Again, the brush current density rating will serve to limit the number of outer brushes used and the maximum possible current. The current density limit of the 93%Ag brushes we've specified is 300 A/in². In practice, total brush ampacity should be strictly matched to the highest I max figure derived that does not exceed the calculated safe ampacity of the rotor disk (as discussed further below) and total pickup brush width should be absolutely maximized,  approaching 100% of the disk's entire circumference,  in all generator (non-motoring) variants.
    Since the base 16-pole circuit resistance yields a figure for I max that is so close to the above-stated ampacity of the rotor disk, we will assume that the tentative 1150-rpm operating speed selected will be very slightly reduced to match that figure and that a 16-pole/48-point pickup brush circuit can therefore be used in this 'prototype' model and in the calculations to follow.
 
    To get a better figure for our model dynamo's actual total circuit resistance Rt, we will now calculate the resistances of the brushes, rotor shaft, and disk.
    Given the OEM specs for brush current density (300 A/sq.in.) and resistivity (2.0 x 10^–6 ohm-cm), we may simply divide the value for I max by 48 to find the ampacity of each negative (pickup) brush and then divide the result by the current density limit to find the contact area required. Therefore, the area of each outer brush An = (1,579 / 48) / 300 = 0.1097 sq.in. or 0.708 cm². Outer brush thickness should be > ½ and < 2/3 of the disk thickness, as the disk's outer edges should be slightly chamfered. So, the width of each pickup brush will be 0.1097 / 0.125" = 0.88". The rotor's circumference is equal to 2πRo = 56.55", the total pickup brush width is 48 (0.88) = 42.24", and so the edge-width 'coverage' ratio is 42.24 / 56.55 or 74.7%.
    Using brushholders which are only 5/8" 'tall', a good minimum brush length L is 1.0" or 2.54 cm. So, each pickup brush's resistance will be (2.0 x 10^–6)(2.54) / 0.708 = 7.175 x 10^–6 ohm, or Rn = ~ 7 μohm. Applying the same method to the 3 positive shaft brushes, at an assigned thickness of 0.375" the area Ap = 3An = 0.3291 sq.in. (2.123 cm²) and width again equals 0.88" or 2.24 cm. With length once again of 1.0", each inner brush resistance Rp = ~ 2 μohm. [As before, it will be important in practice to use the heaviest and shortest brush shunts (buss bar leads) feasible.]
    The rotor shaft alloy that is greatly to be preferred is CDA17200 1.9% beryllium copper, with volume resistivity of 7.733 x 10^–6 ohm-cm but the highest tensile strength of any copper-base alloy. To figure a liberal resistance for the rotor shaft, we'll once again allocate 5" or 12.7 cm as its "electrical" length by mounting the positive brushes inboard of the bearings and drive coupling. The axial end area of the shaft is equal to π( ¾")² = 1.767 sq.in. = 11.4 cm², and the shaft resistance Rs = ~ 9 μohm.
    The rotor disk's greatest electrical resistance is expressed through the edge of the 1½"-dia. central shaft hole, the circumference of which is 4.7124". Given a thickness of 0.187", this inner edge area then equals 0.8812 sq.in. or 5.685 cm². The volume resistivity of pure copper is 1.724 x 10^–6 ohm-cm, and the disk's radial conduction length L is equal to Ro – ¾" = 8.25" = 0.210 m = 21.0 cm. Accordingly, we find the resulting maximum possible rotor disk resistance Rd = ~ 6 μohm.
 
