Calculator formulas
Every formula behind the LowVoltKit calculators, on one sheet.
How to use this guide
Each section matches a calculator and ends with a worked example you can check against it.
| Symbol | Meaning | Unit |
|---|---|---|
| V | Voltage | volts (V) |
| I | Current | amps (A) |
| R | Resistance | ohms (Ω) |
| P | Power | watts (W) |
| L | One-way cable length | feet (ft) unless noted |
| R/kft | Conductor resistance per 1,000 ft | Ω |
| n | Conductors paralleled per leg | count |
| f | Lens focal length, or frequency | mm, or Hz |
Field estimate only. Manufacturer data, the authority having jurisdiction (AHJ), and current applicable code govern material and whether a circuit is energized. See the disclaimer.
Ohm's law and power
Every power calculation in this guide reduces to these four relationships.
V = I × R I = V ÷ R R = V ÷ I
P = V × I P = I2 × R P = V2 ÷ R I = P ÷ V
| Convert | Formula |
|---|---|
| mA to A | mA ÷ 1,000 |
| W to A | W ÷ system V |
| Series resistance | R1 + R2 + R3 … |
| Parallel, identical conductors | R ÷ n |
| Batteries in series | Voltage adds, Ah stays the same |
| Batteries in parallel | Ah adds, voltage stays the same |
Voltage drop
Drop is the round trip: two legs of the same gauge, so the one-way length is doubled.
Vdrop = (2 × L × I × (R/kft)) ÷ (1000 × n)
Vdevice = Vsupply reading − Vdrop % drop = (Vdrop ÷ Vsupply) × 100
Pass window. Built from the nominal rating, not the meter reading: Vnominal × (1 ± tolerance). At ±10%, 12 VDC is 10.8 to 13.2 V and 24 VDC is 21.6 to 26.4 V.
Room left.
Lmax = ((Vsupply − Vmin) × 1000 × n) ÷ (2 × I × (R/kft))
Imax = ((Vsupply − Vmin) × 1000 × n) ÷ (2 × L × (R/kft))
Corrections to the table value.
RT = R20 × [1 + 0.00393 × (T°C − 20)]
- Stranded class B: table value × 1.02. Fine-strand or tinned: × 1.08.
- Paralleling spare conductors divides loop resistance by n (doubling up in 18/4 halves it).
- AC at these gauges and lengths is treated as purely resistive.
- Measured drop above calculated: implied loop resistance = measured drop ÷ I. Extra resistance points to a bad splice, damaged conductor, CCA or aluminum, or a longer run than entered.
| AWG | Solid copper Ω/1,000 ft at 20 °C | Stranded class B (× 1.02) |
|---|---|---|
| 24 | 25.67 | 26.18 |
| 22 | 16.14 | 16.46 |
| 20 | 10.15 | 10.35 |
| 18 | 6.385 | 6.513 |
| 16 | 4.016 | 4.096 |
| 14 | 2.525 | 2.576 |
| 12 | 1.588 | 1.620 |
| Location | Temperature |
|---|---|
| Conditioned space | 68 °F / 20 °C |
| Attic or rooftop | 120 °F / 49 °C |
| Outdoor cold | 0 °F / −18 °C |
Worked example. 24 VDC supply, 0.5 A maglock, 150 ft of 18 AWG stranded: 2 × 150 × 0.5 × 6.513 ÷ 1,000 = 0.98 V drop, 23.02 V at the device. That is inside 21.6 to 26.4 V. Maximum length is 2.4 × 1,000 ÷ (2 × 0.5 × 6.513) = 368.5 ft.
Low voltage load and supply sizing
Size the supply to peak current times a headroom factor, one supply and one voltage at a time.
Irow = qty × draw drawA = mA ÷ 1000 or W ÷ Vsystem
Ipeak = Σ Icontinuous + Σ Imomentary Idemand = Ipeak × 1.30
Loading % = (Idemand ÷ Isupply rating) × 100 Ptotal = I × Vsystem
Icharging = (Ah × 1.2) ÷ recharge hours
- Continuous: maglocks, readers, controllers, anything energized while secure.
