2. Deductions from
meter, viz., glass tube 36 in. long,
4 in. diameter, and liquid mercury.
(b) Perform the experiment.
(a) Hg is free to fall out, but does not,
therefore must be supported.
(b) Only pressure acting is that of air,
therefore air supports column 33 in.
(c) Therefore column of Hg 33 in. high =
column of air, entire height.
3. Further Deduc-
f(a) Only a liquid would transmit pres-
sure from basin.
(6) A lighter liquid for barometer needs
a longer tube.
(c) Therefore Hg chosen, because a
liquid which neither freezes nor
boils easily and because of its den-
I (a) If weight of
creases, Hg rises.
If weight of air de-
creases, Hg falls.
(b) Weight of air increases
when atmosphere is
cold or dry, and de-
creases when hot and
(c) The last because watery
vapour is less dense
(d) Heat not allowed to
make any difference
corrected to freezing-
point of water.
(e) If moisture in air, rain
Therefore low baro-
meter means rain, high
barometer means fine
Long foretold, long last,
Short foretold, quick past.
2. How height is measured ;
Here merely point out scale affixed in inches.
(27 or 28 rather low, 29, 30 high reading.)
Summary : The barometer is an instrument used to
1 86 Notes on Herbartian Method
measure weight of air which varies with temperature and
quantity of moisture present. In general air is able to sup-
port 30 inches of mercury. Principle discovered by Torricelli
in 1643. Indirectly it tells us what sort of weather to
I. Open lesson by a few questions as to the pressure of
fluids, lead class to distinguish downward pressure and
weight. Elicit fact that atmospheric air is a fluid and
obeys laws of fluids. Ask (or give) derivation of word
" barometer," and make class see it is to measure the weight
of the air.
II. Presentation. As in matter.
III. Call attention to the fact that till 1643 nothing save
the hollow spheres of Galileo had shown the air to have
weight. Lead class to connect date with early years of
Civil War in Charles I. in history, and so make use of
association to fix date. Allow class to examine the three
things needed, viz., Hg, tube, and basin or cistern Tell
them the experiment to be done is exactly the same as that
done by Torricelli : hence name. Draw from them that
Hg is a metal, but in a liquid form, and is very dense and
heavy in comparison to bulk. Let a few of the class feel
weight of a small quantity. Dense bodies have particles
closely packed, hence great cohesion and little adhesion.
Elicit fact by questions that on this account mercury will not
wet sides of tube. Show that this does away with friction and
leaves it free to move.
When tube is filled and inverted, fix attention of class
well on preliminaries by questions something as follows :
How long did we say the tube is ? How much is filled with
Hg ? Is there any air inside ? Can the air press on the
top (after inverted) ? If I remove my finger what would
happen ? Why ? What is pressing on surface of Hg in
cistern ? What property does Hg possess in regard to this
pressure ? If I put the open end of the tube under the
surface and remove my finger the two portions of Hg will
be in connection, and consequently be one mass of fluid.
The Mechanical Powers 187
Now when that happens what pressure will be communicated
to the Hg in the tube ? Will the pressure over all the sur-
face be transmitted ? Can it be ? How much will be ?
(Here refer to pressure of hydraulic press.) Only the
surface of Hg in tube, therefore if we had a much smaller
or a much larger cistern it would make no difference.
About how many inches has the Hg fallen ? Why does not
the rest fall? Show class that the tube does not touch
bottom of cistern. If Hg does not fall, what keeps it up ? By
similar questions draw from class the deductions written in
IV. Have complete barometer and show scale in inches.
Tell what is considered low reading and high.
Draw from class that heat expands air, so part resting
on Hg is not so heavy, also watery vapour lighter than air,
so, if present, takes place of air, therefore column not so
heavy. Elicit behaviour of Hg in tube, and conclude by
showing how this indirectly tells us the weather to expect.
V. What is the barometer ? Why is Hg chosen ? What
would be the effect of using water ? Why is the reading
low before rain ? What does a sudden fall portend ?
LESSON ON THE MECHANICAL POWERS.
Class Oxford Seniors ; age, 17 to 18 years. Time Fifty minutes.
Aim To exercise the reasoning powers of the pupils and lead them to
have an intelligent and practical knowledge of the three mechanical
powers applied to ordinary uses.
