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- zulugurl
- Topic Author

Hi guys!

I am in serious need of some help - it seems I keep making the same mistake - which leads me to believe I have learned something wrong!

In doing Bob's cyber exam,I came across this question:

Echo loading 18 - 21

An Echo aircraft has an empty weight and moment index of 1970 and 480 respectively.

It is loaded as follows:

Forward compartment 50kg

Row 1 Pilot & Passenger

Row 2 2 Passengers (assume 77kg all people)

If loading is not changd and no fuel added to the auxillary tanks, determine the max fuel that can be added to the mains without exceeding the Forward Limit of the COG

93 / 159 / 261 / 76

Can someone please show me how to do this so that I can get them right!

PS: I got 159????

I have my exam soon and really don't want to fail again!

Thanks guys,

I am in serious need of some help - it seems I keep making the same mistake - which leads me to believe I have learned something wrong!

In doing Bob's cyber exam,I came across this question:

Echo loading 18 - 21

An Echo aircraft has an empty weight and moment index of 1970 and 480 respectively.

It is loaded as follows:

Forward compartment 50kg

Row 1 Pilot & Passenger

Row 2 2 Passengers (assume 77kg all people)

If loading is not changd and no fuel added to the auxillary tanks, determine the max fuel that can be added to the mains without exceeding the Forward Limit of the COG

93 / 159 / 261 / 76

Can someone please show me how to do this so that I can get them right!

PS: I got 159????

I have my exam soon and really don't want to fail again!

Thanks guys,

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- bobtait
- Offline

- Posts: 1929
- Thank you received: 101

This problem cannot be solved by using a flow chart because if you are adding weight to get to the forward limit you cannot know which forward limit you want until you know what weight you are going to add. [catch 22]. Therefore you have to **use the C of G envelope.**

**Step 1. **

Do a load sheet with the present load. You will find the total weight is 2328 kg and the total moment index is 568.6.

**Step 2**

Plot 2328 and 568.6 on the envelope [not easy I know but do it as accurately as you can]. Call this Point A.

**Step 3**

Add 100 kg of fuel in the mains. New total is 2428 kg and 586.4 moment index. Plot this point and call it Point B.

**Step 4**

Now join point A to point B and note where it cuts the forward limit on the envelope. What weight corresponds to that point? It's about 2420. So the fuel that can be added is about 92 kg. [First answer].

*To prove you answer [if you want to], Check the C of G position with the 93 kg added to the mains. 2421 and 585.1 - 2417 mm.*

Now check the forward limit at this weight of 2421 kg. It is also 2417. So the present C of G is also the forward limit.

Do a load sheet with the present load. You will find the total weight is 2328 kg and the total moment index is 568.6.

Plot 2328 and 568.6 on the envelope [not easy I know but do it as accurately as you can]. Call this Point A.

Add 100 kg of fuel in the mains. New total is 2428 kg and 586.4 moment index. Plot this point and call it Point B.

Now join point A to point B and note where it cuts the forward limit on the envelope. What weight corresponds to that point? It's about 2420. So the fuel that can be added is about 92 kg. [First answer].

Now check the forward limit at this weight of 2421 kg. It is also 2417. So the present C of G is also the forward limit.

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- zulugurl
- Topic Author

Thank you Bob!

I hope I get one just like this in my exam now!

I hope I get one just like this in my exam now!

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- zulugurl
- Topic Author

Hi Bob,

This is another one i got wrong on the cyber exams:

Can you please tell me what is meant by 'specific' air range?

and walk me thru this one too.

Question 9 Cyber 1

An aircraft cruising at 10,000 ft density height in a 20kt headwind has an IAS of 175kt

If the fuel flow is 20 USg per hour, the specific air range is closest to:

â€¢ 0.10 kg / nm

â€¢ 0.27 kg / nm

â€¢ 0.29 kg / nm

â€¢ 0.31 kg / nm

Thank you

This is another one i got wrong on the cyber exams:

Can you please tell me what is meant by 'specific' air range?

and walk me thru this one too.

Question 9 Cyber 1

An aircraft cruising at 10,000 ft density height in a 20kt headwind has an IAS of 175kt

If the fuel flow is 20 USg per hour, the specific air range is closest to:

â€¢ 0.10 kg / nm

â€¢ 0.27 kg / nm

â€¢ 0.29 kg / nm

â€¢ 0.31 kg / nm

Thank you

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- zulugurl
- Topic Author

I might as well ask ytou about this one too! - please

The following data refers to a charter flight in an Echo:

Weight of empty aircraft plus pilot 2109kg

Fuel on board at take-off 300 litres

Flight fuels to first intended landing 110 litres

Fixed reserve 15 Gallons

Taxi allowance 3 gallons

The maximum weight of baggage (ignoring balance) that may be added is:

â€¢ 521 kg

â€¢ 629 kg

â€¢ 320 kg

â€¢ 481 kg

Being a charter flight, do we always allow for the 15% variable reserve even tho it is not mentioned by the examiner?

thanks again!

The following data refers to a charter flight in an Echo:

Weight of empty aircraft plus pilot 2109kg

Fuel on board at take-off 300 litres

Flight fuels to first intended landing 110 litres

Fixed reserve 15 Gallons

Taxi allowance 3 gallons

The maximum weight of baggage (ignoring balance) that may be added is:

â€¢ 521 kg

â€¢ 629 kg

â€¢ 320 kg

â€¢ 481 kg

Being a charter flight, do we always allow for the 15% variable reserve even tho it is not mentioned by the examiner?

thanks again!

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- bobtait
- Offline

- Posts: 1929
- Thank you received: 101

Yes. The exam stipulates that the company fuel policy is to adopt the recommendations of the CAAP. So any charter flight carries a variable reserve of 15%.

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- bobtait
- Offline

- Posts: 1929
- Thank you received: 101

Specific air range is the inverse of ground nautical miles per gallon [or kg or litre]

So instead of dividing the ground speed by the fuel flow, you divide the fuel flow by the ground speed. If Specific Air Range is mentioned then you simply use TAS instead of ground speed. So in this question you would convert 20 gallons to kilos then divide that by the TAS. That gives you kg/air nm.

So instead of dividing the ground speed by the fuel flow, you divide the fuel flow by the ground speed. If Specific Air Range is mentioned then you simply use TAS instead of ground speed. So in this question you would convert 20 gallons to kilos then divide that by the TAS. That gives you kg/air nm.

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- zulugurl
- Topic Author

Good morning Bob,

Thank you for your help with these - had my exam yesterday and passed! YES!

one more to go - Air Law.... and then it's mine! damn, it has been a hard slog - but so worthwhile. Should have done this thirty-something years ago when I started flying!

Thanks again for the support!

Cheers,

Judy

Thank you for your help with these - had my exam yesterday and passed! YES!

one more to go - Air Law.... and then it's mine! damn, it has been a hard slog - but so worthwhile. Should have done this thirty-something years ago when I started flying!

Thanks again for the support!

Cheers,

Judy

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- bobtait
- Offline

- Posts: 1929
- Thank you received: 101

Mate! You've made my day....

That's just fantastic. Wish I could be there to raise a glass with you!! Congratulations. I'm sure you'll get Air Law without too much trouble.

Keep having fun......

That's just fantastic. Wish I could be there to raise a glass with you!! Congratulations. I'm sure you'll get Air Law without too much trouble.

Keep having fun......

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- Richard
- Offline

Congratulations Judy!!!! Brilliant stuff

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