Minimum Separation on crossing traffic

without vectors

 

 

We apply vectors for separation only if we have previously evaluated the existing separation of traffic and we consider it as insufficient or 'tight'. This is why before using vectoring we have to be able to assess how much the future minimum spacing between traffic will be. This is more difficult to judge, although easier to solve, with crossing traffic and especially when the tracks are crossing at right angles. There are means to measure precisely the separation between aircraft  before the conflicting point , as with say a mouse marker or a ralling ball bearing/distance measurement and even try to evaluate their future one by simply measuring their present distances from the conflicting point , two planes will have. But ....this will not be their future minimum separation. The thing is that their distance difference is kept almost unaltered until the first plane arrives at the conflicting point and then ...it starts decreasing to reach the 70% only of this difference .

In the past many people were rather surprised by this natural event or , worse , some were sure that the future separation has been secured if the first plane crosses the conflicting point since their distance difference was kept steady until that moment  

 

The article below not only proves that this is wrong but will also propose

a simple rule of thumb to estimate the minimum separation after the crossing point

 

 A Mathematical study 

 Suppose that we have 2 aircraft A and B of equal speeds on tracks forming exactly an angle of 90 degrees . These aircraft will move on the axis OX and OY . The separation S of the two aircraft is always the distance between A and B or S = AB . We will compare this value against the distance difference OA-OB which is their initial distance difference from the common crossing point as a function of their displacement x :    

 Let us assume that aircraft A and B move with equal speeds and so each one is displaced by the same amount X along their tracks that cross at a normal angle . Their equivalent distances before the conflict point O, are D1 and D2.  We will measure their separation S while the aircraft are at some intermediate position displaced already by X . According to their new positions we have from the triangle OAB:

 

[1]                   S(x) =  ((D1-x)2 + (D2-x)2 )(1/2)

 

and S is a function of X since it is due to the displacement X that the positions A and B change .

 To find the maximum or minimum of the function S(x) we must find where its first derivative versus x becomes zero , or the same :  

[2]        dS/dx = 0  = > d((D1-x)2 + (D2-x)2 )(1/2)  / dx

  = > (1/2)((D1-x)2 + (D2-x)2 )(-1/2).((D1-x)2 + (D2-x)2 ) = 0

  The expression ((D1-x)2 + (D2-x)2 )(-1/2) can not be zero  "xeR , so for ds/dx =0 , it requires that :

            => ((D1-x)2 + (D2-x)2 ) = 0

            =>( D12-2D1x+x2 + D22-2D2x+x2) = 0

            => ( -2D1 +2x - 2D2 + 2x ) = 0

            => 4x =  2 ( D1+D2)

[3]        => x = ( D1+D2) /2

 

That shows that a minimum , or maximum value is reached when x , the displacement from the initial positions , is the mean average of the initial distances D1 and D2 . To find S at this x=(D1+D2)/2 we substitute [3] in [1] , so that :

[4]                   S(x=(D1+D2)/2)) =  ((D1-(D1+D2)/2)2 + (D2-(D1+D2)/2)2 )(1/2)

                                               => ( (D1-D2)/2)2 +((D2-D1)/2)2)(1/2)

you may note that (D1-D2)2= (D2-D1)2                

so that              S = (2 . ( D1-D2)2 / 4 )(1/2)

                        S= ( (D1-D2)2/2)(1/2)=(D1-D2)/(2)(1/2))=

[5]                   S=(D1-D2)/2 , or since 1/2 = 1/1.414=0.707=>

                                                S=0.707(D1-D2)

 

remember that D1-D2 was their initial distance difference from the common point and that it finally becomes at the critical point  the 0.7 ( 70 % ) of this initial value , or the same it is reduced by 0.3 or 30 % .

 So a rule of thumb should stipulate that :  

The final minimum separation at crossing tracks of 900    is 30% LESS    than the one expected before the crossing point

 

Since 0.7(D1-D2)/2 < ( D1 - D2 ) there is no doubt that S(x) is decreasing until that point but it can be verified mathematically by studying the sign of the derivative at any previous point say xo<(D1+D2)/2 or the same xo=(D1+D2)/2 -e ,   " 0<e<(D1+D2)/2  

 

[6]        dS(xo)/dx =d((D1-x)2 + (D2-x)2 )(1/2)  / dx

 

= > (1/2)((D1-xo)2 + (D2-xo)2 )(-1/2).((D1-xo)2 + (D2-xo)2 ) = 0

again the  (1/2)((D1-xo)2 + (D2-xo)2 )(-1/2) > 0 "xeR so that the sign of ((D1-xo)2 + (D2-xo)2 ) will decide the sign of dS(xo)/dx :

dS(xo)/dx =-2D1xo-2D2+4xo = - 2 ( D1+D2 ) + 4xo=-2(D1+D2)+4(D1+D2)/2-e)

