CP, PNR or PSR and ROA
An important aspect of flight planning is the calculation of the action to be taken in the event of a diversion or an emergency. The decision to be made is whether with the available fuel and knowledge of the wind velocity, it will be preferable to return to base, divert or continue to the destination, and indeed which of these options is feasible.
CRITICAL POINT & POINT OF SAFE RETURN or THE POINT OF EQUAL TIME (PET)
An important aspect of flight planning is the calculation of the action to be taken in the event of a diversion or an emergency. The decision to be made is whether, with the available fuel and knowledge of the wind velocity, it will be preferable to return to base, divert, or continue to the destination, and indeed which of these options is feasible. This chapter will describe the various decision points, which can be determined at the flight planning stage and the methods by which they can be calculated.
(a) The point of equal time (PET), previously known as the critical point (CP) is the point along track at which it will take equal time, in the prevailing conditions and specified configuration, to reach either of the two nominated points, which do not necessarily be on the same track. Note that the calculations depend only on time not on the fuel available. The exact position of the PET can be determined by the calculation, by plotting or by using the flight progress charts.
(b) The purpose of determining the PET is to enable the Captain to make a sound, rational decision when confronted with unforeseen circumstances. Examples of these ranges from the serious illness of a passenger requiring hospitalisation as soon as possible, to an engine failure necessitating an emergency landing or a pressurization failure compelling a descent to continue the flight at a lower level.
(c) A PET can be predetermined for any circumstance or configuration between the departure and destination airfields. In the event of the emergency accounted before the PET is reached, the aircraft should turn around and return to the destination point; however if it occurs after the PET has been reached the aircraft should continue to the destination point.
(d) A PET is normally calculated before flight for the all engines–operating and the one engine in-operative condition. There are two types of PET, the single track case, often referred as the simple case and the multi-track case.
The single track, constant TAS, constant wind velocity case.
(a) This is the simple case of a PET and the understanding of this is the basis of the multi track solution where the factors are more variable. (Refer figure 1.)
Assume that A is the departure aerodrome, referred to as Home, and that B is the destination aerodrome referred as On. Then the time taken to reach either A or B from the PET is the same. Time is equal to distance divided by the ground speed. Therefore, the distance from the PET to departure aerodrome divided by ground speed home (H) is equal to the distance from the PET to the destination aerodrome divided by the ground speed On (O).In the following figure D is the total distance from A to B; X is the distance from A to PET; H is the ground speed home from PET to A; and O is the ground speed on from PET to B.
Critical Point : Single Track
Assume D = Total distance
X = Distance to CP
O = G/S out (with Reduced TAS)
H = G/S Home(with Reduced TAS)
G = G/S out (with normal TAS)
P = The CP
The time from PET to A = X / H
The distance from PET to B = ( D- X)
The time from PET to B = (D-X) / O
By definition, therefore,
X / H = (D-X) / O
XO = H (D-X)
XO = (DH-XH)
XO + XH = DH
X (O+H) = DH
X = DH / (O+H)
:. Time to CP = X/G = DH/ G(O+H)
(b) The above derived equation is the PET formula. To calculate the time taken to reach PET from the departure point, the distance to the PET, X, is divided by the ground speed out from the departure point. The result will be the time in hours and decimals of an hour. This may be converted to minutes by multiplying by 60.The ground speed out from the departure point and onward from the PET will be the same if all engines are operating, the cruise altitude is the same and the wind velocity is the same.
(c) For example, given a wind velocity of 240/50; track 270 (T); TAS 300 kts.The distance from the departure to the destination is 1000 nm.Calculate the distance and time to all-engines-operating PET. For obtaining the solution the calculations are as follows.
(i) Calculate the ground speed for track 270(T) and for track 090 (T).On track 270(T) it comes to 256 kts and on track 090(T) it is 340 kts.
(ii) The distance to the PET from the departure point using the formula X=DH / (O+H) is 1000 * 340 / (256 + 340) = 570.47 nm.
(iii) The time to the PET = (570.47 / 256) * 60 = 2 hr 13.7 mts.
(d) In the above derived solution, it has been assumed that all engines are operating and that the altitude and ambient temperature are all the same in both directions from the PET. Therefore, the TAS will be same in both directions. If, however, the PET is for one- engine- inoperative configuration or for the pressurization failure case then the outbound ground speed on the onward ground speed will be different. The speed to be used in the formula to determine the distance to the PET must be the ground speeds for the case under consideration. Thus, if the PET is to be for the one-engine –inoperative case the ground speeds to be used in the PET formula must be those with one-engine- inoperative. However, the ground speed to be used to calculate the time taken to reach the PET from the departure point is always the all-engines –operating ground speed from the departure point.
The multi -track, variable TAS, variable wind velocity case.