    Finally, we can now make the best possible projection of the rotor circuit Rt in milli-ohms as follows:
 

    As we saw before, the electrical 'hardware' really has little impact on the total rotor circuit resistance of a Faraday disk generator – if properly designed – since the figure we just derived with fair effort differs from the quick estimate we made earlier by only 1 micro-ohm! The rotor current developed at a preferred operating voltage of 0.3978 (as derived above) would then be equal to 0.3978 / 0.000251 or 1,588 amps, since I = V/R. Unfortunately, it can be shown using rather involved numerical anlysis of existing copper wire data that the safe ampacity of the 0.187"-thick rotor disk is 'only' 1,579 amps (as mentioned above) around its inner circumference. Our revised figure for the "nominal" output current will then be equal to the calculated safe rotor ampacity = nom. Io = 1,579 A.
    preceding subsection   top of page

Output Power:  The new value for max. Vo = 0.3963 volts, and the corresponding maximum rotor speed is then 1143 rpm or f = 19.05 rps (since f = Vo / 2πrRaB). And so, the maximum allowable output power is equal to (0.3963)(1,579) or max. Po = 625.8 watts, and the new maximum brush speed is 5,386 fpm. The pickup brush width will not have to be increased, and so the edge-width coverage ratio is still 74.7%.
[See this same section of the Induction Dynamo Analysis for reference.]     preceding subsection   top of page

Input Torque & Power:  Although it may seem that we now have good final figures for operating speed and output power, we have yet to determine if the rotation speed we just derived is acceptably close to that of an off-the-shelf electric motor (as a practical source of input torque) which will provide adequate start and run torque at a given available horsepower rating. This will not be nearly as difficult as it might sound, with the aid of a simple yet indispensable electric motor formula that relates speed (rpm), torque (T), and power (Hp):  T = (Hp x 5252) / (rpm) , where the constant 5252 is equal to 33,000 ft.lbs./min./Hp divided by 2π radians/rev., and T is the torque in ft.lbs. [The equivalent metric expression is:  T = P / ω , where ω = 2πf , P is power in watts, and T is torque in N-m.]   preceding subsection   top of page
 
For the 'classical' case:
  [i] primary full-load counterforce = Fa = B I Ra = (.1383)(1,579)(.180) = 39.31 N
  [ii] primary back-torque = Ta = Fa (r) = (39.31)(.133) = 5.23 N-m = 3.86 ft.lb.
  [iii] neg. brush counterforce (ea.) = Fnb = pkA = (4)(0.2)(.125 x .88) = 0.088 lb. (run)
  [iv] pos. brush counterforce (ea.) = Fpb = pkA = (4)(0.2)(.375 x .88) = 0.264 lb. (run)
  [v] total brush retarding torque = Tb = 48(Fnb)(Ro) + 16(Fpb)(¾") = 41.18 in.lb. = 3.43 ft.lb.
  [vi] total load torque T = Ta + Tb = 3.86 + 3.43 = 7.29 ft.lb. (treating the ~ 0.5% rolling losses as negligible).
 
    As stated earlier, there are a number of reasons for selecting a permanent magnet or shunt-wound [wound field] DC drive motor in this application, since they will operate on normal AC line power and an economical variable speed control is usually also available. This feature is especially desirable in cases where the tentative operating speed falls somewhere between standard electric motor speeds. However, in the case of a statorless Faraday generator with a large moment of inertia, it may be permissible (and less expensive) to use an AC capacitor-start motor if it so happens that a standard-speed motor can be used – since they are designed to start under heavily-loaded conditions. Although they may briefly draw over 300% of normal running amps to do so, a capacitor-start motor can be used in certain instances to effect a corresponding reduction in generator starting time.
    Referring again to the motor selection and ordering guide, we find that a 1.5 Hp 1150 rpm PM motor (with 180vdc armature; FL amps = 7.2) is available [GE 5CD125TP001B] that develops full-load torque of 6.85 ft.lb. (82.2 in.lb.). Even though this is somewhat less than the 7.29 ft.lb. that would be needed to operate our model dynamo at its maximum capability, it must be remembered that the motor's output torque will climb as its speed is reduced – and it may be that just enough torque will be available very near the maximum 1143 rpm operating speed we desire.
    The most straightforward way of determining that point on the selected motor's output power / torque curve where its operating speed is maximized for this particular application's input torque requirement is by repeated spreadsheet calculations to assemble tabular data, from which the best 'solution' of such a complex covariable problem is obtained when the net motor torque developed (Tm) just exceeds the total load torque (Tn) required at the reduced speed.
    Starting at 1125 rpm and calculating output voltages in descending 5-rpm increments, the following optimal resolution was found at 1100 rpm (f = 18.333 rps):
 