((a) Work. That which is done by force.
(b) Power. Name given to the applied
(c) Mechanical. Relating to a machine.
A contrivance by which
a force applied at one
(d) Machine. ^ point produces work or
Notes on Herbartian Method
2. Use of Machines.
3. Mechanical Advantage :
When the power applied is less than the resistance.
Mechanical Powers Enumerated and Described :
LEVER OF FIRST CLASS.
^ /Is a rod movable about a fixed
point, called a fulcrum.
g j Parts of: Two arms and ful-
LEVER OF FIRST CLASS.
From simple experiment with different weights on a bar
used as lever deduce Conditions of Equilibrium, i.e., that
the product of the power into its distance from fulcrum =
that of weight into distance.
Examples of : See-saw, common balance, poker, wheel-
2. The Pulley : Small wheel with grooved rim fixed to a block,
or suspended by a cord; forces acton ends of a cord round it.
Deduce ; Mechanical advantages in above two illus-
3. Inclined Plane.
The Mechanical Powers
In raising weights, the power is dimi-
Advantage in I nished according as the slope is
General Terms.i smaller and the length or distance
'Train or horse and cart ascending moun-
Familiar Inclined path, instead of steps, in garden.
Examples. Wheel barrow along a plank, etc.
Principle applied to machines : The
screw, the wedge.
Simple examples throughout lesson. Apply principle of
work done remaining the same.
Of (i) definitions in lesson ; (2) description of three
190 Notes on Herbartian Method
Apply knowledge to lessening the difficulty in the follow-
ing practical instances :
1. Opening a nailed-down case.
2. Raising cargo on board ship.
3. Ascending Mount Cenis by rail.
I. Introduce lesson by referring to such expressions as
" One has not the power to do some work," and again,
" One does it mechanically ". What do we mean by power
and mechanical ? Elicit that power is the strength or force
by which we do some work. Ask definition of work, and
examples. Show that it often takes the form of resistance,
e.g., raising a weight. Lastly, ask for examples of machines.
Lead class to see that such simple contrivances as pair of
scissors, etc., are machines. Question as to the force applied
in each case, and the work done, and show that in all of them,
however complicated or simple as the case may be, there is
this in common, that the force is applied at one point and work
done or resistance overcome at another.
Ask the uses of machines, e.g., (i) a weight drawn up
an inclined plane instead of raised vertically, (2) a bicycle,
(3) a poker. Point out the advantage in each case. Con-
trast the advantage of (2) and (3), the latter merely applying
the power at a more convenient point while the former pro-
duces greater motion. Tell class that the principles on which
all machines are based may be seen in their simplest forms
in the three mechanical powers, the lever, the pulley, and the
II. i. Lever. Ask class how men sometimes move a
heavy block, using an iron bar (draw diagram). By a simple
experiment before the class, first without fulcrum, then with
one, draw from pupils the necessary parts of a lever. Give
name (Fulcrum}. Ask what are the forces at work, as
shown in diagram, and represent them as P, W and F.
Give examples of the common balance, and ask which is the
The Mechanical Powers 191
power, which the weight, and where the fulcrum. Treat the
see-saw in the same way. Ask what we notice about the
relative lengths of the arms of the see-saw when children of
same weight are on and when they are unequal. What
conclusion is to be drawn ? By supposing different distances
and weights elicit in general terms the conditions of equili-
brium. Lastly, consider the wheel-barrow as a lever, and
show that both arms may be on same side of the fulcrum.
2. Pulley. Call attention of class to little wheel of window
cord. Ask descriptions and draw diagram, also ask what
use it is. Show that, as in Fig. i, the mechanical advan-
tage is merely in changing the direction and lessening
friction. Now draw Fig. 2. Ask how the pulleys differ
here. How are the movable pulleys supported ? To find
out what mechanical advantage there is here. Suppose
window equals weight 8 lb., how is it held up ? Then each
string bears half its weight, i.e., the beam one and the second
pulley the other. Again, this pulley is supported. How ?