(a) To determine the position and time to reach the PET for a multi- track case it is necessary to calculate the ground speed and leg time , both outbound and inbound, for each leg of the for the configuration being considered.Then,by the process of elimination, it is possible to determine on which leg the PET is located. When this has been determined, the PET can be exactly located on that leg by the application of formula already discussed.
(b) For example given the following completed flight plan, determine the distance and time to reach the PET from A.
|LEG||GROUND SPEED(KTS)||DISTANCE (NM)||TIME (MIN)|
(c) To arrive at the solution (refer fig.2) draw a straight line diagram of the route showing each of the turning points. This is to show the time between each of the turning points outbound (above the line) and the times inbound (above the line but in the opposite direction). Not all of the times are required, only those needed to effect a balance of times needs to be shown. Bear in mind that the time onward to the destination must equal the time home to the departure point.
(d) As shown in fig.3 insert time D to E first as 43.4 min. Now insert time B to A as 43.2 min.
(e) Next as shown in fig.4 insert time C to D as 37.1 min. Thus the total time from C to E is 43.4 + 37.1 = 80.5 min. Now insert time C to B as 62.6 min. The total time from C to A is, therefore 43.2 + 62.6 = 105.8min
(f) Clearly, the time from C to A does not balance the time from C to E. Refer fig.5 an interim balance point must be introduced to equal the smaller time. This point is labeled as Z. Now Z to A is equal to C to E as 80.5 min. Therefore, the portion Z to C has to be time balanced by the PET formula.
PET Multi Leg Calculation
(g) The total distance D in the PET formula is the distance Z to C in fig.6. The ground speed on from the PET is the ground speed on leg B to C.The ground speed home from the PET is the ground speed on leg C to B. The distance Z to C is equal to the time difference (105.8 – 80.5) = 25.3 min.multiplied by the ground speed on the leg in which it fell in the direction of the arrow (i.e., ground speed CB = 230 kts).Therefore, distance ZC = (230 / 60) * 25.3 = 97 nm.
(h) In the PET formula, X = DH / O+ H,the distance X is the distance from Z to the PET ,which always falls in unbalanced portion of the route, in this case ZC,and X is always to the left off PET. In this example X = 97 * 230 / (260+ 230) = 45.5 nm.In addition, the time from Z to the PET = (45.5 / 260)* 60 = 10.5 min.
(j) To determine the distance and time to the PET from A it is necessary to calculate the distance from B to Z = BC – ZC = 240 – 97 = 143nm.The time from B to Z is equal to the distance divided by the outbound ground speed on that leg (i.e., (140 + 260)* 60 = 33 min.). Therefore, the distance from A to PET = 180+143+ 45.5 = 368.5 nm and the time from A to PET is 46+33+10.5 = 89.5 min.
(a) The critical point represents that point on track from which it will take equal time to proceed to destination or return to base. If the aircraft is off track however the critical point loses its significance and must be replaced by a critical line. For a straight line track as in Fig. 7, in still air, the perpendicular bisector of track represents the equal time line back to A or on to B. thus for an aircraft well of track (at C) sit would be quicker to return to A than proceed to B in still air.
(b) To be valid this line must be modified for the effect of wind. The time for the aircraft to fly from the still air critical line to either A or B in still air at the reduced TAS is calculated. The critical line is then moved upwind by a distance equal to the still air time multiplied by the wind speed.
(c) In the example the still air critical line is at 500 nm. The reduced TAS is 260 kts and the wind velocity is 060/60. The still air time to fly from the critical line to either A or B is 115 minutes (1.923 hrs). Thus the critical line must be moved upwind (i.e. in the direction 060° T) by 1.923 x 60 nm = 115 nm (Fig 8).
(d) The assumption in this solution is that the distance from any point on the still air craft line to A or B is the same (i.e. 500 nm in this example). This assumption becomes less valid as distance off track increases but the errors induced are unlikely to be significant unless the track error is large and the route relatively short.
Critical Point Between Three Airfields.
(a) The discussion in the foregoing paragraphs has considered only the case where the options available are proceedings to the destination or returning to base. More commonly there will be a third option of diverting to an off-track airfield. Fig 9 shows the still air situation where between. A and L it will be quicker to return to base, from L to N it will be quicker to divert to C, and beyond N it will be quicker to proceed to B. Thus L and N represent two critical points. The best method of finding the positions of L and N is graphical and is based on the critical line solution.
(b) The method is illustrated in Fig,9 AC and BC, joining the departure and destination airfields to the diversion are drawn, and the perpendicular bisectors of these lines (LM and NO) are constructed to cut the track, AB, at L and N. As L is equidistant from A and C and N is equidistant from B and C, L and N are the still air critical points.
(c) To account for the wind effect the point L and N are moved upwind, in the same manner as constructing a critical line, by an amount equal to the wind speed multiplied by the time to fly from L and N respectively to C at the reduced TAS. Critical lines are drawn through the ends of the wind vectors, parallel to LM and NO, and where these cut the track represent the critical points.