- Momentary: fail-secure strikes and other devices energized only while released.
- The 1.30 headroom factor is an allowance for charging and growth, not a code requirement. Charging current is a separate line and is not added into demand a second time.
Worked example. Four maglocks at 500 mA = 2.0 A continuous. 2.0 × 1.30 = 2.6 A demand. A 3 A supply carries it at 87% loading; a 2.5 A supply does not.
AC to DC conversion
A meter reads RMS, a filter capacitor charges to the peak, and each diode in the current path costs a fixed drop.
Vpeak = VRMS × 1.4142 VRMS = Vpeak × 0.7071
VDC, filtered = Vpeak − (Ndiodes × Vd)
VDC, unfiltered full wave = 0.6366 × Vpeak − (Ndiodes × Vd)
VDC, half wave = 0.3183 × Vpeak − Vd
VRMS required = (VDC required + (Ndiodes × Vd)) ÷ 1.4142
Vunloaded secondary = VRMS × (1 + regulation %)
Vripple, p-p ≈ Iload ÷ (fripple × C)
| Item | Value |
|---|---|
| Diodes in path, bridge | 2 |
| Diodes in path, center-tap or single diode | 1 |
| Silicon diode drop | 0.7 V |
| Schottky diode drop | 0.3 V |
| Ripple frequency, full wave | 2 × line (120 Hz on 60 Hz) |
| Ripple frequency, half wave | line (60 Hz) |
Ripple valley = VDC − Vripple. A device that needs a minimum DC can fall out of spec at the valley even when the average looks fine.
Worked example. 24 VAC through a silicon bridge and a filter capacitor: 24 × 1.4142 = 33.94 V peak, minus 1.4 V of diode drop = 32.54 VDC with no load. At 10% regulation the unloaded secondary is 26.4 VAC, which reads about 35.9 VDC.
Backup battery and UPS sizing
A panel battery is sized in amp-hours at the 20-hour rating; rack equipment is sized in watts and watt-hours.
Panel battery (12 or 24 VDC)
Ahrequired = (Iload × hours) ÷ rate factor Ahdesign = Ahrequired × 1.25
Icharging available = Isupply rating − Iload
recharge hours = (Ahinstalled × 1.2) ÷ Icharging available
runtime hours = (Ahinstalled × rate factor) ÷ Iload
| Discharge time | Rate factor (fraction of 20-hour Ah) |
|---|---|
| 20 h | 1.00 |
| 1 h | 0.60 |
Other durations: use the battery datasheet's discharge table. Capacity also falls with age and cold.
- Batteries in series add voltage, not capacity. Two 12 V 7 Ah batteries are a 24 V 7 Ah bank.
- A 7 Ah battery is 7 Ah over 20 hours and closer to 4 Ah over one hour.
- If available charging current is zero or negative, the supply cannot recharge the battery under load.
Equipment UPS
VA = (Wload × headroom) ÷ power factor
Whbattery = (Wload × hours) ÷ (efficiency × DoD × derate)
Whusable = Wpublished × hourspublished runtime ≈ Whusable ÷ Wload
Inverter efficiency defaults to 0.85. The chart-point estimate is a straight line; runtime falls faster than linear at higher loads, so the manufacturer's runtime chart governs.
Worked example. 24 VDC panel at 3.75 A for 1 hour: 3.75 × 1 ÷ 0.60 = 6.25 Ah, × 1.25 = 7.81 Ah. An 8 Ah battery covers it; a 7 Ah does not. On a 4 A supply, 0.25 A is left for charging, so a 7 Ah battery takes 7 × 1.2 ÷ 0.25 = 33.6 hours to recharge.
Video bandwidth
Start from the H.264 table value at 30 fps, then scale for frame rate, codec, and smart codec.