Therefore, pulley has only quarter of weight to bear and the
force is quarter the weight. What would be the effect of add-
ing another pulley ? There is a third case of pulleys, often
used and to be seen in ship rigging. Draw diagram 3. Treat
this case as the former to find the mechanical advantage,
and compare it with Figs, i and 2. In both cases deduce
that the work done remains the same by showing that
window is raised only quarter of the distance pulley moves
in second case and one-fifth in third.
3. Inclined Plane. Refer to the descent to the electric
railway. Why is it so made ? Ask examples from class
of where the inclined plane is used to raise weights im-
possible otherwise. Deduce from their own experience the
advantage of it, and how it is increased according to the
incline. Ask if principle of work being the same applies
here, and why ? Show a screw and ask which of the three
mechanical powers acts in it. Recapitulate the matter briefly,
and ask pupils how they would apply their knowledge prac-
tically to the instances mentioned in application.
Notes on Herbartian MetJiod
LESSON ON THE PRINCIPLE OF
Class Age, 15 to 17. Time Forty minutes. Previous Knowledge
Laws of pressure of fluids. Aim To exercise the reasoning powers
of the pupils in teaching them the principle and leading them to deduce
its application to floating bodies.
Story of Archimedes and Hiero of Syracuse very briefly
(a) Weigh two cylinders
out of water.
(b) Immerse solid in water
and weigh the two
(c) Lastly, fill hollow cylin-
-^y " \ der with water.
1. After (b) found lighter.
2. After (c) weight same as before.
Apparent loss of weight after (b).
c _ we ight of equal volume of water to solid.
This loss | = we i g ht O f water displaced.
Principle stated : A solid immersed in a fluid loses as much
of its weight as is equal to the weight of the fluid displaced.
Proof of the Principle :
N N ABCD = body in equilibrium.
Therefore all pressures are
But side pressures are equal
to opposite, therefore counter-
Downward pressure = column
of water AN N B + weight of body .
The Principle of Archimedes 193
Upward pressure = column of water CNND.
Therefore ANNB + weight = column of water CNND.
Take away column ANNB from each and weight =
column of water CABD.
Take away column ANNB from each and weight =
column of water equal to weight of water displaced.
1. Draw from class how Archimedes tested the gold
crown of Hiero, or how purity of any substance could be
2. Why some bodies float and others sink.
Of the principle and how it is demonstrated.
" A ship is said to draw more water in a river than at
sea." What is the reason ?
I. Introduce lesson by briefly telling class the story of
Archimedes. Who he was. One of the most renowned of
Ancient mathematicians, who lived in Syracuse (ask where it
is, and what now Called) under the tyrant Hiero, about the
beginning of third century B.C. What Hiero asked him to
do. To test purity of gold crown without injuring it. His
perplexity. How and when he got the clue to his work.
Leave Archimedes and return to the question of the crown
at the end of lesson, when class will be able to follow up
his clue to solving the difficulty.
II. Now perform the experiment i as described in matter.
Weigh an object out of water and then in water. Ask what
difference is noticed. Again do the same with some object of
same weight but differing in bulk. Compare results, and
deduce that when the bulk is greater the difference of the
weight in and out of water is greater, finally perform ex-
periment with balance. Elicit results and draw the con-
clusions from class and state them in form of principle.
194 Notes on Herbartian Method
Let class apply knowledge of the pressure of fluids to
account for this fact and prove the principle. Suppose a
cylinder ABCD is immersed in water and is in equilibrium,
what do we know of the pressures on all sides ? Consider
them separately. What pressure acts on sides, and how ?
equal and opposite. What is the downward pressure of
the water ? Is there any other downward pressure besides ?
(Weight of body.) What counterbalances this down-
ward pressure ? What is the upward pressure due to ?
Take away the column ANND common to both. What
pressure is equal to the weight ? Call attention to fact that
this is only true when body is in equilibrium or at rest.
Return to experiment, and apply the above and explain that
the amount, as body is lightened in water, depends on its
III. Also refer to floating bodies, and ask class to account
IV. Recapitulate the principle and show how it is demon-
strated. Conclude lesson by returning to Archimedes and
the crown, and elicit from class how he discovered the fraud.
And lastly, give the question in application, and let class
discover the explanation and tell it later.
LESSON ON A PROBLEM IN GEOMETRY.