Factors affecting the PET.
(a) In still air conditions or wind at 90˚ to track giving an equal effective head wind component in both directions, the CP will be exactly half way between the two bases being considered. The factors affecting the position of the PET are as follows.
(i) The total distance between the two bases.
(ii) The wind components-the PET will always be along track and upwind of the mid point.
(iii) The TAS –reducing the speed will increase the significance of the wind and so cause the PET to move further along track into the wind.
Generally CP is needed for an eventuality of loss of power in flight due to engine failure or any such emergency. Therefore Reduced TAS is extracted from Aircraft Operational Manual for Calculation of H and O, where as G is calculated with normal TAS.
CP is not dependent on Fuel or Endurance.
POINT OF SAFE RETURN (PSR) OR POINT OF NO RETURN (PNR)
The Point of Safe Return (PSR), earlier referred to as PNR is that point furthest removed from base to which an aircraft can fly and still return to base within its safe endurance. PNR is normally calculated on long flights where the options are limited to landing at the departure or destination airfields. The (safety) fuel reserve is stipulated as part of operating instructions or company policy. In essence, safe endurance is the Fuel on Board (FOB) less the fuel reserve divided by the fuel consumption. PLE or the ‘Prudent Limit of Endurance’ is the endurance available after deducting the quantity of the reserve fuel from the fuel on board.
The point of safe return (PSR).
(a) The point of safe return (PSR), previously known as the point of no return (PNR), is the furthermost point from which an aero plane can safely return to the departure point using all the fuel available, but excluding the safety reserve. The precise position of the PSR can be determined by the calculation or by using the flight progress chart.
(b) The purpose of determining the PSR is that in the event of an emergency, when no diversion or alternate aerodrome is available, the captain can rapidly determine whether returning to the departure aerodrome is a viable option. If the emergency occurs before the PSR is reached then it is, but once the aero plane is beyond this point, it is not.
(c) Usually the PSR is calculated for the all engines operating configuration. It could be calculated for the one engine inoperative condition, assuming all the engines are operating until PSR is reached and that the return journey is with one engine inoperative. However, the difference between the two is not very much and the assumption that an engine fails at PSR is a most unlikely event. Therefore, it is normal to determine only the all-engines-operating configuration PSR. However there are two cases, as with the PET; the simple, single-track case in which the TAS, altitude and fuel flow remain constant, and the multi-track case, in which the TAS, fuel flow and altitude are all variable.
(d) The greatest distance from the departure point to the PSR occurs in still air. Any wind component, no mater whether it is head or tail, will cause the PSR to be closer to the departure point than in still-air position. The greater the wind component the nearer is the PSR to the departure point.
The Single-track, Constant TAS, Constant Altitude, Constant fuel flow case.
(a) The simple case assumes that the TAS, altitude and the wind velocity remain constant throughout a track along a single track. By definition, the time taken to reach the PSR from the departure aerodrome plus the time taken to reach the departure aerodrome from the PSR is equal to the safe endurance of the aero plane. Time is equal to the distance divided by ground speed. In figure 10, X is the distance of the PSR from the departure aerodrome; O is the ground speed out from the departure aerodrome is the ground speed home to the departure aerodrome; and E is the safe endurance of aero plane in hours.
PNR/ PSR Formula (For Single Track):
Assume E or P = Total safe endurance in hours
T = Time to PNR in hours
O = G/S out
H = G/S Home
X = Distance to PNR
Then: Total Safe Endurance (E) = Time out + Time Home
X/O + X/H = P or E
∴ POH = X ( H+O)
∴ X = POH/ (H + O)
∴T (Time to PNR/ PSR) = PH/ (H+O)
(b) The time to reach the PSR from the departure aerodrome is equal to the distance X divided by the ground speed out, which is assumed to be same as that onward from the PSR. Therefore, time to PSR = EOH / O (O+H) = EH / (O+ H)
(c) For example given wind velocity of 240 / 50; track 270˚ (T); TAS 300 kts, safe endurance 8hr. Calculate the distance and time to reach the PSR.
(d) To arrive at the solution first calculate the ground speed for track 270 ˚ (T) and for the track 090 ˚ (T). On track 270˚ it is 256 kts and on track 090˚ it is 340 kts.
(e) The distance to the PSR from the departure point using the formula is
EOH / (O+ H) = 8 * 256 * 340 / (256 + 340) = 1168.3 NM.
(f) The time to the PSR from the departure point using the formula is
EH / (O + H) = 8 * 340 / (256 + 340) = 2720 / 596 = 4.56 HR = 4hr 33.6 min.
The multi- track, variable TAS, variable altitude, variable fuel flow case.