Mbpscamera = MbpsH.264 @ 30 fps × (fps ÷ 30)0.75 × codec factor × smart factor
Mbpsgroup = Mbpscamera × lenses × qty Mbpslink = Σ Mbps × (1 + headroom)
MbpsMJPEG = (KBframe × fps × 8) ÷ 1000
GB/hour = (Mbps × 3600) ÷ (8 × 1000) GB/day = (Mbps × 86400) ÷ (8 × 1000)
| Factor | Value |
|---|---|
| H.264 | 1.00 |
| H.265 | 0.55 |
| Smart codec / dynamic GOP | × 0.60 |
| 1080p, medium activity, H.264, 30 fps | 4.0 Mbps |
| MJPEG 1080p, medium quality | 130 KB per frame |
| Network headroom | 20%, 25%, or 30% |
| Link | Cable for a 100 m channel |
|---|---|
| 100 Mbps, 1 Gbps | Cat5e or better |
| 2.5 Gbps | Cat5e or better |
| 10 Gbps | Cat6A (Cat6 only to about 37 to 55 m) |
Utilization = traffic with headroom ÷ link speed. Figure recording (cameras to recorder), local viewing, and remote viewing (WAN upload) as three separate segments.
Worked example. 20 cameras at 1080p, 30 fps, H.264, medium activity: 20 × 4.0 = 80 Mbps, or 100 Mbps with 25% headroom. That is 80 × 86,400 ÷ 8,000 = 864 GB per day. On H.265 it is 80 × 0.55 = 44 Mbps.
Video storage and RAID
Storage is daily volume times retention, plus overhead, figured per camera group and summed.
GB/day = ((Mbps × 86400) ÷ (8 × 1000)) × recording share
TBraw = Σ (GB/day × retention days) ÷ 1000 TBusable required = TBraw × (1 + overhead)
TiB = TB × (10004 ÷ 10244) = TB × 0.9095
retention delivered = (TBusable array ÷ (1 + overhead)) ÷ (TB/day)
| Recording mode | Recording share |
|---|---|
| Continuous | 1.00 |
| Motion | percent of the day writing (40% = 0.40) |
| Schedule | (hours per day ÷ 24) × (days per week ÷ 7) |
| RAID | Usable capacity (n drives of size d) | Minimum drives |
|---|---|---|
| RAID 0 | n × d | 2 |
| RAID 1 | d (mirrored pair) | 2 |
| RAID 5 | (n − 1) × d | 3 |
| RAID 6 | (n − 2) × d | 4 |
| RAID 10 | (n ÷ 2) × d | 4 |
A hot spare adds one physical drive that holds no data. Default overhead for file system and VMS database is 10%. Drives are sold in decimal TB; the operating system reports TiB.
Worked example. 864 GB per day for 30 days = 25.92 TB raw. With 10% overhead, 28.5 TB usable required (25.9 TiB). Six 8 TB drives in RAID 6 give (6 − 2) × 8 = 32 TB usable, which holds 32 ÷ 1.1 ÷ 0.864 = 33.7 days.
Field of view and DORI
Pixel density sets what a camera can do at a distance; the same pixels spread over a wider scene as distance grows.
HFOV = 2 × arctan(sensor width ÷ (2 × f))
VFOV = 2 × arctan(tan(HFOV ÷ 2) × 9 ÷ 16)
scene width = 2 × D × tan(HFOV ÷ 2) = (D × sensor width) ÷ f
scene height = scene width × 9 ÷ 16
px/ft = horizontal px ÷ scene widthft px/m = px/ft × 3.2808
Dlevel = horizontal px ÷ (thresholdpx/ft × 2 × tan(HFOV ÷ 2))
fneeded = (threshold × sensor width × D) ÷ horizontal px
Use the 9/16 ratio for 16:9; use 3/4 for 4:3. With a datasheet angle, use it directly in place of the arctan calculation. Default sensor width is 5.37 mm (typical 1/2.8 in).
| IEC 62676-4 level | px/m | px/ft | Meaning |
|---|---|---|---|
| Detect | 25 | 7.6 | Something is there |
| Observe | 63 | 19.2 | Can follow what it is doing |
| Recognise | 125 | 38.1 | Probably someone known |
| Identify | 250 | 76.2 | That is the person |
Mounting geometry (flat, level ground; h = mounting height):
near edge = h ÷ tan(tilt + VFOV ÷ 2)
center = h ÷ tan(tilt)
far edge = h ÷ tan(tilt − VFOV ÷ 2)
The far edge is unbounded when tilt − VFOV/2 is at or above horizontal. License plates are not a DORI level; they need far more density plus correct lighting, angle, and shutter.
