This is a transcript of the handwritten lab notes produced by
Tesla in Colorado Springs between June 1, 1899 and Jan. 7, 1900, and on Long
Island between June 2, 1900 and September 18, 1901. The book starts out
by giving reader a day by day
account of the nine months in Colorado that followed Tesla's decade-long investigations
in New York City into the
fundamental principals of wireless transmission. The account picks up
again about five months after Tesla's return to New York at a time when
work on the Wardenclyffe Plant had just begun. The notes provided
here drop off in the middle of 1901, a point in time when the Tesla
Tower had not
yet been completed and preliminary testing had not yet begun.
RESEARCH OF NIKOLA TESLA IN LONG ISLAND LABORATORY, Aleksandar
Marinčić, 1986 for a summary Tesla's Long Island Notes for the
entire period June 2, 1900 to January 29, 1906.
. . . investigation outside and
such use. The synchronized coil was wound in one instance around a drum
10" in diam. and about 4 feet high from the ground and carrying on
top a board for placing the box with the instruments and supporting a
light rod for air or capacity wire. In another form of apparatus the
synchronized coil was wound on drum of 2 foot diam. and 18" high,
which was supported on a tripod of photographic outfit. These two
connections illustrated in Diagrams 1. and 2. were found best suitable.
The small condenser around secondary s comprised only a few sheets of mica
and tinfoil sufficient to let the currents of a frequency of 50,000 per
sec. pass through easily.
Sept. 18, 1899
Experiments were resumed with all transformers in place, high speed break
and connection in multiple arc of West. Transformer. The object was to
further test the intensity of the vibrations produced particularly without
spark. The connection was as in diagram. It was though that in this
arrangement, which was dwelt upon before, the disturbances were produced
more economically than when using a spark discharge. The experiments fully
confirm this. In the tests the capacity of the two balls of 18" diam.
did not very materially derange the adjustment and period of the circuit. This is to be expected; as for the secondary the capacity was far too smaIl and on the other hand the independent vibration of the extra coil
could not be materially interfered with since the condenser formed by the
two balls and zinc plate allowed free passage of currents to earth. Now
the important thing was to decide whether it is better to make length of
extra coil one half or one quarter of wave as before. This to be
thoroughly investigated. The working was excellent with 1/4 wave length.
the opposite pole when receiving station is supposed to be located over
an area covered by half of the wave. An example will aid in giving
an illustration of what can be reasonably expected in applying the
principles for the above purpose. Suppose the capacity C =
10,000 cm. This can be obtained by a roof on insulated supports.
If the roof be say in the shape of a sphere (this will serve in the
present example) it would have to have a radius of 100 meters. But
since as demonstrated in many tests the capacity of these insulated
terminals increases with elevation about 1/2 % per foot, by elevating
sphere 400 feet only 33 meters radius will be all that is needed.
Assume now that 100 000 per second to be adopted as frequency and that
we expend on the generator 100 H.P. How much under best conditions
can we expect to get at receiving station? 100 H.P. roughly = 75
000 Watts. This enables us to get pressure to which the insulated
sphere or capacity C can be charged by this amount of energy.
Let the pressure be p then we have: 1/2 p2 10000/9x1011 . 200000 = 75000
and p2 = 67500000 = 6750 x 104.
From this p = 100 6750 = 8200 Volts roughly. This means to say
that we could charge the terminal periodically (200000 times per second)
up to a pressure of 8200 Volts.
Evidently with such small pressure the effect at distance would be
qualitatively very small but we can, without using more power increase
the pressure many times by resonating action.
For the moment will be useful to consider what might be done by working
with the above pressure of 8200 Volts only. According to law of
density referred to before the charge distributes itself over the
surface of Earth as illustrated and the density of each zone is
inversely as the surface of the zone.
If we use 100 000 vibrations per second the wave length will be about
1.86 miles. Hence 1/2 wave 0.93 miles. Taking for simplicity
density uniformly distributed over the area or zone we may estimate it
to be 2/3 of the maximal.
Let a = area of sphere or terminal of transmitter,
A = area of zone - the last on opposite pole where density greater.
