Not without knowing something about the circuit. For example if it's a simple resistive load then it would be:
\$(12V \ \div 18V)^{2} \times 4.5W = 2W\$
\$2W \div 12V = 0.1666A\$
Or
\$ 4.5W \div 18V = 0.25A \$
\$ 18V \div 0.25A = 72\Omega\$
\$ 12V \div 72\Omega = 0.1666A\$
\$ 12V \times 0.1666A = 2W \$
However, it's probably going to be more complex than the above. Depending on the circuit, it may be non-linear, draw the same or even more current at the lower voltage.
For another example take a switching regulator, in your example say you have a 100 percent efficient switcher (don't exist, but for easy calculations) delivering 5V (from the 18V input) to the rest of the circuit. 4.5W at 5V is 900mA. So the 18V at 250mA is turning into 5V at 900mA.
If we drop the input voltage to 12V, then to supply the required 4.5W the switcher will automatically draw more current at it's input in order to deliver the 4.5W at it's output. So 4.5W / 12V = 0.375A. The input voltage has dropped, but the current has actually increased. This can be seen as a negative impedance at the switcher input.
With more details about the circuit a reasonable estimation could be arrived at, but I think the easiest way would be simply to run it from 12V and measure the current with a multimeter.
Solution Summary:
The term Crest and Peak seem to be used interchangeably both by learned bodies [tm] such as ASTM and IEEE and by some Holiday Meter tester manufacturers. While it may seem obvious that "crest" and "peak" are synonyms, the discussions re waveform and risetimes make this not 100% certain. But ...
Some HM test manufacturers actually state that they perform Crest voltage testing.
If there is a difference it is subtle, but even if it is not solely a matter of terminology, it is clear from the literature available that use of a crest detecting meter will achieve your aims when used consistently.
For the inquiring minds that want to know [tm].
According to ASTM (who should know)
A "holiday" is a small fault or pinhole that permits current
drainage through protective coatings on steel pipe or polymeric
precoated corrugated steel pipe.
A "holiday detector" is a highly sensitive electrical device
designed to locate holidays such as pinholes, voids, and thin
spots in the coating, not easily seen by the naked eye. These are
used on the coatings of relatively high-electrical resistance
when such coatings are applied to the surface of materials of
low-electrical resistance, such as steel pipe.
This is based on what the net tells me:
This document Standard Test Methods for Holiday Detection in Pipeline Coatings
- ASTM, Designation: G 62 – 87 (Reapproved 1998) provides a brief (3 page) but succinct description of the procedures to be carried out for holiday testing pipeline coatings. You can expect some procedural differences between pipes and general surfaces but the same principles apply.
Also 1990 version
The document os old but what it says seems to be basically unchanged and knowing it exists should allow nwer versions to be located.
The document uses the terms peak and crest 3 times, each, always together and always interchangeably. viz
6.3 Peak or Crest Reading Voltmeter—A kilovoltmeter
capable of detecting a single pulse and holding it long enough
for the meter circuits to indicate.
footnote 4: The sole source of supply of a suitable peak or crest reading voltmeter known to the committee at this time is Itt-Jenning ...
9.1 The instruments shall be standardized with respect to
voltage output in accordance with the manufacturer’s instructions, using a peak or crest reading voltmeter. This is used more
commonly with Method B where voltage may vary from test to
test but can also be used for verification of the voltage on a
Method A test.
This SPY Jeepmeter Holiday Meter tester document says
- The SPY JeepMeter series of Crest voltmeters are instruments
designed to provide accurate and reliable measuring of the crest
value of high voltage non-sinusoidal waveforms. The output
waveform of holiday detectors typically has this type of shape and
therefore prevents the use of conventional RMS voltmeters
This interesting IEEE document
IEEE Standard Requirements for
Instrument Transformer uses the terms Crest and Peak interchangeably - sometimes in the same paragraph. eg page 74
- a) Measure the crest open-circuit secondary voltage, V1 [see part a) of Figure 31], using a
high-impedance crest reading voltmeter, oscilloscope, or calibrated gap. Increase the primary
current gradually from zero to the maximum continuous-current rating or until the crest voltage
reaches 3500 V, whichever occurs first. Maintain the primary current for 1 min, and record the magnitude of the peak voltage. If 3500 V crest is not exceeded by this test, then the information in item b) should be followed
Page 44 - 45 here is very relevant:
EVALUATION OF CIVIL WORKS METAL STRUCTURES
Jeep Holiday meter tester
Several HMs
More
HM testers - Jeep
If there IS a difference, which is not certain given the above, it's not obvious what "crest meter" is meant to imply in the current context. Can you provide instrument brand and model. The term "crest" is notably absent from discussions of holiday meter testers by people such as jeep, who use the term "peak", but this does not mean that they are or aren't synonymous.
From a quick skim through the available internet material I'd say that consistency of test method is as or more important than instrument used. It should be easy enough to trial your crest meter in a range of situations and see if its reading seems to reflect your expectations.
Best Answer
Please fasten your seatbelts - our flight today will fly over a bit of the math to get from power (dBm) to field strength (V/m).
The most pragmatic way to think about the "strength" of any RF signal is in terms of how much power it can deliver to the terminals of a practical receiving antenna.
Ultimately the laws of physics dictate that for a given receiving antenna, at a given frequency, there is a simple linear relationship between the square root of the power at the receive antenna terminals and the strength (in Volts/meter) of the RF field:
\$Field\ Strength\ = k\cdot\sqrt {P_{rx}}\ \ \ \ \cdots (1)\$
You can think of the factor \$k\$ as the "conversion factor" of that antenna at that specific frequency.
That's it - once you know your antenna conversion factor \$k\$ and the received power (in Watt), you simply calculate the field strength using the above formula.
So, to answer your questions: 1. How does it work? Primarily due to the inverse square law of radiators and aperture. 2. Yes, it's possible. 3. Once you know k for your antenna, use Equation (1) above to calculate RF field strength - see example below.
Example:
The \$k\$ factor for an ideal half-wave dipole, which has a gain of 1.643 (2.14dB) can be found in literature - \$k\$ is essentially the square root of (\$120\pi\ \div \$Antenna Effective Area).
\$k = \sqrt{120\pi} \cdot \sqrt{\frac{4\pi }{1.643\cdot\lambda^2}} \approx\frac{53.7}{\lambda}\$
Let's say we are receiving 0dBm (.001W) at the terminals of our half-wave dipole at 100 MHz. (\$\lambda\$=3m) Using Eq. (1) we see
Field Strength = \$\frac{53.7}{\lambda} \cdot \sqrt{.001\ }\$ = 0.566 V/m
TLDR: Convert dBm to Watt, get square root, multiply by k.