Radar Location

  • Thread starter Thread starter Guest
  • Start date Start date
I'm just having the JRC1000 I bought at the boat show fitted. I thought long and hard about interior/exterior mounting and in the end bought a spare mounting bracket (35ukp, ouch) so I can mount it inside and out depending on what I'm doing.

Regards,

Peter
http://www.mistressofmourne.com
 
Are you absolutely sure on this formula.

Don't understand reason for Sq root

To achieve 24 miles range:-

1.23 x Rt (total height) = 24

RT (total height) = 24/1.23 = 19.51

total height = 19.51 x 19.51 = 380

This sounds rather high to me
 
Yes the formula is correct I think that I possibly did not explain it very well. The formula is 1.23 x (sq rt height ae + height target) in feet, and the answer is in miles. Therefore if you have a scanner that is 10ft high it will see a target that is 15ft high at a range of 1.23 x (sq rt 10 + 15) which = 1.23 x (sq rt 25) which = 1.23 x 5 which is 6.15 Miles.

This formula takes into account for the fact that Radio Waves bend when they travel over water and is good for both Radar at a frequency of approx 9.5Ghz and VHF at 156 Mhz.
 
I think that square root must be wrong. The approximation works on the basis that sin(x) is approximately x for small values of x. Therefore range is roughly proportional to the combined height of observer and target.
 
The Square root is an integeral part of the formula for this approximation and was taken from my copy of Skolnecs the radar Bible. and I have been using it in Radar trials on ranging for the last 25 years.
 
when you next go out sailing climb the mast to the position where a rader would be located and check out the motion. You're going to travel a lot more than a rader located on pole at the stern.
 
Think your formula is ok and it seems to assume about 16% refraction. But I dont think you can simply add the two heights before taking the square root. In your 10 and 15 ft example you would have to calculate the radar horizon for the two heights separate and then add the two distances. Like:

1.23*sqrt(10) = 3.89
1.23*sqrt(15) = 4.76

3.89 + 4.76 = 8.65 as opposed to 6.15
 
this is not the case otherwise if you had a scanner that was 20ft high looking at a target 5ft high you would get a different range again yet the horizon should be the same.
1.23*sq rt 20 = 5.5
1.23*sq rt 5 = 2.75
2.7 + 5.4 = 8.25
I have used this formula for many years and have never had any problems, it is a good rule of thumb for radar range although it does not and cannot take into account for variables such as annoprop (ducting in radio) which can double or even halve radar range without any warning.
 
There is a boat owner in my yard who has mounted his JRC by fixing it to his backstay, ie no pole. It's sort of lashed with a couple of extra stabilising wires. He says that it works fine. I can press him for more details if you like.
 
It sounds very good idea. I would appreciate more information please.
 
You must log in or register to reply here.

Share this page

Top