DATA SHEET

The sections in this data sheet lettered A thru L are:

• A Table 1
• B Solutions to problems 1-1 and 1-2
• C Wind Power Equation Derivation
• D Air Density
• E Estimating Wind Speed
• F Wind and Altitude
• G General Power Estimates
• H Finding Average Windspeed
• I Usable Average Energy
• J KWH/Month Needed
• K Measuring Aeroturbine Efficiency
• L Cube/Average

SECTION A - TABLE 1

The following table will enable you to estimate the full load 'shut-down' speed of your S-rotor, as described in Chapter 4, Section G. The aeroturbine described in the power figures is 27 ft.², such as one would be if it were made out of 55-gallon drums. Alternator/gearing efficiency is assumed to be 50%.

TABLE # 1

WINDSPEED		WATTS @ 20% EFF.		WATTS 30% EFF.
in MPH	in (MPH)3	IMPELLER	ALTERNATOR	IMPELLER	ALTERNATOR
A	B	C	D	E	F
32	32768	560	280	840	420
37	50653	840	420	1260	630
40	64000	1120	560	1680	840
44	85184	1400	700	2100	1050
47	103823	1680	840	2520	1260
49	117649	1960	980	2940	1470
50	125000	2240	1120	3360	1680
53	148877	2520	1260	3780	1890
55	166375	2800	1400	4200	2100
57	185193	3080	1540	4620	2310
59	205379	3360	1680	5040	2520
60	216000	3640	1820	5460	2730

 

SECTION B-----SOLUTIONS TO PROBLEMS

Problem #1-1 take equation #1-8 and plug in A = 40 square feet, V == 15 MPH. What is P?

Problem #2-1 take the answer from problem #1, and, assuming the same alternator and gearing efficiencies we were just using, figure out energy losses for power at the impeller, when it is made into electricity.

Interestingly enough, it you generate power from a wind device, and know the wind speed and have some clear idea of how much power you generate, you can make the same assumptions about alternator/gearing efficiency, and come up with the assumed efficiency of your wind device... But that's only included here for those brave and gallant souls who can absorb this much Information at one sitting...)

The solution for problem 1-1 is:

P = 0.0006137 AV³

P = 0.0006137 (40) (15)³

P = 82.8495 watts

 

The solution for problem 1-2 is:

P= (0.90) (0.50) (83) P = 37.35

 

Now, you may be wondering why we used 830 watts (power at the impeller) to derive our answer to problem #2, 373.5 watts (power in watts after conversion to electricity). The reason is simple. Even though our data, as presented, could give us an answer to several decimal places (828.495), in the real world of hammers and nails, we can never hope to achieve such accuracy in measuring wind speed, so there is no point in thinking of our answers as accurate. Besides, if your computer happens to look a great deal like a pencil, the figuring is easier. The answer we would have gotten if we had decided to pay attention to all those decimal points is 372.822, not a great deal different from our answer anyway.

The point is, these calculations are based on a great deal of estimation anyway. If they happen to give answers that agree with your actual power out (if you can measure that figure), then you've probably made a lucky guess (or six) somewhere along the line. Otherwise, don't worry too much about it. Estimate low on efficiency, and high on needs, then pray for wind.

SECTION C - WIND POWER EQUATION DERIVATION

This is the tough one. Hang on, 'cause we're going-for a ride.

Power, in any way it's measured, is spoken of as a force times a distance, divided by a period of time, or, for example,

P=

(F x D)
---------
T

or:

P =

(X feet) (Y pounds)
-----------------------
Z seconds

Now, we obviously can't multiply a foot times a pound and divide by a second in the real world. Such terms are only abstract conveniences, and it seems to be a misunderstanding of this distinction which confuses the practical but non-mathematical folks, and causes the physics boys to lose touch with the world of dirt, sow bugs and tree forts.

This distinction we make when we say 'red.' ' Red' is 'a color,' and we use it as a noun, or a word which represents an object, but we never have 'red' all by itself, we have a red sky, or a red box, but 'red' is only a convenience. It's the same way with measurements like 'foot-pounds.' They don't exist, except in connection with other things, such as a blue '54 Dodge, going 42.7 miles per hour (62.77 ft./sec). Important point, 'nuff said.

