color to the larger

sector.’

 

Finally, if the sectors are equal, the bands must also be