conversions & calculations
TRANSCRIPT
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Kilogram (kg) Grams (g) Pounds (lb) Ounce (oz)
kilograms (kg) 1 1000 2.20462 35.2740
grams (g) 0.001 1 0.00220462 0.0352740
pounds (lb) 0.453592 453.592 1 16
ounces (oz) 0.0283495 28.3495 0.0625000 1
FLOW CONVERSION TABLE
Liters persecond
Liters perminute
Liters perhour
Cubic feetper second
Cubic feetper minute
Gallons (US)per second
Gallons (US)per minute
Milliongallons (US)per day
(l/s) (l/m) (l/hr) (CFS) (CFM) (GPS) (GPM) (MGD)
l/s 1 60.0000 3600.00 0.0353157 2.11894 0.264179 15.85077 0.0228251
l/m 0.01666667 1 60.0000 5.88594e-4 0.0353157 0.00440299 0.264179 3.80418e-4
l/hr 2.77778e-4 0.01666667 1 9.80990e-6 5.88594e-4 7.33832e-5 0.00440299 6.34031e-6
CFS 28.3161 1698.963 101937.8 1 60.0000 7.48052 448.831 0.646317
CFM 0.471934 28.3161 1698.963 0.01666667 1 0.1246753 7.48052 0.01077195
GPS 3.78531 227.118 13627.10 0.1336806 8.02083 1 60.0000 0.0864000
GPM 0.0630884 3.78531 227.118 0.00222801 0.1336806 0.01666667 1 0.001440000
MGD 43.8114 2628.68 157721.1 1.547229 92.8337 11.57407 694.444 1
VOLUME CONVERSION TABLE
Liter Cubic cm(cc)
Cubic meters Gallons (US) Cubic in. (cuin).
Cubic ft. (cuft)
Acre ft Cu ft per secday (CFSD)
Liter 1 1000.028 0.001000028 0.264179 61.0255 0.0353157 8.10736e-7 4.08746e-7
cc 9.99972e-4 1 1e-6 2.64172e-4 0.0610237 3.53147e-5 8.10713e-10 4.08735e-10
cu meters 999.972 1e6 1 264.172 61023.7 35.3147 8.10713e-4 4.08735e-4
Gal.(US) 3.78531 3785.41 0.00378541 1 2310.1336806 3.06888e-6 1.547229e-6
cu in. 0.01638661 16.38706 1.638706e-5 0.00432900 1 5.78704e-4 1.328521e-8 6.69796e-9
cu ft. 28.3161 28316.8 0.0283168 7.48052 1728 1 2.29568e-5 1.157407e-5
Acre ft 1.233447e6 1.233482e9 1233.482 325851 7.52717e7 43560.0 1 0.504167
CFSD 2.44651e6 2.44658e9 2446.58 646317 1.492992e8 86400 1.983471 1
MASS CONVERSION TABLE (WEIGHT)
FLOW, VOLUME & MASSCONVERSION TABLES
EXAMPLETo convert pounds into ounces, nd the row that is labeled lb. Look across to the column labeled ounce. The number at that intersection is16. Multiply pounds by 16 to convert to ounces. 22 * 16 = 352 Therefore 22 pounds is the same as 352 ounces.
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RADIANT EXPOSURE CONVERSION TABLE
J m-2
Wh m-2
cal cm-2
kcal m-2
BTU ft-2
langley erg cm-2
J m-2 1 2.778e-4 2.388e-5 2.388e-4 8.810e-5 2.388e-5 1000
Wh m-2 3600 1 0.08598 0.8598 0.3172 0.08598 3.600e6
cal cm-2 4.187e4 11.630 1 10 3.688 1 4.187e7
kcal m-2 4187 1.1630 0.1 1 0.3688 0.1 4.187e6
BTU ft-2 1.1351e4 3.153 0.2711 2.711 1 0.2711 1.1351e7
langley 4.187e4 11.630 1 10 3.688 1 4.187e7
erg cm-2 0.001 2.778e-7 2.388e-8 2.388e-7 8.810e-8 2.388e-8 1
1 langley = 1 cal cm--2
IRRADIANCE CONVERSION TABLE
W m-2 mW cm-2 kW m-2 cal cm-2 min-1 BTU ft-2 h-1 erg cm-2 s-1 W cm-2
W m-2 1 0.1 0.001 0.0014331 0.3172 1000 100
mW cm-2 10 1 0.01 0.014331 3.172 1e4 1000
kW m-2 1000 100 1 1.4331 317.2 1e6 1e5
cal cm-2 min-1 697.8 69.78 0.6978 1 221.3 6.978e5 6.978e4
BTU ft-2 h-1 3.153 0.3153 0.003153 0.004519 1 3153 315.3
erg cm-2 s-1 0.001 1e-4 1e-6 1.4331e-6 3.172e-4 1 0.1
W cm-2 0.01 0.001 1e-5 1.4331e-5 0.003172 10 1
1 cal cm-2 min-1 = 1 langley min-1 1 W m-2 = 1 J s-1 m-2
TEMPERATURE CONVERSION TABLE
C = (F - 32) / 1.8 F = (C * 1.8) + 32
C = K - 273.15 F = R - 459.67
K = C + 273.15 R = F + 459.67
CONVERSION TABLE USE
Find the units you have in the left column. Look across the table to nd the units you wish to convert to. The value that is contained in that rowand column is the multiplier that converts from the units you have to the units you want.