Class Age, 14 to 16 years. Time Half an hour. Aim To exercise
reasoning power of class in leading them to apply their previous knowledge
of Geometry to solve the problem, thus to strengthen their powers of
1. Problem : Something is to be done.
2. Enunciation : To construct a parallelogram equal to a
given triangle having one angle equal to a given angle.
3. Definitions : Triangle and parallelogram.
4. Two kinds of Equality : Identical and in area.
5. Relation between parallelogram and triangle.
A Problem in Geometry
1. Given : A triangle ABC and an angle D.
Required : To construct a parallelogram equal to a
triangle ABC having one angle equal to angle D.
(a) Bisect BC at E.
(b) At E make angle CEF equal to
Points. J(c) Through A draw AFG parallel to
(d) Through C draw CG parallel to
Then shall EFGC be required parallelogram. Join AE.
Proof: Triangle ABE = triangle AEC, therefore triangle
ABC = double triangle AEC. But parallelogram EFGC =
double triangle AEC. Therefore parallelogram EFGC =
triangle ABC, and it has angle FEC = angle D (Construc-
What is required in this problem ? What are the three
points in construction ? Go through the whole construction
and proof again with assistance of class.
Erase work on board and require class to write out the
proposition with different letters and a triangle and angle of
BLACKBOARD SKETCH : As in Presentation and Asso-
196 Notes on Herbartian Method
1. Introduce lesson by asking distinction between Problem
and Theorem. Read enunciation. To which class of pro-
position does it belong ? Why ? What are we asked to
do ? What is a triangle ? What is a parallelogram ? In
how many ways may triangles and parallelograms be equal ?
Which propositions treat of parallelograms equal in area ?
What theorem do you know which connects areas of triangles
with those of parallelograms on same base ?
II. i. Read enunciation again, and get class to put it in
form of given and required, and write this on blackboard.
2. Construction. You have just said that the parallelo-
gram is double the triangle. Now in this proposition we
want the parallelogram to be equal to the triangle. How
much of it will it then be double of? (half). How then
can we find half the triangle ABC ? If I bisect base BC at
E and join EA, what do we know of the two triangles ?
Why ? Now on what conditions will a parallelogram be
double of triangle AEC ? (On same base and between same
parallels.) Can any one suggest how to fulfil these con-
ditions ? Suppose we draw parallel to BC a line through A
and one through E parallel to AE. Have we fulfilled the
conditions for having a triangle equal to a parallelogram ?
(Yes.) But have we done all that was asked ? (No.) What
is not done ? (Angle = angle D.) Clearly then we must
try again. How do we make an angle equal to a given
angle ? Now our parallelogram is going to be on EC.
Where then must the given angle be ? (At E or C.) Now
try the construction again. Draw from class points i, 2
and 3, and ask again if parallelogram EFGC answers all
III. What is noticeable about proofs of problems ?
Generally easy and apparent. We want to prove parallelo-
gram = triangle ABC. What is equal to triangle AEC ?
How do you know ? Is any other figure connected with it ?
(Parallelogram double.) What proposition proves this ?
Pythagoras' Theorem 197
But you have said that triangle ABC = double triangle
AEC, and parallelogram is also double of triangle AEC.
What is the conclusion ? And what about angle = to D ?
Then we may write Q.E.F.
IV Recapitulation, j Ag .
V. Application. j
LESSON ON PYTHAGORAS' THEOREM.
Class Oxford Junior Division. Time Half an hour. Aim To
exercise judgment and reason of the class in deducing proof of Pro-
i. Enunciation. In a right-angled triangle the square
described on the hypotenuse is equal to the sum of the
squares described on the other two sides.
((a) Right angle.
(c) Right-angled triangle.
2. Definitions to
1. Analysis of Enunciation.
f Right-angled triangle.
Given -j Square described on hypotenuse.
I Squares described on sides.
Required to be C Square on hypo- J sum of square on other
proved. \ tenuse = (^ sides.
On BC describe square BDEC, and on BA, AC describe
the squares BAGF, ACKH.
Through A draw AL parallel to BD or CE. Join AD,
Notes on Herbartian Method
(a) Because each of angles BAG and BAG is a right
angle, therefore CA, AG are in same straight line.