Wind power is measured in many different units, such as horsepower (550 ft.lb./sec., or 550 ft.lb. sec.^-1), watts (. 7367 ft.lb. sec.^-2), and so on, depending on the application for which you wish to use the wind power, and the units of measurement commonly available.

The basic wind-power equation, in fact the basic equation for estimating extractable power from any moving fluid mass (water being the other common 'moving fluid mass'), is:

Equation #1-1: P = 1/2 ρAV³E

where:

P - power

ρ = rho, the air density

A = area (as explained below)

V = velocity, and E = total efficiency

The equation for estimating power in any particular moving body, fluid or not, is:

1 2

Equation #1-2: P = 1/2 MV²

where:

P = power

M - mass, and

V = (once again) velocity

Now, to get from equation #1-1 to equation #1-2, we take each term in turn. M (mass) of the wind, equals density times volume

Equation #1-3: M = ρVol.

and the volume of air we are considering is found by multiplying the speed (V) of the wind, times the 'silhouette area', of the S-rotor (height times width),

Equation #1-4: Vol. = AV

since we need three dimensions to have a volume. In other words the 'silhouette area1 of the S-rotor gives us 2 dimensions, and the distance the wind travels per unit time (velocity) gives us the third dimension. Substituting into equation #3, we get:

Equation #1-5: M = ρ AV,

then plugging this expression back into equation #2, we get:

Equation #1-6: P = (ρAV) V²

or, even better,

P = 1/2 ρAV³

This gets us much closer. This equation describes the power in the wind (which we called A.T. wind energy), but it does not describe the power we can extract from the wind. In order to derive this, we need to multiply by T.M. wind energy (59.3%, or 0.593), and, as well, by the efficiency of the S-rotor, or whatever wind device we are studying. If we assume 20% efficiency for the S-rotor, then taking equation #1, and substituting, we find:

Equation #1-7: P = 1/2 ρAV³ (.593) (.20)

or,

P = 0.0593 ρAV³

But, what of P(rho), the air density constant? That doesn't show up as a variable in the final wind energy equstion therefore, we must have assumed that it was a constant (the same in all cases) and simply made it part of the number which appears in the equation.

In fact, that was our assumption. And we're in good company, since even the 'big boys' make similar assumptions. The reasons are several:

(1) with 'funky' equipment (translates: any wind-speed measuring device costing less than $400-$1000) the margin of error in wind-speed 'estimation' is bound to be more than the effect any change in the value of ρ is going to have. A 10% margin of error in V becomes a 30% margin of error in V³.

(2) While it is true that ρ has an effect on Indicated Air Speed (IAS), versus True Air Speed (TAS), the equations which fully cover the very synergistic effects of these and other

*Note: for a prop-type, the principle is similar depth of the blades is not considered. The area (A) of a prop-type is pi times the radius squared (A =pi R²) variables are complex and many. After you know enough to be able to design an airplane wing, then you can adjust for ρ .

(3) the term "exact science' is another abstraction. Einstein said that 'everything is relative,' and so a more exact term is 'relatively exact science.' Wind is invisible,as is thought. We measure the effect of wind, just as we see the effect of thought. But in neither case can we deal directly with the thing itself.

(4) wind is tremendously variable beyond the capacity of calculus to really 'describe.' It may vary in speed 10 MPH in 1/2 second, making a series of peaks and valleys on a recording anemometer (wind speed indicator). This makes it next to impossible to talk about 'wind speed, " even at any given moment we always deal with some kind of 'average' wind speed!

We could go on and on, but the point is made: accuracy is not a practical goal. Ballpark figures are good enough for government work, and they are good enough for us.

So, our value for ρ will be 0. 0024 slugs per cubic foot, or otherwise;

ρ = 0.00 24 ft.⁴sec.^-2*

Substituting this into equation #1-7, we find:

P = (0. 0593) (0. 0024) AV³, or, P= 0.0001423 AV³

But, all the values for our variables (P, A, and V) are also variable meaning, P is in watts, A is in square feet, and V is in miles per hour. In order to get everything working together, we have to multiply miles per hour by 1.46 to convert into feet-per-second (which gives us answers in foot-pounds per second) and then we have to multiply that by 1.36 to convert to watts. Back where the numbers dance and sing, that means:

P = (0.0001423) A (1.47V)³ (1.36)

Equation #1-8: P = 0.0006137AV³

(If you have a calculator, you may get a slightly different constant don't worry it's close enough.)

