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Old April 10th, 2015, 02:29 PM
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Default TNE/T4 Sensor collision avoidance range & T4 A/P/J values

Morning all,

In my attempt to go through T4 QSDS 1.5e the Sensors section indicates the Basic Sensor suite, my page count is 10, is only designed for collision avoidance with Sensor Combat Statistics USD of A0 P2 J0.

Does anyone have an idea what the range in km and/or hexes would be collision avoidance?

If I have guessed right A0 means that the active sensors range is less than 30,000 km, which using TNE FF&S Mk I Mod 0 would be 3,000 km and the P2 is the passive sensor's range of maximum of 60,000 km.


The tables in T4 QSDS 1.5e appear to have been filled in using TNE FF&S Mk I Mod 0, which I used to reconstruct the T4 QSDS 1.5e TL 9 Basic Sensor package.

From The T4 QSDS 1.5e Standard Sensor Systems Table, the page that also includes the Communicators background text, the TL-9 Basic system has the following statistics: Note Td = displacement tons (sorry the subscript and superscript features don't like me today). I am to do two entries one for m^3 and the other in Td when working through TNE FF&S Mk I Mod 0.

The basic sensor system is composed of a radar set for the active system and a high-resolution thermal set for the passive system per T4 QSDS 1.5e with the table statistics as:

TL-9; Description: Basic; Min (Hull) L: 5 m; Volume: 1.1 Td; Power: 3.2 MW; Cost: MCr13.5; Area: 45 m^2; Crew: 1.1; USD A0 P2 J0.

Going to my worn copy of TNE FF&S Mk I Mod 0 Book II Chapter 5 Electronics B. Sensors p. 50 the Radar table has three ranges to select from at TL 9 which are 3,000, 30,000, and 60,000 km.

Of the three the range that I believe will return an A0 is 3,000 km. Per the Book III ranges are converted from distance to hexes by dividing the distance in km by 30,000 km which is the size of one space hex. 3,000 km divided by 30,000 returns 0.1 hexes or rounding to the nearest whole number is zero (0).

Both the active and passive systems calculate statistics for a processor and an antenna.

Starting with the TNE FF&S Radar Table, p. 50, a TL-9 processor with a 3,000 km guessing that means an USD rating of A0 has a volume = 15 m^3 = 1.0714 or 1.1 Td. The volume in Td matches the table without the HRT included.

The HRT table, p. 52, has a TL 9 processor with a range of 60,000 km, which I believe reflects the USD of P2, has a volume = 1.5 m^3 without the antenna array.

The combined radar and HRT processors volume =16.5 m^3 or 1.1786 Td or after rounding to the first decimal place = 1.2 Td.

The HRT antenna volume = HRT Antenna Area x 0.05 fixed array or 0.1 folding array.

Antenna Area = 80 m^2 x TL-9 mod 0.5 = 40
Antenna Diameter determines if the array is fixed or folding based on the hulls unmodified hull length a 100 Td hull is 14 m long and the 60,000 km HRT antenna has a diameter of 10 m. The hull is longer than the antenna diameter making the antenna a fixed array using the modifier of 0.05.

HRT antenna volume m^3 = 40 x 0.05 = 2 m^3.

The 15 m^3 radar processor + 1.5 m^3 HRT processor + 2 m^3 HRT fixed antenna array = 18.5 m^3 or 1.3214 or 1.3 Td.

I have also checked T4 SSDS by David Golden and the minimal package uses a TL-9 radar of 3,000 km and a TL-9 HRT with a fixed array with a range of 30,000 km.

By my calculations the TL 9 radar is 15.8 m^3 and the TL 9 HRT is 1.1 m^3 for a total of 16.9 m^3 or 1.2071 Td or 1.2 Td.

Of course my quick calculations including the computers, radio, and flight avionics to the T4 QSDS 1.5e basic sensor systems returned a volume of 31.9 m^3 which is 0.7 higher than T4 SSDS by David Golden.

Have I started the calculation for the radar and HDT correctly?

If I have then the rest of my calculations for power, cost, and area do not match.

Also does anyone know hoe T4 QSDS 1.5e calculated the crew requirement?
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