As above assumed the radius r of sphere may be 33 meters or about 100
feet. The radius of polar cap = 0.93 miles 4900 feet approx.
a/A = 4r2/R2 = 4(r/R)2 = 4(100/4900)2 = 4492 = 1/600 about.
Under the conditions and with reference to above law, since the quantity
of electricity on polar cap is exactly equal to the quantity on sphere C
the density of charge on the polar cap will be 1/600 that on sphere.
From what has been said above, however only 2/3 of this value or 1/900
can be taken. If Q the quantity of electricity both on sphere and
polar cap, D density on sphere, d density on cap we have: D = Q/a d =
Q/A D/d - A/a. On the other hand Q = CP = 10000/9 x 1011 . 8200 =
8/105 . Hence D = 8/105a . Here a should be taken in cm square. Now a =
4pi x 33002 = 140 x 106 cm sq. approx.
Therefore D = 8/105 x 140 x 106 = 8/144 x 1012 = 4/7 x 1012 and d = 4/63
x 1014 as absolute values of D and d respectively.
From the law of density and obvious considerations it follows that the
e.m.f. impressed at polar cap will be in the same ratio smaller as the
density is smaller or it will be 8200/9= 8 Volts [i.e. 9.1 Volts - Ed.].
As the polar cap gets all the energy of the transmitter less of course
the frictional and medium losses which from my tests seem very
insignificant, we could theoretically recover all the energy if we could
utilize the whole of the successive charges of the cap. We could
do so if we would utilize a large capacity C1 capable of taking up the
whole charge. The capacity C1 should be such that C1p = CP. Hence
C1 = C P/p = 900C. Undoubtedly this capacity could be obtained by
vertical wires for instance, distributed over the polar cap at suitable
distances. The currents passing into the wires could be converted
to suitable tension as usually is in my system. The secondaries
could however be connected to advantage only when the currents [are] of
same phase. This would probably necessitate the arrangement of the
wires in concentric stations at which this condition would obtain.
But this difficulty is not a serious one. A wire even then of say
200 feet height might have, with consideration of increase by elevation
about 500 centimeter and to make up a capacity of 900 x 10000 cm as
would be required we would want 900 x 10000/500 = 18000 vertical wires
or other kind of terminals. If wires were used about 680 miles
would be needed, but they could be thin as the pressure would be small
and the air loss consequently negligible. In addition to this the
secondaries would require about as much wire.
The amount of copper seems large -- too large in fact for practical
purposes but it seems only so because we have assumed but 100 H.P.
energy expenditure at transmitter.
Suppose we would install a plant with 10000 H.P. that is one giving 100
times the energy before assumed. Then, since the e.m.f. at
transmitter will be greater in proportion to the square root of the
power we would get at transmitting station an e.m.f. of 82000 Volts
instead of 8200 as before, and then the polar cap would be charged to 80
instead of 8 Volts.
Here I come to a momentous question: Can the transmitter convey all its
energy to the polar cap or can it merely convey the same quantity of
To be sure the energy in a vibrating system as a wire is conserved but
will the Earth behave identically? This is of enormous importance.
For if all the energy is conserved then the polar cap will receive all
the 10000 H.P. of the transmitting station and of the 18000 wires
circuits or terminals each will receive 10000/18000 H.P. = 416 Watts!!!
But -- if the transmitter can do no more than to convey the same
quantity of electricity which it sets in movement to the polar cap then
under the conditions presently considered the energy collected at the
polar cap will be small. Namely we could only employ a capacity
900 times as great as that of the transmitter and could only charge it
to 80 Volts. To use a greater capacity -- even if we could --
would seem of no avail since, the quantity of electricity being fixed,
we would not be able to charge the capacity to 80 Volts. The
maximum of energy we could collect then would be: 1/2 x 900x10000/9x1011
. 802 x 200000 = 516 x 1013/9x1011 = 51600/9 = 5733 Watts or we could
get only 5733/750 = 7.6 H.P. approx.
