# Using ACN Tables

Thread Starter

Join Date: Nov 2009

Location: PK

Posts: 187

**Using ACN Tables**

Hi

I have 2 queries regarding the ACN tables.

Q.1. Determining the ACN if the mass of aircraft is different from that given in the ACN tables?

e.g. An A/C XYZ (in ACN Tables) has a Mass of 350,000 and a corresponding ACN of 90.

If an A/C which is of same type as above (i.e. XYZ) has a different mass value i.e. 340,000, then can we adjust the ACN value considering it a linear relation like:

340/350 = 0.97 so that the ACN becomes 90 x 0.97 = 87.3

or the relation is not linear?

Q.2. Where or How is "Load on one main gear (given in %)" used in routine calculations?

Thanks

I have 2 queries regarding the ACN tables.

Q.1. Determining the ACN if the mass of aircraft is different from that given in the ACN tables?

e.g. An A/C XYZ (in ACN Tables) has a Mass of 350,000 and a corresponding ACN of 90.

If an A/C which is of same type as above (i.e. XYZ) has a different mass value i.e. 340,000, then can we adjust the ACN value considering it a linear relation like:

340/350 = 0.97 so that the ACN becomes 90 x 0.97 = 87.3

or the relation is not linear?

Q.2. Where or How is "Load on one main gear (given in %)" used in routine calculations?

Thanks

Join Date: Jan 2000

Location: FL410

Posts: 842

Q1) The tables give ACN values for two weights, one at the maximum total weight authorized and the other at the operating weight when empty. If an aircraft is operating at an intermediate weight, the ACN value can be calculated by a linear variation between the limits. Extrapolation is not permissible.

Thread Starter

Join Date: Nov 2009

Location: PK

Posts: 187

ok perhaps I didn't submit enough data, let me give an example (as given in Jeppesen manual):

Following is the formula to calcute the ACN.

This is the ACN table with data for the specific type.

Note the empty mass in the table is 160481 whereas in the example it is 157850. ACN empty corresponding to 160481 is 25 but one used in the example is 24.

So just wanted to confirm that if the aircraft you are working on has different maximum and empty masses from those that are published in the table then is it ok to determine the corresponding ACN by a linear relation.

e.g. considering the above example the empty mass of the a/c is quite close to the one published in the table, that's why ACN is just 1 less than published. What if the empty mass was 150,000 (as it actually is in many cases). Would the ACN in that case be:

150000 / 160481 = 0.93

0.93 x 25 = 23 (ACN empty)

Thanks

Following is the formula to calcute the ACN.

This is the ACN table with data for the specific type.

Note the empty mass in the table is 160481 whereas in the example it is 157850. ACN empty corresponding to 160481 is 25 but one used in the example is 24.

So just wanted to confirm that if the aircraft you are working on has different maximum and empty masses from those that are published in the table then is it ok to determine the corresponding ACN by a linear relation.

e.g. considering the above example the empty mass of the a/c is quite close to the one published in the table, that's why ACN is just 1 less than published. What if the empty mass was 150,000 (as it actually is in many cases). Would the ACN in that case be:

150000 / 160481 = 0.93

0.93 x 25 = 23 (ACN empty)

Thanks

**Prof. Airport Engineer**

Join Date: Oct 2000

Location: Australia (mostly)

Posts: 726

Haroon,

An over-simplification of the ACN system, useful because it does explain it well, is that on a strong runway, the ACN is simply twice the wheel load in tonnes. So your 777, with 12 main gear wheels, at 300 tonnes MTOW, has 25 tonnes per single wheel which is an ACN of 25 x 2 = 50. And a linear relation of weight to ACN is fine.

The runway [pavement system] is modelled as a linear elastic layer system. That means that if the wheel load is 25 tonnes and the response is 50, then a wheel load of 28 tonnes means a response of 56.

Ok – it is not quite as simple as that. The load on the main gear wheels is a bit less because some load is taken by the nose gear. Then if you have a weaker subgrade (more so as you move weaker and down to Very low CBR = 3%), the runway pavement gets pretty thick and you find that the loading from some of the other wheels in the gear start to load the pavement under your wheel, so their loads have to get added to this wheel load. You end up with an “equivalent single wheel load” which adds all these contributions together, and which is why for weak subgrades, the ACN is now pretty much double what it was for strong subgrades, say 100. No need to understand that in detail, because for a given subgrade CBR, the relation of weight to ACN is linear and you can use the tables to calculate it.

For your question about Jeppesen, the explanation is that the ACN is just a straight-line graph with ACN vs aircraft weight. And it is a linear relation. The graph doesn’t change for a given aircraft and undercarriage layout. It can be interpolated. It can be extrapolated. Since aircraft weights change with time and variant (but provided that the gear layout stays the same), the graph still applies and you can look up or calculate the ACN. Most published tables tend to be out of date as aircraft weights creep up with time. Here is my example of the 777 graph.