ExampleTo convert 60 degrees C to degrees F, multiply by 1.8 and add 32. (60C * 1.8) + 32 = 140F
RADIANT EXPOSURE, IRRADIANCE& TEMPERATURE CONVERSION
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PoutPin
Pout1 * 10-3
UPLINK EIRPOne of the more important rules for communicating with satellites is touse the proper power (Effective Isotropically Radiated Power or EIRP)when transmitting to the satellite. Uplink EIRP is a combination of powertransmitted by a transmitter and gain added to the transmitter by anantenna minus any cable losses. Uplink EIRP is referenced after theantenna contribution before any free space loss is encountered. First,a quick review of a decibel is in order.Decibel (as referenced to 50 ohm systems for satellitetransmitters)
A decibel (as applied to system gain or loss) is defined by theequation:dB = 10 * Log10 ( )
(Pin and Pout represent the power into and out of a system.) A dBm is a decibel relative to a milliWatt. It is dened by the decibelequation with Pin set at 1*10-3 (1 milliwatt) or .....
dB = 10 * Log10 ( )
The dBm equation is used to convert a power in watts to dBm asshown in the following chart:
Power Output in Watts Power Output in dBm
0.001 0.000.01 10.00
0.1 20.00
1.00 30.00
2.00 33.01
3.00 34.77
4.00 36.02
5.00 36.99
6.00 37.78
7.00 38.45
8.00 39.03
8.50 39.29
9.00 39.54
10.00 40.00
12.00 40.79
15.00 41.76
18.00 42.55
20.00 43.01
25.00 43.98
To calculate the EIRP of a system, use the following equation: (Thisis considered the uplink EIRP and is not necessarily the powerreceived by the satellite.)
EIRP(dBm) = Transmit Power(dBm) + Antenna Gain(dB) - CableLoss(dB) - Connector Loss(dB)
EXAMPLE: A transmitter outputs 8.5 watts with an 11 dB gainantenna, cable loss is 1.2 dB, and connector losses are 0.25 dB. Assume that the antenna is directly pointed at the satellite. Theuplink EIRP is : EIRP = 39.29 dBm + 11.0 dB - 1.2 dB - .25dB = 48.84 dBm EIRP
WARNING: NESDIS requires that under all service conditions,with any platform directed to the GOES series satellite that theuplink EIRP never exceed +50 dBm.
THE DECIBEL AS APPLIED TO 600 OHMAPPLICATIONSTelephone, private line or twisted pair applications, are standardizedon 600 Ohm impedance. Telephone companies supply line signallevel information typically in terms of dB and dBrnc and the customermay at times have the need to measure the level of the signals in
or out of his modem with a voltmeter or oscilloscope. The followingequation from above is:
dB = 10 * Log10 { } Using the following substitution:
Power ={ } =Then:
dBm = 10 * Log10{ } = And, nally, since
Vp =
then:
dB = 10 * Log101 { }
Power 1 * 10-3
Vrms2R
(Vp * .707)2600
(Vp * .707)2600
1 * 10-3(Vp * .707)2
.6
Vp/p2
(Vp * .707)22.4
UPLINK EIRPTELEPHONE SIGNAL EQUATIONS
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Extremely powerful transmitters may be rendered useless if thetype of antenna chosen, height of antenna from ground, or path lossconsiderations are not taken into consideration. Due to the manyvariables within the early design of a system, the best approach is toperform a link analysis determining the overall margin of the link, andfrom this, decide where the cost effective tradeoffs may be made.