Now, as we said before, this equation assumes that our aeroturbine is 20% efficient. For translating this constant into other efficiencies, we need only multiply the constant 0.0006137 by that new efficiency, divided by the 20%, or, at 70% efficiency, our new constant is:

C = 70/20 (0.0006137), or,

C = 0.0021479,

Note: for you diehards, a table relating ^ to altitude is included Table # 2. We also explain why P is in slugs/ft.^, assuming you are interested, in the next section.

and substituting into equation #1-8, we get:

P = 0.00215 AV³,

at 70 % aeroturbine efficiency.

SECTION D----AIR DENSITY

The energy in the wind is directly related to both its mass and its velocity (see equation #1-2, earlier in this Data Sheet).

However, its mass depends upon its density times its volume. For the details of this, refer to Section A of this Data Sheet here we are only concerned with mass and air density.

Mass is not the same as weight. Most of us are not used to making this distinction, and so we are not generally aware that the same object may have a different weight depending on where it is, even at different spots on the earth's surface.

This may not at first make sense, but consider that while a mass of 6 pounds generally weighs (approximately) 6 pounds on the earth, it will only weigh approximately 1 pound on the moon, or 0 (zero) pounds in deep space. Yet, in all these cases, its mass remains constant, and also its inertia, which is the tendency of a body to remain at rest (or moving in a particular direction at a particular velocity) unless acted upon by an outside force.

A general physics' text will make this all much clearer if it isn't now, but those of you who read Science Fiction will be aware that a 5-ton girder in deep space can crush a man against a space station with its inertia alone, as surely as it would if it fell on someone on earth (and thus used its weight to accomplish the same end.)

So, inertia is the key. The air holds or gives energy relative to its inertia, which itself is relative to its mass. Therefore, air density is generally expressed in slugs (a measure of mass), per cubic foot. The metric system has a different measure of density, since it has a different measure of mass and distance. But the principle is the same.

'Density* is more correctly called 'mass density,' and it is found by weighing a certain amount of air (in this case a cubic foot), and dividing by the local value for the acceleration of gravity.

Gravity, in acting on mass, determines weight. If you have ever experienced the force of acceleration (as in an automobile) you will probably be aware that it 'feels' the same as gravity 'feels.' That is, it forces you in a direction opposite of the acceleration, as (in a car) against the car seats.

Gravity is indistinguishable, in its effects, from acceleration. So, it is measured in the same terms, that is, a distance per unit time (speed) per unit time. When we are traveling at a constant speed (say 10 feet per second) we are moving thru a distance in a certain time.

When we are accelerating, so that every second we are going 10 feet-per-second faster than the second before, then we say we are gaining a velocity of 10 feet-per-second every second, or 10 feet-per-second per second, or 10 ft. per sec.² , or 10 ft.sec.^-2 a measure of acceleration.

When we divide the weight of a cubic foot of air by the local value of the acceleration of gravity (which hovers around 32 ft. sec.^-2), we get (at 760 mm air pressure and 59.9°F.) a value of 0.002378 ft.^-4lb. sec.² , since we divide pounds per cubic ft. (lb. ft.^-3) by an acceleration in feet per sec. (ft. sec.²), or (ft.^-2lb.)

 

(ft.^-3lb.)
-----------
ft. sec.^-2

or

(ft.^-3) (ft.^-1sec.⁺²)

or

ft.^-4 lb. sec.²

When we put these values into our equation for wind, we find that:

where:

P = CρAV³

C is our constant (accounts for efficiency and conversion factors)

ρ is measured in lb. ft. ^-4sec.²

A is measured in ft.²

V is measured in ft. /sec.