Worked example. 4 mm lens, 5.37 mm sensor, 1920 px: HFOV = 67.7°. At 30 ft the scene is 30 × 5.37 ÷ 4 = 40.3 ft wide, 1920 ÷ 40.3 = 47.7 px/ft (Recognise). Identify reaches 1920 ÷ (76.2 × 1.3425) = 18.8 ft; Detect reaches 187.7 ft.
Conduit fill
Fill is total cable area over conduit internal area, held to the NEC Chapter 9 Table 1 limit for the cable count.
Acable = π × (OD ÷ 2)2 = 0.7854 × OD2
fill % = (Σ (qty × Acable) ÷ Aconduit) × 100
cables remaining = ((Aconduit × limit) − Σ A) ÷ Acable
| Cables in conduit | NEC Ch. 9 Table 1 limit |
|---|---|
| 1 | 53% |
| 2 | 31% |
| 3 or more | 40% |
| Trade size | EMT total area (sq in) | EMT at 40% | PVC Sch 40 total area (sq in) | PVC Sch 40 at 40% |
|---|---|---|---|---|
| 1/2 | 0.304 | 0.122 | 0.285 | 0.114 |
| 3/4 | 0.533 | 0.213 | 0.508 | 0.203 |
| 1 | 0.864 | 0.346 | 0.832 | 0.333 |
| 1-1/4 | 1.496 | 0.598 | 1.453 | 0.581 |
| 1-1/2 | 2.036 | 0.814 | 1.986 | 0.794 |
| 2 | 3.356 | 1.342 | 3.291 | 1.316 |
| 2-1/2 | 5.858 | 2.343 | 4.695 | 1.878 |
| 3 | 8.846 | 3.538 | 7.268 | 2.907 |
| 3-1/2 | 11.545 | 4.618 | 9.737 | 3.895 |
| 4 | 14.753 | 5.901 | 12.554 | 5.022 |
Source: NEC Chapter 9 Table 4. Confirm against the edition your AHJ enforces.
- Jam ratio. Three same-size cables can wedge at a bend when conduit ID ÷ cable OD falls between 2.8 and 3.2.
- Total bends between pull points are commonly held to 360°.
- Use the top of the cable's published OD range unless the datasheet is in hand.
Worked example. 24 Cat6 at 0.250 in OD: 0.0491 sq in each, 1.178 sq in total. In 1-1/2 in EMT that is 1.178 ÷ 2.036 = 57.9%, over the 40% limit. In 2 in EMT it is 1.178 ÷ 3.356 = 35.1%, which passes.
Cable tray and J-hook capacity
Check area and weight separately; whichever reaches its limit first governs, and above a ceiling that is usually weight.
Area
Atray = width × loading depth AJ-hook = π × (opening ÷ 2)2
Aallowed = Apathway × design fill max cables = Aallowed ÷ Acable
bundle depth = Σ Acable ÷ width
Weight
lb/ft = Σ (qty × lb/kft) ÷ 1000 lb per hook = lb/ft × hook spacingft
% of tray rating = (lb/ft ÷ rated lb/ft) × 100 kg/km = lb/kft × 1.488
- Design fill is a design choice (25%, 40%, or 50%), following TIA and BICSI practice and the tray rating. NEC tray fill tables are for power conductors.
- Many installers cap a bundle near 24 cables and dress bundles side by side.
- Support spacing is commonly 4 to 5 ft, independent of the ceiling grid.
- J-hook catalog cable counts govern over the geometric figure.
