CPL Theory - ETP & PNR Calculations
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CPL Theory - ETP & PNR Calculations
I was wondering if anyone can shed some light on Critical Point (Cp)/Equal Time Point (ETP) & Point of No Return (PNR) with doing simple calculations?
I understand the theory behind it;
The PNR is a fuel consideration and the ETP/CP is a time consideration.
The ETP will always move into the wind and with PNR any wind component, either headwind or tailwind, will reduce the distance to the PNR relative to the nil-wind position.
Later on during the year I will be sitting the CPL Theory test and need to have good understanding for doing the calculations.
Thanks
AW
<img src="confused.gif" border="0">
I understand the theory behind it;
The PNR is a fuel consideration and the ETP/CP is a time consideration.
The ETP will always move into the wind and with PNR any wind component, either headwind or tailwind, will reduce the distance to the PNR relative to the nil-wind position.
Later on during the year I will be sitting the CPL Theory test and need to have good understanding for doing the calculations.
Thanks
AW
<img src="confused.gif" border="0">
Join Date: Jul 2001
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Equal time point depends upon groundspeed outbound from (lets say) point A to point B and groundspeed back inbound to point A.
Lets say groundspeed outbound A-B is 360knots and inbound B-A is 300knots. Distance A-B is 1800nm.
Now assume the equal time point is distance X from A, at this point the time to return to A is equal to that time to carry on to B. So the calculation involves simple time distance formula :
Time = Distance/Speed
So X/300 = Time to return to A from point X
and (1800-X)/360 = Time to carry on to point B
By equating these two we get.
(X/300) = (1800-X)/360
lets expand brackets
X/300 = (1800/360) - (X/360)
now multiply through by 360
X(360/300) = 1800 - X
group X on LHS
X(1 + 360/300) = 1800
Therefore :
X = 1800/(1+360/300)
= 818.18 nautical miles
I hope you followed. In general then you can write the ETP formula as :
X = D/(1+O/H) where D is distance A-B
O is groundspeed outbound
H is groundspeed home to point A
Or you may see it written as :
X = DH/(O + H)
Which is simply obtained by multiplying the numerator and denominator of RHS
of formula above by H.
Bit long winded I know but hope this helps.
[ 09 January 2002: Message edited by: The Greaser ]
[ 10 January 2002: Message edited by: The Greaser ]</p>
Lets say groundspeed outbound A-B is 360knots and inbound B-A is 300knots. Distance A-B is 1800nm.
Now assume the equal time point is distance X from A, at this point the time to return to A is equal to that time to carry on to B. So the calculation involves simple time distance formula :
Time = Distance/Speed
So X/300 = Time to return to A from point X
and (1800-X)/360 = Time to carry on to point B
By equating these two we get.
(X/300) = (1800-X)/360
lets expand brackets
X/300 = (1800/360) - (X/360)
now multiply through by 360
X(360/300) = 1800 - X
group X on LHS
X(1 + 360/300) = 1800
Therefore :
X = 1800/(1+360/300)
= 818.18 nautical miles
I hope you followed. In general then you can write the ETP formula as :
X = D/(1+O/H) where D is distance A-B
O is groundspeed outbound
H is groundspeed home to point A
Or you may see it written as :
X = DH/(O + H)
Which is simply obtained by multiplying the numerator and denominator of RHS
of formula above by H.
Bit long winded I know but hope this helps.
[ 09 January 2002: Message edited by: The Greaser ]
[ 10 January 2002: Message edited by: The Greaser ]</p>
Join Date: Oct 2000
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Why not try the simple graphical way for CP. Draw an equal time line between your two chosen airfields. Where it intercepts the track, measure the distance to either airfield, using your diversion TAS calculate the time, then apply your wind vector to that point for the diversion time; then draw a CP line parallel to your equitime line.
You can do the same for PNR by extending your track to the limit of endurance and applying the same graphics.
You can do the same for PNR by extending your track to the limit of endurance and applying the same graphics.