When we plug these kinds of units into the equation, this gives us:

P = (ft. ^-4lb. sec.²) (ft.²) (ft. sec. ^-1)³

P = (ft.^-4lb. sec.²) (ft.²) (ft.3sec.^-3)

P = ft.lb. sec.^-1, or

P - foot-pounds per second.

is affected by elevation in feet [(ρ a) (0. 0024)] according to the table below:

TABLE # 2

ELEVATION IN FEET	ρ a	ELEVATION IN FEET	ρ a
0	1.000	5500	.849
500	.985	6000	.840
1000	.971	6500	.823
1500	.957	7000	.811
2000	.943	7500	.798
2500	.929	8000	.786
3000	.915	8500	.774
3500	.902	9000	.762
4000	.888	9500	.750
4500	.875	10,000	.739
5000	.862

SECTION E: ESTIMATING WIND SPEED

As we have already pointed out, wind is almost never directly measured, rather, it is averaged over greater or lesser periods of time, and, as well, the amount of estimation involved is greater or lesser.

It is recommended that you buy some kind of wind-measuring device (anemometer). There ;i several good but more or less inexpensive anemometers made by Dywer see Data Sheet Three (Sources) in this chapter.

If you don't have access to an anemometer of any kind, Table # 3, which follows, will allow you some kind of wind-speed estimation by phenomena:

TABLE # 3

SPEED RANGE	PHONOMENA	DESCRIPTION
0	Smoke rises vertically	Calm
1-3	Direction of wind shown by drift of smoke, but not by wind vane.	Light air
4-7	Wind felt in the face; leaves of trees rustle; wind vane moves easily	Light breeze
8-12	Leaves and small twigs in a constant motion; wind extends a light flag.	Gentle breeze
13-18	Raises dust and loose paper; small branches are moved.	Moderate breeze
19-24	Small trees in leaf begin to sway; crested wavelets form on lakes and ponds.	Fresh breeze
25-31	Large branches in motion; telegraph wires whistle; difficult to use an umbrella	Strong breeze
32-38	Whole trees move in wind; walking difficult	Moderate gale
39-46	Breaks twigs and branches off trees Generally impedes progress	Fresh gale
47-54	Outdoor flower pots and house tiles or slates are removed	Strong gale

By the way, if you should chance to build an anemometer (or buy one), which needs calibration, an excellent method of doing this was described in an article in 'The Amateur Scientist* section of Scientific American for October, 1971.

SECTION F: WIND AND ALTITUDE

Wind, as a fluid medium, tends to be slowed down by obstacles near the ground, as well as the ground itself. In general, it can be said that wind at some altitude is relative to wind at 10 feet by the formula developed by Hellman:

Equation #7-1: V_h = V₁₀ (0. 2337 + 0. 656 log₁₀ (h + 4. 75))

where:

V_h = velocity at height h (MPH)

V₁₀= velocity at 10 feet (MPH)

h = height of desired estimation, in meters (log₁₀ means 'log to the base 10')

 

For example, at 200 meters, when V₁₀ = 10 MPH,

 

V₂₀₀ = 10 MPH (0.2337 + 0.656 log₁₀ (200 + 4.75)) V20o

V₂₀₀ = 10 MPH (0.2337 + 0.656 (2.3096)) V200

V₂₀₀ = 17.5 MPH

 

(If you don't understand logarithms, consult any mathematical reference.)

But, actually, that's rather complex and not as useful as it might be, since it assumes that there are no obstacles to be overcome (or to get above) except the ground.

Best advice is get as high as you can, keep your bowels clean, and trust in the Lord. Towers are mostly a matter of budget, anyway.

SECTION G: - GENERAL POWER ESTIMATES

This section centers around a table which gives windspeed (V in column A), its cube (V^ in column B), the impeller watts per-equare-foot at each windspeed (Watts/ft. ^ column C), and the watts per-square-foot of impeller when translated into electricity at the alternator (Watts/ft. column D assumes 30% efficiency of transmission & translation).