(a) The multi- track PSR case can be solved only by completing a full flight and fuel plan for both the outbound and inbound legs of the entire route. From the total fuel available, the sum of the fuel required for each leg both outbound and inbound, starting at the departure point is subtracted until the leg is reached where there is insufficient fuel available to complete the return leg. The PSR occurs on this leg. The fuel available the last turning point to the PSR is then divided by the sum of the outbound and inbound gross fuel flows on that leg to determine its exact location. The gross fuel flow is the fuel flow in Kg per hour divided by the ground speed in kts.This produces the number of Kgs of fuel used per nautical mile.
(b) For example given the following table and fuel available 20,000Kg calculate distance and time to PSR.
|LEG||G/S (Kts)||Fuel Flow(Kg/hr)||Leg Distance (nm)||Leg Time (hr)||Fuel Used(Kg)|
(c) To arrive at the solution firstly calculate the fuel requirement from point A to all the other points and back to A as shown below (refer figure 11).
(i) Fuel required from A to B and back to A = 2376 + 2250 = 4626 Kg.
(ii) Fuel required from A to C and back to A = 4626 + 2526 +2493= 9645 Kg.
(iii) Fuel required from A to D and back to A = 9645+3830 + 3840= 17315Kg.
(iv) Fuel required from A to E and back to A = 17315+2843+2768 = 22926Kg
(d) The fuel available is insufficient to reach E and return to A .However, there is enough fuel to reach beyond D and to return to A .The distance that the PSR is beyond D now has to be determined. The fuel available from D to the PSR and to return to D is 20000 – 17315 = 2685 Kg. To determine the distance that it will enable the aero plane to travel outbound and to return to D, the fuel available must be divided by the total gross fuel flow for both legs.
(e) The gross fuel flow on from D = fuel flow D to E divided by the ground speed from D to E = 2900 / 255 = 11.37 Kg / nm. For the return leg, the fuel flow E to D must be divided by the ground speed E to D = 3100 / 280 = 11.07 K g / nm. Therefore, the total gross fuel flow is equal to 11.37 + 11.07 = 22.44 Kg / nm.
(f) The distance from D to the PSR is therefore, equal to 2685 / 22.44 = 119.7 nm and the time from D to the PSR = 119.7 / 225 = 0.532 hr.
(g) The total distance from A to the PSR = 180 + 220 + 300 + 119.7 = 819.7 nm and
the total time from A to the PSR = 0.72 + 0.81 + 1.28 + 0.53 = 3.34 hr = 3 hrs 20.4 min.
FACTORS AFFECTING PNR/PSR:
1) Maximum distance to PNR is achieved in nil wind conditions. Any wind will reduce the range.
2) Greater the fuel greater the PNR. If 10% fuel is increased, PNR distance will increase by 10% provided all other conditions remain same.
3) If fuel is carried in excess, the PNR may be beyond the destination.
RADIUS OF ACTION
Radius of Action abbreviated as (ROA) is defined as the maximum distance that an aircraft can travel out from a datum point along a specified track before returning to the same or a different datum point within a specified time.
The time so specified, also called the patrol time, may be dictated by either fuel considerations or operational requirements. The specified track that the ac requires to maintain is referred to as the patrol track. When the ac returns to the start point on completion of the patrol, it is a case of ROA to a fixed base. In contrast, when the ac departs a certain base and lands at a different base or the same base (such as an aircraft carrier) on completion of the patrol, it is considered a case of ROA to a moving base.
It has the following uses.
(i) To find out whether a target is within reach from the operating base or not.
(ii) To choose an alternate base for landing after carrying out the task over a target.
(iii) After a patrol to find the time to turn and to reach a moving base at a specified time.
ROA – FIXED BASE
In calculations of ROA, the critical factor is the patrol time. This may be considered analogous to the safe endurance associated with Point of Safe Return (PSR). When the ac is on patrol:
Time (out) + Time (back) = Patrol time
Let X represent the maximum distance away from departure base that the ac can reach before returning to base within the patrol time, designated by P, O and H have their standard connotations of ground speed “Out’ and ground speed ‘Home’.
Time (out) = X / O
Time (back) = X / H
X / O + X / H = P
Transposing of the terms yields the value of X.
X = POH/ O+H
Hence, Radius of action is the product of total patrol time, ground speed out and ground speed home divided by sum of ground speed out and ground speed home. In the calculation of Radius of action the ground speeds calculated are at full TAS and not at reduced TAS as in calculation of PET.
An important aspect of flight planning is the calculation of action to be taken in the event of emergency or diversion. The decision to be made is that with the available fuel and knowledge of wind velocity, will it be preferable to return to base or continue to the destination and indeed which of these options is viable and feasible. The knowledge and calculation of these actions is made easy with the calculation of point of equal time, point of safe return and radius of action before the flight is undertaken so as to save the valuable time in the air.