Each wire or circuit will then receive only 5733/18000 = 0.3 Watts or a
little more an amount of energy extravagantly large for operating
automatic devices even such things as clocks but otherwise inadequate. .
Sept. 3, 1901
General considerations relative to transmission of energy by conduction
through Earth with my system
The problem in any given case is to displace a certain quantity of
electricity per unit of time. This quantity should be the largest
possible with the means at disposal. Here then is the first
practical limit - the quantity given. Calling it Q we have
Q = Pc. From the equation it follows that for a
fixed value of Q, the greater c the smaller will be the
potential P. This in itself imposes no limitation as to
P, in fact it may be sometimes better to work with a small P
and make c very large. But we must remember that the
density imposes a limit. Now the density D = Q/s = Pc/s,
s being the surface of the free terminal. Evidently then if
D stands for the maximum practical density which it should be an
object to attain within a reasonable limit, then p will be all
the greater the greater c/s [i.e. the smaller c/s].
This means to say that to get high pressure c should be small.
Logical conclusion is that sphere best to employ. It should be
furthermore as high in the air as possible. The sphere has least
metal, largest pressure or effect, the system is rigid. Following is
useful to remember:
D = Pc/s for insulated terminal and d =
pC/S for Earth.
Now S = 4π r2
C = R hence:
Sd/C = p = d
· 4π R now pR = d · 4π R2 but pR
= Pc = Q.
Therefore Q = dS and Q = Ds. From this
D : d = S : s
Transmission of energy for purposes of "Welttelegraphie"
Assume roof as now proposed with plates of 1 meter curvature. The
active area may be taken as equal to s1
, u being mean circumference of toroidal surface s2.
As now planned
u = 112 feet, r2
= 6 feet, r1
= 96 feet, h = 34 feet. From this
= 6.28 x 96 x 34 = 20500 feet square approx.,
= 3.14 x 6 x 112 = 2100 "
" " .
Surface active s1
= 22600 square feet about.
If we make half-spherical pieces about 1 meter curvature each would take
the space of about 9 square feet, hence we would have 22600/9 = 2500
such half spheres. The capacity of one sphere of diameter as above
would be at small height 100 cm. Of half the sphere it would be 50
cm but as a matter of fact we can not count on more than 25 cm each.
Suppose this value be nearly correct at small height we should have
capacity of total structure on roof 2500 x 25 = 62500 cm and at a height
of 200 feet certainly not less than 100000 cm.
On the basis of above estimate if I were to adopt 50000 per second I
1 / 5 x 104
= 2π / 103
/ 9 x 105
L and (3 / 2π x 50)2
= L = 9/105
Henry = 90000 cm.
Now with 50000 per sec. λ/2 = 1.86
miles = 9820 feet approx.
Equatorial belt would be: 8000 x 5280
x π x 9820.
If we assume the roof to be in the form of a sphere of same capacity it
would be of a radius = 1000 meters. The surface of this would be
taking feet 3280 x 3280 x π x 4.
The ratio of the small to the large surface (equatorial belt) would be:
3280 x 3280 x π x 4 / 328 x 328 = 8000 x 5280 x π x 9820 / 20 x 5280 x
This ratio would also be the ratio of the pressure on the roof to the
impressed pressure on ground. Suppose we had on roof only 500000
Volts we would have on equatorial zone:
328 x 328 x 500000 / 528 x 982 x l000 = 328 x 328 x 500 / 2 x 528 x 982
= 52 Volts nearly.
This would be the impressed e.m.f. and could be easily raised to a 200
times higher value.
The current in transmitting terminal would be:
I = ECω = 500000 x 105 /
9 x 1011 · 50000 x 6.28 = 17500 Amp.
For the time of closure - certainly too great.
Sept. 18, 1901
Following results may be confidently expected with smaller tower 200
feet and terminal roof of cheap construction as last designed. The
roof will comprise a single platform with spherical bodies of large
curvature on rim. The construction of latter will be given in
detail. The platform 20 meter diameter, 15 round surfaces on top
and 15 on bottom as shown in sketch below. It is difficult to
estimate in advance the capacity of the structure with precision but an
approximate idea may be obtained. . . .