The table allows anyone to calculate power at the impeller and alternator in watts, without using equation #1-8. The new, much simpler equation, is:

Equation #7-2: P = AC_v

where:

P= power, in watts, at the impeller

A - area of your aeroturbine

C_v = reading in column C for the wind speed in question

For example, at 34 MPH, an aeroturbine of 40 square feet will develop:

P = (40) (24), or

P - 960 watts at the impeller

Using equation #1-8, the power would be:

P = 0.0006137 (40) <34)³, or,

P= 964.835

Not a great deal of difference. These figures assume 20% aeroturbine efficiency, and can be translated to other efficiencies by multiplying by the ratio of the new efficiency to 20, or:

P=E_n/20 AC_v

where

E_n = new aeroturbine efficiency

(Note: Power in watts out of the alternator is found by substituting D_v for C_v in equation #7-2. This gives power in watts of electricity, assuming 30% efficiency of gearing/alternator.)

At 50% aeroturbine efficiency, E_n/20 = 50/20, or 2.5

 

TABLE #4

A	B	C	D	A	B	C	D
V	V3	Watts ft.*	Watts ft.2	V	V3	Watts	Watts ft.*"
						ft.2
6	216	0.13	0.044	30	27000	17.0	5.5
8	512	0.31	0.10	32	32768	20	6.6
10	1000	0.61	0.20	34	39304	24	8.0
12	1728	1.1	0.35	36	46656	29	9.4
14	2744	1.7	0.56	38	54872	34	11
16	4096	2.5	0.83	40	64000	39	13
18	5832	3.6	1.2	42	74088	45	15
20	8000	4.9	1.6	44	85184	52	17
22	10648	6.5	2.2	50	125002	77	25
24	13824	8.5	2.8	55	166375	100	34
26	17576	11.0	3.6	60	216000	130	44
28	21952	13.0	4.4	65	274625	170	56
				70	343000	210	69

SECTION H: - FINDING AVERAGE WINDSPEED

'Average wind speed' refers to the speed of all winds averaged out for the time of measurement. If a weather station has continuous monitoring of windspeed, then the 'average wind-speed1 for that day for that station would be the average of all windspeeds recorded for that day; or, if they are measured at the rate of one a minute, then 'average windspeed' for that day for that station would be all those recorded speeds divided by 1440 (the number of minutes in 24 hours).

However, it is expensive to record wind continuously, and so a great many weather stations only record at 7:30 AM, 1:30 PM, 7:30 PM, and 1:30 AM Washington, D.C., time. These readings are taken for only a few minutes, and thus the 'average wind speed* for that station for that day may not be very representative of 'true' average wind speed.

These considerations are important since the weather station nearest you may be your best source of information on the 'average wind speed* for your area. But, these records may be inaccurate for other reasons as well.

If the wind speed is taken at ground level, this will not be the same 'average wind speed' as one which is averaged from readings taken at 50 feet in the air. Wind speed reading will vary because different instruments are used in different places, because there are obstacles in the way (trees, etc.), because there is no standard wind-speed calibration, etc., etc.

Further, wind varies with the site, and may average significantly more at one particular spot on a piece of property than it averages at another.

Once again, the only thing all these words mean is do it if you feel it. That's nearly as good an indicator as any, because you're bound to be dealing with a certain amount of 'seat-of-the-pants' estimation, anyway.

Find out what 'average wind speed' for the nearest weather station is, and find out what that phrase means to them where and how they take their readings, etc. then figure out a way to guess or 'know' if your average is higher or lower.

Below 8 MPH average wind speed, consider 3 times carefully before investing much money in a wind/electric system. From 8-10 MPH average, look twice; from 10-12 MPH average, move on it; and from 15-20 MPH average, why don't you already have a wind/electric system?

SECTION I: - USABLE AVERAGE ENERGY

The extremely vague 'exact science' of wind finds its pinnacle here. The usable average energy in your wind system will vary according to the average wind speed and patterns, the time of year, the size and efficiency of your aeroturbine and alternator, the amount of your storage capacity and its efficiency, and the variables of your electric transfer system and its appliances.

If we make a half-dozen assumptions, and cross our fingers, we could say that it might be that perhaps the figures below would be realized, maybe.

This table assumes that aeroturbine efficiency is 20% (of T.M.), negligible output below 10 MPH, 25 MPH rated windspeed, alternator puts out rated amps at 14V, alternator/gearing is 50% efficient over the whole range of wind speeds, and, further, that all factors are 'balanced ' batteries are 'correct' in number and capacity, etc.