Atten (decibels) = 36.6 + 20 * log F(MHz) + 20 Log D(miles)
1. Use the nomogram to determine the attenuation in decibels or
use the equation:Atten (decibels) = 36.6 + 20 * F(MHz) + 20 Log D(miles)Example: A 30 mile radio line of sight distance at 174 MHzwill equate to Atten = 36.6 + 20 * log(174) + 20 Log(30) =110.95 dB
2. Next, verify that the distance in miles between the transmitand the receive antennas is acceptable, given the installedheight of the antenna on a tower at each end. This willconrm that the curvature of the earth will not interfere withthe radio wave propagation. Each sight under considerationneeds to be investigated for obstructions between thetransmitting antenna and the receive antenna. For example,mountains, major buildings, heavy wooded forests, etc. allmay have a substantial impact to the quality of the linkespecially when rain, ice, and snow are covering the above
obstructions. Use the following nomogram to verify antennaheight or distance in miles between stations.3. Determine the level of reliability desired within a communication
link in terms of percent reliability and read the correspondingvalue of fading/multipath losses:
Percent Reliability (%) Fading Margin (dB)
90 10
99 20
99.9 30
99.99 40
LINE OF SIGHT (LOS)RADIO LINKS
4. Calculate the path margin using the following equation:Link Margin =
[Transmitter Power] + [Transmit Antenna Gain] +[Receive Antenna Gain] - [Cable Loss at Transmitter] -
[Cable Loss at Receiver] - [Attenuation Loss] - [Fading Margin] -[Receiver Sensitivity]
Example: Using the following typical equipment specicationscoupled with the desired reliability of 99.9% over a 10 mile path: Radio Transmit Power (174MHz) 4 Watts (+36 dBm) Radio with Receiver Sensitivity -105 dBm Transmit Antenna Gain +7 dB Receive Antenna Gain +3 dB Tx Cable Loss (RG-8) .5 dB Rx Cable Loss (RG-8) 1.2 dB Attenuation over 10 Miles (above) 110.95 dB Desired Data Reliability 99.9% (Fading Margin = 30 dB) Link Margin = 36 + 7 + 3 - .5 - 1.2 - 110.95 - 30 - (-105) Link Margin = 8.35 dB
FINAL COMMENTS1. Always use the lowest loss cable possible (or within budget) forantenna cables. Typically, remote sites have short cable lengths
but base stations most often have over 100 feet of cable betweenthe transceiver and antenna. In this common application, standardRG-8 cable is suitable for the remote antenna cables but the basestation may benet from use of a low loss rigid antenna cablefor the long distance between the antenna and the transmitter.Quite often, 3 dB of loss may be recovered through the use oflow loss cable instead of increasing the transmitter output powerby a factor of two.
2. The above link analysis is calculated on the basis of a clear Lineof Sight path meeting the above nomogram requirements. Ifdense foliage, trees, buildings, or other obstructions are betweenthe transmitter and the receiver, then much more margin will berequired to compensate for the poor weather conditions.
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ASCII CHART
Key Stroke ASCII DEC HEX Octal Key Stroke ASCII DEC HEX Octal Key Stroke ASCII DEC HEX Octal
^@ NULL 0 0 0 + + 43 2B 53 V V 86 56 126
^A SOH 1 1 1 , (comma) , 44 2C 54 W W 87 57 127
^B STX 2 2 2 - - 45 2D 55 X X 88 58 130
^C ETX 3 3 3 . . 46 2E 56 Y Y 89 59 131
^D EOT 4 4 4 / / 47 2F 57 Z Z 90 5A 132
^E ENQ 5 5 5 0 0 48 30 60 [ [ 91 5B 133
^F ACK 6 6 6 1 1 49 31 61 \ \ 92 5C 134
^G BELL 7 7 7 2 2 50 32 62 ] ] 93 5D 135
^H BS 8 8 10 3 3 51 33 63 ^ ^ 94 5E 136
^I HT 9 9 11 4 4 52 34 64 _ _ 95 5F 137
^J LF 10 A 12 5 5 53 35 65 ` ` 96 60 140
^K VT 11 B 13 6 6 54 36 66 a a 97 61 141
^L FF 12 C 14 7 7 55 37 67 b b 98 62 142^M CR 13 D 15 8 8 56 38 70 c c 99 63 143
^N SO 14 E 16 9 9 57 39 71 d d 100 64 144
^O SI 15 F 17 : : 58 3A 72 e e 101 65 145
^P DLE 16 10 20 ; ; 59 3B 73 f f 102 66 146
^Q DC1/Xon 17 11 21 < < 60 3C 74 g g 103 67 147
^R DC2 18 12 22 = = 61 3D 75 h h 104 68 150