Worked example. A 12 in × 4 in usable tray is 48 sq in. At 50% fill that allows 24 sq in, about 488 Cat6 at 0.0491 sq in each. At about 28 lb/kft that bundle is 13.7 lb/ft. On J-hooks at 4 ft, each hook carries about 55 lb.
Fiber link loss budget
Total loss is fiber length times attenuation, plus every mated pair and every splice.
lossdB = (Lkm × dB/km) + (pairs × dBpair) + (splices × dBsplice)
power budget = PTx − PRx sensitivity margin = power budget − lossdB ≥ 3 dB
km = m ÷ 1000 = ft ÷ 3280.84
| Fiber | Wavelength | Attenuation (dB/km) |
|---|---|---|
| Multimode OM3 / OM4 | 850 nm | 3.5 |
| Multimode OM3 / OM4 | 1300 nm | 1.5 |
| Single-mode OS2 | 1310 nm | 0.4 |
| Single-mode OS2 | 1550 nm | 0.3 |
| Component | TIA maximum (dB) | Typical field (dB) |
|---|---|---|
| Mated connector pair | 0.75 | 0.3 |
| Fusion splice | 0.3 | 0.1 |
| Mechanical splice | 0.3 | higher than fusion |
- Each patch panel the link passes through is one mated pair. A switch-to-switch link through two panels is two pairs.
- Certification is judged against the TIA maximum. Typical values show what a clean run often reads.
- Dirty end faces are the most common cause of a failed test. Clean and inspect before troubleshooting.
Worked example. 300 m of OM4 at 850 nm, two pairs, one fusion splice: (0.3 × 3.5) + (2 × 0.75) + 0.3 = 2.85 dB maximum. On typical values: 1.05 + 0.6 + 0.1 = 1.75 dB.
Subnet and CIDR (IPv4)
The prefix sets the block size; the first address is the network, the last is the broadcast, and everything between is usable.
total addresses = 2(32 − prefix) usable = total − 2
network = IP AND mask broadcast = network OR wildcard wildcard octet = 255 − mask octet
Exceptions: a /31 has 2 usable addresses on a point-to-point link (RFC 3021); a /32 is a single host.
| Prefix | Mask | Wildcard | Total | Usable |
|---|---|---|---|---|
| /22 | 255.255.252.0 | 0.0.3.255 | 1,024 | 1,022 |
| /23 | 255.255.254.0 | 0.0.1.255 | 512 | 510 |
| /24 | 255.255.255.0 | 0.0.0.255 | 256 | 254 |
| /25 | 255.255.255.128 | 0.0.0.127 | 128 | 126 |
| /26 | 255.255.255.192 | 0.0.0.63 | 64 | 62 |
| /27 | 255.255.255.224 | 0.0.0.31 | 32 | 30 |
| /28 | 255.255.255.240 | 0.0.0.15 | 16 | 14 |
| /29 | 255.255.255.248 | 0.0.0.7 | 8 | 6 |
| /30 | 255.255.255.252 | 0.0.0.3 | 4 | 2 |
| RFC 1918 private range | Block |
|---|---|
| 10.0.0.0 to 10.255.255.255 | 10.0.0.0/8 |
| 172.16.0.0 to 172.31.255.255 | 172.16.0.0/12 |
| 192.168.0.0 to 192.168.255.255 | 192.168.0.0/16 |
Keep static device addresses outside the DHCP pool, and confirm the plan with IT before addressing cameras or controllers.
Worked example. 192.168.1.0/27: mask 255.255.255.224, 25 = 32 addresses. Network 192.168.1.0, broadcast 192.168.1.31, usable 192.168.1.1 through 192.168.1.30 (30 hosts).
PoE budget and distance
The switch budget is spent at the class PSE figure, not the sum of device datasheets.