Once again even though the table may look impressively accurate these are only ballpark estimations. The efficiency of the alternator (assumed 50%) might well be lower at lower wind speeds, and higher at higher wind speeds.

To convert these average usable power figure estimates into power figure estimates for more efficient units, multiply by the new efficiency divided by 20. For instance, for a 50% efficient aeroturbine (all other parameters as described above) and a nominal amperage rating of 80, with a 10 MPH average wind speed for a certain month, we find:

est. KWH
------------
month

= (50/20) 67

or

est. KWH
------------
month

= 167.5

 

( NOTE: This is a 'questionable' use of Table # 5. In essence, in order to achieve this higher accumulated power out, the aeroturbine would have to use an 'altered' alternator - see Chapter 2, sec. A- or a larger alternator.)

The same conversion expression can be used to estimate a new area for a more efficient aeroturbine. For instance, if we have an aeroturbine of 70% efficiency (all other parameters as described), and we live in an area which experiences 12 MPH average wind speed for the month which concerns us, and if our aeroturbine should produce about 100 KWH for that month with a 90A alternator, then reading the chart, we find that at 20% efficiency, our aeroturbine should be 263 ft. 2. Instead of multiplying by the ratio of new efficiency to 20, we divide:

A_{70 =}

263 ft.²
-------------
(70/20)

or

A_{70 =}75 ft.²

 

Thus the table can provide us estimates for all different situations.

TABLE #5

AMPERAGE RATING OF ALTERNATOR	AREA OF S-ROTOR (FT2)	MONTHLY AVERAGE WIND SPEED
		6	8	10	12	14	16
30	88	8	17	25	33	41	49
35	102	10	20	30	39	48	57
40	117	11	22	33	44	55	65
45	131	12	25	38	50	62	74
50	146	14	28	42	55	69	82
55	160	15	31	46	60	76	90
60	175	16	34	50	66	83	98
65	190	18	36	54	71	90	106
70	207	19	39	58	77	97	115
75	219	20	42	63	83	104	123
80	234	22	45	67	88	110	131
90	263	24	50	75	99	124	147
100	292	27	56	83	110	138	164
110	321	30	61	92	121	152	180
120	351	33	67	100	132	166	196
130	380	35	73	108	143	180	212

SECTION J: - KWH/MONTH NEEDED

If you are familiar with the simple electrical formulas used in Chapter 2, this should be duck soup:

KWH=

(W₁H₁) + (W₂H₂) + (W_nH_n)
-----------------------------
1000

where:

KWH = Kilowatt hours per month

W₁ = Wattage of appliance one

H₁ - Number of hours per month appliance one is used

W₂= Wattage of appliance two

H₂ = Number of hours per month appliance two is used

W_n = Wattage of appliance 'n'

H_n = The number of hours appliance 'n' is Used per month

To use the formula, go around your house and get the wattage off of each electric appliance, light bulb, etc. anything which uses electricity and, as well, estimate the number of hours that appliance is used per month. Wattage or amperage is usually stated somewhere on the appliance. For instance, you may have a reading light of 60 watts that you turn on in the evening for 1/2 hour.

This would be (30) (0.5) or 15 hours per month use, at 60 watts. To take a more complex example, here at Earthmind we have a toaster which we use in the morning to make toast. We don't use it every morning, but for the purposes of illustration, let's say we do.

Further, let's assume we make 4 pieces of toast per morning. Since the toaster takes 2 slices of bread at once, this means the toaster is punched down twice, cycles, and pops up. This takes 2.5 minutes (I timed it), and so that's 5 minutes per morning for two cycles of the toaster and 4 pieces of toast. 5 minutes is l/12th of an hour, (use each day), so we use the toaster \Ts) 2*5 kurs Per month. (By the way, a toaster is an 'I R1 machine.)

On the bottom of the toaster it says '110-120 volts, 7.7 amps.' This is (120 x 7.7) 924 watts. At 2.5 hours per month, that's (924 x 2.5) 2310 watt-hours per month. Dividing by 1000 to get KWH per month, we get 2.3 KWH per month.