^S DC3/Xoff 19 13 23 > > 62 3E 76 i i 105 69 151
^T DC4 20 14 24 ? ? 63 3F 77 j j 106 6A 152
^U NAK 21 15 25 @ @ 64 40 100 k k 107 6B 153
^V SYN 22 16 26 A A 65 41 101 l l 108 6C 154
^W ETB 23 17 27 B B 66 42 102 m m 109 6D 155
^X CAN 24 18 30 C C 67 43 103 n n 110 6E 156
^Y EM 25 19 31 D D 68 44 104 o o 111 6F 157
^Z EOF/SUB 26 1A 32 E E 69 45 105 p p 112 70 160
^[ ESC 27 1B 33 F F 70 46 106 q q 113 71 161
^\ FS 28 1C 34 G G 71 47 107 r r 114 72 162
^] GS 29 1D 35 H H 72 48 110 s s 115 73 163
^^ RS 30 1E 36 I I 73 49 111 t t 116 74 164
^_ US 31 1F 37 J J 74 4A 112 u u 117 75 165
Space SPACE 32 20 40 K K 75 4B 113 v v 118 76 166
! ! 33 21 41 L L 76 4C 114 w w 119 77 167
34 22 42 M M 77 4D 115 x x 120 78 170
# # 35 23 43 N N 78 4E 116 y y 121 79 171
$ $ 36 24 44 O O 79 4F 117 z z 122 7A 172
% % 37 25 45 P P 80 50 120 { { 123 7B 173
& & 38 26 46 Q Q 81 51 121 | | 124 7C 174
39 27 47 R R 82 52 122 } } 125 7D 175
( ( 40 28 50 S S 83 53 123 ~ ~ 126 7E 176
) ) 41 29 51 T T 84 54 124 Delete DEL 127 7F 177
* * 42 2A 52 U U 85 55 125
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The power budget is an analysis of how much power a data collectionsite requires. Analysis is required to determine how long adata recorder or Remote Telemetry Unit (RTU) will operatefrom the battery without recharging and what size solar panel(or charging source) should be used. The Model 8200A DataRecorder is used in the following example.The 8200As power requirements vary with the task it is
performing. A power budget is determined by calculating howmuch time the 8200A spends in each of its tasks and how muchpower is used. The following list shows the power used by the8200A in some typical tasks. Please note power consumptionis approximate: Quiescent (basic model): 0.25 mA
Transmitting GOES: 3500 mA Quiescent GOES: 10 mA
Transmitting LOS: 2500 mA Quiescent LOS: 30 mA
Telephone OFF HOOK: 50 mA Measuring: 5-30 mA
To determine the power needed by a site, sum the powerrequired by each of the tasks, taking into account the relative% of time dedicated to each task. A convenient way to do thisis to make a table (or spreadsheet) listing each of the tasks,current consumption, and times.Example: GOES 8200A collecting data every 15 minutesand transmitting once every 4 hours.
TASK CURRENT PERCENT TIME AV.CURRENT
COLLECTING 30mA * 5 sec/900 sec = 0.2mA
(5 sec to collect data every 15 minutes)
TRANSMITTING 2500mA * 45 sec/14400 sec = 10.9mA
(45 second transmission every 4 hours)QUIESCENT 10mA * 100% (always) = 10mA
TOTAL AVERAGE CURRENT = 21.1 mA
TOTAL AVERAGE POWER (current * 12VDC) = 253milliwatts
NOTE - ADD IN THE POWER REQUIRED BY SENSORS.
It is necessary to estimate the amount of time spent collectingand transmitting data, as well as the power required forcollecting data. It is best to obtain these numbers using actualmeasurements of power consumption for an operating 8200A.Once a value for the average consumption is determined,record it and use it as a reference when troubleshootinga station. A site that shows a marked change in powerconsumption warrants a closer look.Once a stations average power usage is determined, two otherimportant calculations should be made: BATTERY LIFE AND SOLAR PANEL SIZE
BATTERY LIFE
Battery life is computed in two steps.1. Compute the theoretical battery life. To do this, dividethe battery capacity by the average power required bythe 8200A, as follows:
THEORETICAL BATTERY LIFEBattery Capacity / 8200 Average Current
Example: Compute the theoretical battery life for a 24amp-hr battery powering an 8200A with average power
consumption of 50 mA.THEORETICAL BATTERY LIFE
24000mA-hr/50mA = 480 hrs2. Compute the actual battery life. Since it is not possible
to use 100% of any battery the actual life will be less.We recommend planning on using 75% of the capacityof a battery. This reduces the theoretical life by 25%. Inour example, the 480 hrs duration would become 480 *0.75 = 360 hours.