Wreserved = Σ (qty × WPSE class) Wremaining = Wbudget − Wreserved
headroom limit = Wbudget × 0.80 % of budget = (Wreserved ÷ Wbudget) × 100
Cable loss on one run
I = WPSE ÷ VPSE Rloop = (2 × Lft × (R/kft)) ÷ (1000 × c)
Wloss = I2 × Rloop Wdevice = WPSE − Wloss
c = conductors per polarity: 2 on two-pair PoE, 4 on four-pair (802.3bt). VPSE = 44 V for Type 1, 50 V for Type 2 and above. Copper: 24 AWG 25.67 Ω/kft, 23 AWG 20.36 Ω/kft.
| Class | Standard | Type | Pairs | PSE output (W) | PD guaranteed (W) |
|---|---|---|---|---|---|
| 0 | 802.3af | 1 | 2 | 15.4 | 12.95 |
| 1 | 802.3af | 1 | 2 | 4.0 | 3.84 |
| 2 | 802.3af | 1 | 2 | 7.0 | 6.49 |
| 3 | 802.3af | 1 | 2 | 15.4 | 12.95 |
| 4 | 802.3at | 2 | 2 | 30.0 | 25.5 |
| 5 | 802.3bt | 3 | 4 | 45.0 | 40.0 |
| 6 | 802.3bt | 3 | 4 | 60.0 | 51.0 |
| 7 | 802.3bt | 4 | 4 | 75.0 | 62.0 |
| 8 | 802.3bt | 4 | 4 | 90.0 | 71.3 |
- The 100 m channel (90 m horizontal plus 10 m of cords) is a data limit, not a power limit. Past it, use a PoE extender, a media converter with local power, or fiber with an injector at the far end.
- Port count and watt budget are separate limits. Some switches allocate by class, others by measured draw; the switch datasheet governs.
Worked example. Eight Class 4 cameras reserve 8 × 30 = 240 W. On a 370 W switch that leaves 130 W at 64.9% of budget, inside the 296 W headroom limit. One 90 m run of 24 AWG: I = 30 ÷ 50 = 0.6 A, Rloop = 2 × 295.3 × 25.67 ÷ (1,000 × 2) = 7.58 Ω, loss = 0.62 × 7.58 = 2.73 W, so about 27.3 W reaches the device.
Unit conversions and EOL resistor color code
| From | To | Multiply by |
|---|---|---|
| Feet | Metres | 0.3048 |
| Metres | Feet | 3.2808 |
| Inches | Millimetres | 25.4 |
| °F | °C | (°F − 32) ÷ 1.8 |
| °C | °F | (°C × 1.8) + 32 |
| lb/1,000 ft | kg/km | 1.488 |
| TB (decimal) | TiB | 0.9095 |
| Mbps sustained | GB per day | 10.8 |
| Watts | BTU/h (rack heat) | 3.412 |
| px/ft | px/m | 3.2808 |
Resistor value = (first band digit, second band digit) × 10(third band). Fourth band is tolerance: gold ±5%, silver ±10%.
| Color | Digit | Multiplier |
|---|---|---|
| Black | 0 | × 1 |
| Brown | 1 | × 10 |
| Red | 2 | × 100 |
| Orange | 3 | × 1k |
| Yellow | 4 | × 10k |
| Green | 5 | × 100k |
| Blue | 6 | × 1M |
| Violet | 7 | × 10M |
| Gray | 8 | |
| White | 9 | |
| Gold | × 0.1 |
| Common EOL value | Bands |
|---|---|
| 1 kΩ | Brown, black, red |
| 2.2 kΩ | Red, red, red |
| 4.7 kΩ | Yellow, violet, red |
| 5.6 kΩ | Green, blue, red |
| 10 kΩ | Brown, black, orange |
The required value on a supervised zone comes from the panel documentation.
Sources
- LowVoltKit calculators: formulas, defaults, and worked examples for all twelve calculators
- LowVoltKit EOL resistor color code
- NEC Chapter 9, Tables 1 and 4 (conduit fill limits and internal areas)
- IEEE 802.3af, 802.3at, 802.3bt (PoE classes)
- ANSI/TIA-568 (fiber component allowances, 100 m channel)
- IEC 62676-4 (DORI pixel density levels)
- RFC 1918 and RFC 3021 (private ranges, /31 links)