  – net Vo = 0.3814 volts, net Io = 1,519.5 amps, and net Po = 579.5 watts ;
  – net Fa = B I Ra = (.1383)(1,519.5)(.180) = 37.826 N ;
  – net Ta = Fa(r) = (37.826)(.133) = 5.031 N-m = 3.71 ft.lb. ; and
  – total load torque = Ta + Tb = 3.71 + 3.43 = Tn = 7.14 ft.lb. [treating the ~ 0.5% rolling losses as negligible].
  – net motor torque developed = Tm = (1.5 x 5252) / 1100 = 7.16 ft.lb. > Tn (@ 7.14 ft.lb.).
 
  – nominal motor efficiency = (Hp x 746) / (V x I) = 1,119 / 1,296 = ~ 86.3%
  – nominal dynamo efficiency = Po / (Hp x 746) = 579.5 / 1,119 = ~ 51.8%
  – combined system efficiency = (0.863)(0.518) = ~ 44.2%
 
  – letting avg. brush k = ½ (0.3 + 0.2) = 0.25,  avg.Tb = (0.25 / 0.2)(3.43) = 4.29 ft.lb.
  – avg. no-load motor torque  Tnl = Tm – avg.Tb = 7.16 – 4.29 = 2.87 ft.lb.
  – no-load motor starting time  min. ts = ωI / Tnl = 2πf [.181 + 2(0.213)] / 2.87 = 24.36 sec.
  – Pr = mπ²R²f ² / ts = (24.715)(9.8696)(0.0491)(336.1) / 24.36 = 4,025.4 / 24.36 = 165.2 watts

  – avg. full-load motor torque  Tf = Tm – (½ Ta + avg.Tb) = 7.16 – (1.86 + 4.29) = 1.01 ft.lb.
  – full-load motor starting time  max. ts = ωI / Tf = 2πf (.607) / 1.01 = 69.23 sec.
  – rotor mechanical power expended = Pr = mπ²R²f ² / ts = 4,025.4 / 69.23 = 58.1 watts

    These starting times are of questionable acceptability even for a PM-DC drive motor, and are far too extended for a capacitor-start motor to be used [since no-load start times longer than about 3 seconds may overload the additional start windings in a capacitor-start motor, damaging or destroying the coils and/or tripping a circuit breaker, even though the effective start times will be lower by ~300%]. Also, it can be seen that the much lower gap flux density in the Ferrite-rotor generator has caused an almost 50% reduction in output power and a 33% drop in efficiency, despite the greatly increased rotor current, as compared to the induction dynamo in the previous example. And, the additional brush drag and far-higher rotor mass have served to cause what might be considered an unacceptably long starting time.
    It should be noted that this statorless generator's actual induction efficiency can be found by simply figuring its torque requirement without considering brush drag. Thus, at 1075 rpm, a ¾-Hp PM-DC motor would be adequate to power the generator at rotor voltage of 0.3727, current of 1,485 amps, and output power of 553.5 watts. Provided both field pieces' outside poles will be wholly keepered at saturation, and at less than the disk's full diameter, real induction efficiency would be 553.5 / 0.75 (746) = 98.9%! [These criteria regarding the essential use of saturated steel end-plates to 'keeper' the outer stator poles derive from the innovative prior-art "closed (flux) path" homopolar generator design of Trombly & Kahn (~1982). A saturated material will support no further passage of flux nor any further external magnetic induction – in the form of eddy current losses, in this case.]
    Purists may also notice that the motor's own armature inertia has not been considered above in the interests of clarity and brevity, having been treated as negligible since it corresponds to just 0.4% of its output torque (by OEM specs) in this particular case. The same consideration also applies to the rotor shaft's miniscule inertia in relation to that of the disk. The non-trivial inertia of the steel end-plates has also not been considered, however, which would act to further extend the start times figured.
 