SOLAR PANEL SIZEThe size of the solar panel needed for the site depends onboth the average power needed and the location of the site.Generally, use a panel that provides at least 10 times theaverage power needed.Example: Size a solar panel for a site with average currentof 50 mA
POWER NEEDED10 * (current * voltage) = 10 * (50mA * 12 volts)
= 6000 mWatts ( 6 Watts)The minimum size panel should have an output of at least6 Watts. A standard 9-Watt panel will work great for thisexample site.Note that the internal charger in the 8200A has a maximumoutput of 0.75 amps or 9 Watts. If a panel larger than 9 wattsis used with the 8200A an external regulator is required. The8210 can accommodate panels up to 20 Watts.
Sutron offers a Microsoft Excel Power Budget Spreadsheetthat computes power consumption, 15 day reserve, andrequired solar panel size. Contact Customer Service for acopy. (703)406-2800.
POWER BUDGET CALCULATION NOTERemember this equation:
POWER = CURRENT * VOLTAGEKeep the units uniform. If current is inmilliamps, then power is in milliwatts.
POWER BUDGET CALCULATIONS
POWER BUDGET, BATTERY LIFE,SOLAR PANEL SIZE
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TELEPHONE DEFINITIONS
0 dBm = 90 dBrndBm = dBrn - 90dBrn = dBm + 90Reference Noise = 1 picowatt = - 90 dBm.Decibels above reference noise= dBrnDecibels above reference noise in a C-message weighting lter = dBrnc (tone1000Hz at 0 dBm or 90 dBrn).Picowatts of noise psophometricallyweighted (pWp) dBmp (tone 800 Hz at-90 dBm) dBmp = dBrnc - 90
This guide determines
the required azimuthand elevation anglesto point an antennaat a geostationarysatellite. It appliesto uplink DCPantennas andreceive site (DRGS)parabolic antennas.The following areneeded: Latitude of
Site Longitude
of Site Equatorial
Longitude of theSatellite (contactSatellite Agencyfor this value)
1. Subtract the sitelongitude fromt h e s a t e l l i t elongitude to determine the relative ground longitude. If the result is negative, the site is Westof the satellite. Remember this for step 3.Example: The satellite is positioned at 135 deg. longitude, and a site is located at 77 deg.longitude, 39 deg. latitude, then
135 - 77 = 58 degrees relative longitude.
2. Look at the nomogram above and nd the position that corresponds to the relativelongitude and the site latitude. From this intersection, read the A curves for azimuth andread the e curves for elevation.Example: Find the point on the nomogram that corresponds to 58 degrees relativelongitude (computed in step 1) and 39 degrees latitude. The values for the elevation andazimuth are:
e = 16 deg A = 68 deg3. Identify what quadrant the site is located in relative to the satellite and use the chart to adjust
the Azimuth. Earth Quadrant True Azimuth
Relative to Satellite Az NW 180 - A NE 180 + A
SW A SE 360 - AThe site is in the NE of the satellite because the latitude is North and the site is East of the satellite.Therefore, add 180 degrees to the A number to obtain the nal azimuth from true North. Thenal values are: e = 16 deg Az = 248 degIf you use a compass for pointing the antenna, take into account the local site magnetic variationfrom true north. This value will vary by location of the site. Be careful when using a compass neametal structures or objects as the readings may be inaccurate causing poor antenna pointing.
ANTENNA POINTING GUIDE*
*Reference : Reference Data for Engineers: Radio, Electronics, Computer, and Communications Seventh Edition; Edward C. Jordan, Editor in Chief, Howard W. Sams & Co. 1986 .
ANTENNA POINTING GUIDETELEPHONE DEFINITIONS
dBm (600 OHM) Volts Peak-to-Peak
10 6.93
9 6.18
8 5.50
7 4.91
6 4.37
5 3.90
4 3.47
3 3.10
2 2.76
1 2.46
0 2.19
-1 1.95
-2 1.74
-3 1.55
-4 1.38
-5 1.23
-6 1.10
-7 0.98
-8 0.87
-9 0.78
-10 0.69
-11 0.62
-12 0.55
-13 0.49
-14 0.44
-15 0.39
-16 0.35
-20 0.22
-25 0.12
-30 0.07
-35 0.04
-40 0.02
-50 0.01