The reading light we mentioned (15 hours per month at 60 watts) equals 900 watt-hours per month, or 0.9 KWH per month. If our household had only a toaster and a reading light, that would be (2.3+0.9) 3.2 KWH per month total for the household.

Look in Chapter 5 for some alternatives to energy-inefficient appliances.

SECTION K: - MEASURING AEROTURBINE EFFICIENCY

Because of the estimates and the number of simultaneous variables involved, this is not an easy or accurate measurement. The basic equation used is derived from information in Section C of this Data Sheet, and is:

Equation #7-3: A.Eff. =

(EI) Ae
--------------
0.000708AV³

where:

A.Eff. - aeroturbine efficiency

E = voltage out of the alternator

I - amperage out of the alternator

Ae = alternator/gearing efficiency

A - aeroturbine area

V = wind speed

If you've read and understood everything which precedes this you're a real trooper! But, as well, if you've read and understood it, you know that the errors in estimation involved here will easily be great enough to give you errors on the order of a factor of ten.

For any given interval or moment, in order to use this equation, you must know (simultaneously; Voltage out of the alternator Amperage out of the alternator Wind speed, and Alternator efficiency

Since alternator efficiency varies widely at its lower RPM's, and only begins to level out at (say) 12-15 hundred RPM's, try to take readings at a wind speed which gives your alternator RPM's above this point.

This RPM/wind speed relationship will depend on your particular setup, but the best day to test in any case is a day of strong steady wind (which will give higher aeroturbine, and thus higher alternator RPM's).

Either get a manufacturer's data sheet on your alternator, or test it for output at a different RPM with a known load, so that you have some idea of alternator efficiency. Otherwise, assume 50% alternator efficiency, and 95% gearing/transfer efficiency (or make other assumptions).

A further simplification would be to assume a constant voltage: With a regulator hooked to the alternator, voltage will be below 14V. If you are generating enough juice so that the regulator begins to cut in and out (it makes a clicking noise) then you 'know' that voltage is approximately equal to 14. In terms of our equation (#7-3), that means:

A.Eff =

(14) (I) (.50) (.95)
-----------------------
0.000708 AV³

A.Eff =

(9390) (I)
------------
AV³

In this case the only simultaneous variables to be considered are V (wind speed) and I (amps). Area is of course a constant, dependingon your unit.

It is advisable to have an over-ride positive shut-down circuit on the aeroturbine, as described in Chapter 4 - Control, in the section describing upper wind speed determination.

Circuits which will allow you to read volts and amps are described in Chapter 4, Section G, Control Circuits.

A problem might be experienced with the 'loading effect' of the alternator, if you are using manual 'cut-in,' and trying to measure output at the lower wind speeds. The aeroturbine might put out a false high reading for a few moments until its excess inertia is converted to electricity, and then 'balances' so that wind power in'equals'electric power out. Best, again, is a day of strong steady winds, but gusts, if they are prolonged (some gusts last less than 1/2 second) for 4-5 seconds, will do, assuming your aeroturbine is responsive enough to these gusts.

Take a number of sets of measurements, find Eff. for each set of measurements, and average these numbers, if some of the answers read far too high or low (as they might with 'funky' instruments and 6 or 8 people scrambling around) throw them out and average the others.

By the way, it will take 6 or 8 people (almost) to read all the instruments (simultaneously!) and write down the proper figures. You might make up sheets with 20 or 30 numbered spaces, and have a 'coordinator' call "Mark!" and the number of the reading ('mark twelve!. . . . mark 13!. . .'), at which time, and in which space each person writes his/her reading.

Gather sheets, do the calculations, and

Good luck

SECTION L: - CUBE/AVERAGE

The cube of an average is always less than the average of the cubes.

Consider an 8 MPH average windspeed. This 'average' reading could occur if the wind blew 4 MPH, for 6 hours, 12 MPH for 6 hours, 6 MPH for 6 hours, and 10 MPH for 6 hours, making a total of 24 hours.

The average windspeed for that (24-hour) day (8 MPH), cubed, is 512. The average of their cubes, (respectively 64, 1728, 216, and 1000), is 752. This is (almost) a 150% increase. In the real world, this percentage difference tends to be even greater.