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Nikola Tesla From Colorado Springs to Long Island, Research Notes: Colorado Springs 1899-1900, New York 1900-1901 (hardcover)
Nikola Tesla; Commentary by Aleksandar Marinčić
595 pages, 44 photos, many line drawings.
ISBN-13: 978-86-81243-44-2
348-CSLI ... $185.00 (scarce)

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This is a typescript of the handwritten laboratory notes produced by Tesla in during his time in Colorado Springs between June 1, 1899 and Jan. 7, 1900 (previously published in 1978 as Nikola Tesla  Colorado Springs Notes  1899-1900, with Commentary by Aleksandar Marinčić), and on Long Island during the period 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 (an in-print account popularly known as "The Colorado Springs Notes").  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 laboratory notes that are provided in this book are limited in that they extend only to the middle of 1901, a point in time when the Tesla Tower had not yet been completed and preliminary testing had not yet begun. 

See RESEARCH OF NIKOLA TESLA IN LONG ISLAND LABORATORY, Aleksandar Marinčić, 1986 for a summary Tesla's Long Island Laboratory notes covering the entire period from 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.

Colorado Springs
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.


Long Island

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 electricity?

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 r
2 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 follows

D : d = S : s


Sept.8, 1901
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 + s2. Now s1 = 2πr1h and s2 = πr2u , u being mean circumference of toroidal surface s2.  As now planned

u = 112 feet, r
2 = 6 feet, r1 = 96 feet, h = 34 feet.  From this

1 = 6.28 x 96 x 34 = 20500 feet square approx.,

2 = 3.14 x 6 x 112 = 2100    "        "         " .

Surface active s
1 + s2 = 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 would have

1 / 5 x 10
4 = 2π / 103 105 / 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 9820

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. . . .

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