= - 2 ( D1+D2 ) + 2 (D1+D2) - 4e = - 4e which is obviously < 0 , since e>0, so that 

 

S(x) is constantly falling from the initial value until the point where  x=(D1+D2)/2

 

This justifies the concern we have about people who rely on the initial distance difference before the crossing point. At least , those who realize it , can now understand how much more deviation they should provide with vectoring to achieve a secure minimum safety separation value. This is the real value of this study

   

But if you insist let us see the results from a practical example, which anyone can work on a piece of paper or even on a radar screen :

Say that D1=OA=30 and D2=OB=20 nm so that initially the distance difference from the crossing point is 30-20=10 NM. At the point where the distance run by each aircraft is (D1+D2)/2 = (30+20)/2=25 nm, that is after each one runs another 25 nm, the situation will be the following : A will be just 5 nm before point O and B just 5 nm , passed the point O. Use your basic Pythagorean Theorem  to calculate the AB or S ( for separation) when the sides OA and OB equal to 5 nm , and you get the AB to be 50 =  practically 7or  30% less than 10. 

  

 

What if not on 90 degrees angle ?  

 

One thing is certain : The difference between the final minimum separation and the one you measure before the crossing point becomes less and less with less  angle and is obviously nil on same tracks . Mathematics , however , will not give you a simple calculation for other angles . Our study though has an answer for all practical purposes with a minimum error. After having advised calculations with angles of  30 and 60 degrees we calculated the drop of the initial difference to be at the critical point of minimum separation , according to the table , below 

 

 

Track angle

Separation decrease after crossing point in %

 

Minimum Separation as a portion of the initial distance difference (D1-D2)

 

90

- 30 %

0,7*(D1-D2)

60

-20 %

0,8*(D1-D2)

30

-10 %

0,9*(D1-D2)

 

The values are more than encouraging ! You don't have to carry a calculus machine with you to be able to understand that there is a gradual fall from 90 to 30 degrees reducing at a rate of 10% for every 30 degress.  OK , you may even say 1% for every 3 degrees, or allow me to say , the same : 

For every 10 degrees on track angle you may consider a reduction of separation by (almost)

3 % !  

 

If you like formulas we can put it that way :

If D1 and D2 are the distances of 2 planes from a crossing point on tracks that form an angle of and we want to calculate how much their future minimum separation Smin will be , then :

Smin = [0,7+0,03*(90-)]*(D1-D2)

Check that for = 90 , Smin becomes the typical 0,7*(D1-D2) value and when = 0 , on same tracks, then Smin = D1-D2 as expected, while when = 60, then Smin = [0,7-0,03*(90-60)]*(D1-D2)=almost 0,8*(D1-D2) as the exact mathematical analysis shows ; well OK the actual value is 0,79 not 0,80 ...but...

Let's see a 'paper' example :

2 planes of equal speed at an angle of 45 degrees are found at 50 and 64 nm before a conflicting point. How much will be their minimum separation if they follow these tracks after the conflicting point 

Smin = [0,7+0,03*(90-45)]*(64-50)=[0,835]*(14)= 11,69 nm 

If you separate with , say , 10 nm , it will only be sufficient to instruct the pilots to maintain their headings locked and resume them safely at about say 1 minute after they pass the conflict point. If , however , their D1 and D2 where , say , 60 and 52 correspondingly then Smin would fall down to  Smin =6,68 and that will need you to act with vectors to achieve another 4 (for safety) miles. 

An example out of a radar simulation

Say that with a radar we watch the same thing happening with 2 aircraft on crossing traffic at 60 degrees . Our EIN3344 measures 15 NM from the conflict point CEN while DLH101 37 . 

It seems 'normal' that people will assume a 'constant' separation of 37-15=22 NM but this is not what our analysis predicts . It does actually foresee a reduction of -20% when the first aircraft EIN3344 will have passed the point by half its present distance to the CEN , or something like 7-8 NM . Still while EIN3344 is on the conflict point the separation is not yet decreased and you may still measure 21 NM ! See the picture below :

But ...when EIN3344 is outbound the CEN by 9 NM 

then the lateral separation is measured to be 18 NM in total agreement with the theory that predicts a reduction of the initial 22 NM separation by a - 20% or 18 NM exactly !! Whhaaouuu !!

The above is not a critical situation but if the initial separation is very close to the bare minimum then you might be surprised that your Conflict Alert function will be activated after the first aircraft has passed the point where you obviously think that the problem does no longer exist !

                                                    

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