H Now angle CBD = angle
FBA, for each of them
is a right angle. Add
to each angle ABC,
then angle ABD = angle
(b) Then in triangles ABD
and FBC, because AB
= FB, BD = BC, and
angle ABD = angle FBC,
therefore triangle ABD
= triangle FBC.
(c) Now parallelogram BL is double of triangle ABD,
because they are on same base, BD, and between
same parallels, BD and AL. And square GB is
double of triangle FBC, for they are on same base
FB and between same parallels FB and CG. But
doubles of equals are equal (Axiom 6). Therefore
parallelogram BL = square GB.
(d) In a similar way by joining AE, BK, it can be
shown that parallelogram CL is equal to square CH,
therefore whole square BE == sum of squares GB,
HC that is, square on hypotenuse BC = sum of
squares described on two sides BA, AC. (Q.E.D.)
Propositions, etc., used in proof.
1. Repetition of proof by class.
2. Each step analysed.
1. Deduce principle.
2. Prove in numbers.
3. Value in finding areas and distances
Pythagoras* Theorem 1 99
I. Write enunciation on blackboard. Make the class
read it. Pick out terms and question as to definition of
right angle, triangle, right-angled triangle, square and hypo-
tenuse. What points have been already proved with regard
to a single triangle ? But this is the first mention of a
II. Read enunciation again, and ask what is given.
Right-angled triangle (therefore important to remember
the right angle). Squares described on lines by what pro-
position ? What is a square ? But what other class of
figures may a square come under ? Why is it a parallelo-
gram also ? Make class notice that these squares may be
called parallelograms too. What is required to be proved ?
Is this a problem or a theorem ? Why ? In mensuration
what name is given to " square on side " ? Area of square ?
And it is the sum of areas which is to equal square on
Now as lesson deals with squares we must make them.
Where? Now draw line AL. What is it parallel to?
(Question the class as to the figures into which the square
BCED is now divided by AL.) Call attention to joining
lines, and deduce what figures they make (triangles), and
with what lines. Notice results obtained.
Point out division of square BCED, and probable
reason of this division. Have we learnt any way of finding
out an equality in area between triangles and parallelograms
already ? So if I could get one of the triangles equal to a
square, and then equal to part of the big square BCED, it
would be easy to prove. Look at the two triangles again
and see if we cannot make them equal in all respects. W T hat
is necessary for this equality ? (See which of the conditions
for equality are most nearly fulfilled.) We have two sides
equal to two sides. Now let us examine the angles. What is
angle FBC made up of, and angle ABD ? Therefore we can
say angle FBC equals angle ABD ; therefore our two triangles
2OO Notes on Herbartian Method
are equal. Now notice the position of these triangles
as regards parallelograms. But am I sure that GAC is
parallel to FB ? What part of it am I certain of? GA,
What part must I prove ? AC. What proposition proves
that two lines are in one and same straight lines, or what
conditions are necessary to bring this about ? But what do I
know about the angles at A ? Therefore GA and AC are
in one and same straight line.
Now we have proved our triangles on equal bases and
between same parallels as two parallelograms. What follows ?
By what proposition ? Therefore doubles of equal are equal
by what axiom ? But how much have we proved so far ?
How much is there still to prove ? Could this be done
easily ? How ? When a similar proof is possible is it
necessary to work it out ? Why not ? Therefore, what may
we say to finish proof ?
III. Association. As in matter.
IV. Recapitulate construction and proof in correct order,
asking references and making class account for them.
V. Refer to area of squares, and ask how to find length
of side. Show in numbers that principle is true. 4, 3, 5, etc.
Show by questioning the class how this may become useful
in arithmetic, finding areas and distances, etc., etc.
LESSON ON A THEOREM IN GEOMETRY.
Class Age 13 to 15 years. Time Half an hour. Aim Exercise
of reasoning to lead class to discover proof of this theorem.
i. Enunciation: "If a side of a triangle be produced
then the exterior angle, is equal to the sum of the interior
and opposite angles, and the three interior angles are
together equal to two right angles."
Angle, exterior angle.
Interior opposite angle.
A Theorem in Geometry 20 1
Given : Triangle and produced side.
l (a) Exterior angle = sum
1. Analysis of ) Required to 1 of interior and opposite