› And in the best-case theoretical model: The following computations are based on the application of the 'Tesla' reduced back-torque ratio to the model 'unipolar' generator in question. As we developed earlier above, in this case that ratio is equal to 1 – 0.747 = 0.253.
 
  [i] primary full-load counterforce = Fa = 25.3% [B I Ra] = 0.253[(.1383)(1,519.5)(.180)] = 9.57 N
  [ii] primary back-torque = net Ta = Fa (r) = (9.57)(.133) = 1.273 N-m = 0.94 ft.lb.
  [iii] total brush retarding torque = Tb = 48(Fnb)(Ro) + 16(Fpb)(¾") = 3.43 ft.lb. (same as before)
  [iv] total load torque Tn = Ta + Tb = 0.94 + 3.43 = 4.37 ft.lb. (treating the ~ 0.5% rolling losses as negligible).
 
    Referring again to the motor selection and ordering guide, we find that a 1 Hp 1150 rpm PM motor (with 180vdc armature; FL amps = 5.0) is available [GE 5CD123GP001B] that develops full-load torque of 4.57 ft.lb. (54.8 in.lb.). In this case, the 1-Hp motor seems to have enough torque for the generator to be operated at the same rotor speed as it was with the 1.5-Hp motor in the classical case, at 1100 rpm (with f = 18.333 rps), thereby nicely illustrating the effective proportional reduction in input power:
 
  – net Vo = 0.3814 volts, net Io = 1,519.5 amps, and net Po = 579.5 watts (from classical case above);
  – net Ta = Fa(r) = 0.94 ft.lb. (from above); and
  – total load torque Tn = 4.37 ft.lb. [treating the ~ 0.5% rolling losses as negligible].
  – net motor torque developed = Tm = (1.0 x 5252) / 1100 = 4.77 ft.lb. > Tn (@ 4.37 ft.lb.).
 
  – nominal motor efficiency = (Hp x 746) / (V x I) = 746 / 900 = ~ 82.9%
  – nominal generator efficiency = Po / (Hp x 746) = 579.5 / 746 = ~ 77.7% (or COP = 0.777)
  – combined system efficiency = (0.829)(0.777) = ~ 64.4%
 
  – avg. starting brush retarding torque avg.Tb = 4.29 ft.lb. (same as before)
  – avg. no-load motor torque  Tnl = Tm – avg.Tb = 4.77 – 4.29 = 0.48 ft.lb.
  – no-load motor starting time  min. ts = ωI / Tnl = 2πf [.181 + 2(0.213)] / 0.48 = 145.7 sec.
  – Pr = mπ²R²f ² / ts = (6.965)(9.8696)(0.0491)(336.1) / 145.7 = 4,025.4 / 145.7 = 27.6 watts

  – avg. full-load motor torque  Tf = Tm – (½ Ta + avg.Tb) = 4.77 – (0.47 + 4.29) = 0.01 ft.lb.
  – full-load motor starting time  max. ts = ωI / Tf = 2πf (.607) / 0.01 = 6,992 sec.
  – rotor mechanical power expended = Pr = mπ²R²f ² / ts = 4,025.4 / 6,992 = 0.58 watts.

Conclusions: 
    As should perhaps have been apparent from the outset of this example, there would be no reason not to use the far-stronger NdFeB magnets in this statorless generator model, since the generator's output power must be proportionally lower if the Ferrite magnets are used – and by all appearances achieving a COP > 1.0 would simply not be possible, unless perhaps a non-statorless dynamoelectric motor stage was very carefully employed [with far greater system size, complexity, and cost].
    Also, in much the same way, there would be no point in rotating the additional mass of the pole-sets (even if using metallic NdFeB magnets!) when the result must be an undesirable and perhaps untenable increase in the starting time and in the net mechanical torque requirement, which works against a best efficiency in disk generator design. With so much misinformation disseminated about the statorless Faraday dynamo variant, it might be tempting to speculate how much of it is truly inadvertent.