ACN chart 2009

In your Jeppesen table with 160,481 kgs mass, and Medium CBR 10 subgrade, they gave an ACN of 25. Checking with the same graph (and software) for the 777-300, 157,850 kgs mass, and Medium CBR 10 subgrade gives an ACN of 24, which is the same as your example. Either numbers are valid for interpolation (or extrapolation). Shucks, the plane could have a gold plated interior and an operating mass empty of 200,000kgs and its ACN at 200,000 kgs mass and Medium CBR 10 subgrade would then be 33. That is still valid for interpolation when used with the MTOW numbers. Let me run the software and throw some more numbers out to check, for medium CBR 10 subgrade: 250t mass = ACN 45, 280t = 53. Empty mass 150,000kg = ACN 23 (as you correctly calculated).

Umm - I also find that working out the interpolation from the tables is not that easy so I either use the many graphs available at Airport Engineering or I use airport pavement design software.

An over-simplification of the ACN system, useful because it does explain it well, is that on a strong runway, the ACN is simply twice the wheel load in tonnes. So your 777, with 12 main gear wheels, at 300 tonnes MTOW, has 25 tonnes per single wheel which is an ACN of 25 x 2 = 50. And a linear relation of weight to ACN is fine.

The runway [pavement system] is modelled as a linear elastic layer system. That means that if the wheel load is 25 tonnes and the response is 50, then a wheel load of 28 tonnes means a response of 56.

Ok – it is not quite as simple as that. The load on the main gear wheels is a bit less because some load is taken by the nose gear. Then if you have a weaker subgrade (more so as you move weaker and down to Very low CBR = 3%), the runway pavement gets pretty thick and you find that the loading from some of the other wheels in the gear start to load the pavement under your wheel, so their loads have to get added to this wheel load. You end up with an “equivalent single wheel load” which adds all these contributions together, and which is why for weak subgrades, the ACN is now pretty much double what it was for strong subgrades, say 100. No need to understand that in detail, because for a given subgrade CBR, the relation of weight to ACN is linear and you can use the tables to calculate it.

For your question about Jeppesen, the explanation is that the ACN is just a straight-line graph with ACN vs aircraft weight. And it is a linear relation. The graph doesn’t change for a given aircraft and undercarriage layout. It can be interpolated. It can be extrapolated. Since aircraft weights change with time and variant (but provided that the gear layout stays the same), the graph still applies and you can look up or calculate the ACN. Most published tables tend to be out of date as aircraft weights creep up with time. Here is my example of the 777 graph.

ACN chart 2009

In your Jeppesen table with 160,481 kgs mass, and Medium CBR 10 subgrade, they gave an ACN of 25. Checking with the same graph (and software) for the 777-300, 157,850 kgs mass, and Medium CBR 10 subgrade gives an ACN of 24, which is the same as your example. Either numbers are valid for interpolation (or extrapolation). Shucks, the plane could have a gold plated interior and an operating mass empty of 200,000kgs and its ACN at 200,000 kgs mass and Medium CBR 10 subgrade would then be 33. That is still valid for interpolation when used with the MTOW numbers. Let me run the software and throw some more numbers out to check, for medium CBR 10 subgrade: 250t mass = ACN 45, 280t = 53. Empty mass 150,000kg = ACN 23 (as you correctly calculated).

Umm - I also find that working out the interpolation from the tables is not that easy so I either use the many graphs available at Airport Engineering or I use airport pavement design software.

*Last edited by OverRun; 30th Sep 2015 at 23:14.*

Thread Starter

Join Date: Nov 2009

Location: PK

Posts: 187

Thanks OverRun

That makes it clear, however I am still not quite clear about the load on one gear i.e. 47.4% given in the table I posted. I dont understand this:

For interpolation from the tables I use excel and probably going to write a code in java script to make an app for my cell phone

Regards

That makes it clear, however I am still not quite clear about the load on one gear i.e. 47.4% given in the table I posted. I dont understand this:

if you have a weaker subgrade (more so as you move weaker and down to Very low CBR = 3%), the runway pavement gets pretty thick and you find that the loading from some of the other wheels in the gear start to load the pavement under your wheel, so their loads have to get added to this wheel load. You end up with an “equivalent single wheel load” which adds all these contributions together, and which is why for weak subgrades, the ACN is now pretty much double what it was for strong subgrades...

Regards

**Prof. Airport Engineer**

Join Date: Oct 2000

Location: Australia (mostly)

Posts: 726

Haroon,

The load on the main gear is calculated at maximum aft CG at the maximum ramp weight. For your 777-300 baseline, this is 94.84% of the aircraft weight. With two main gear legs, this is 94.84/2 = 47.4% load on one gear.

Turning to your quote, ACN is numerically equal 2 times the equivalent load in tons on a single wheel with a standard tire inflation pressure of 12.5 bar (181.3 PSI) that would require the same pavement thickness as required by the aircraft for 10,000 coverages. To get this concept to work for multi-wheel gears, ACN is based on equivalent single wheel load (ESWL), which is intended to be the single wheel load which will cause pavement damage equal to that caused by a multi-wheel gear. In practice the ESWL is defined as the load on a single tyre which will produce the same maximum deflection at subgrade level as the multi-wheel load. The ESWL is a function of pavement depth. The current method for calculating the ESWL uses the single-layer Boussinesq expression for deflection beneath, and at offsets from, a uniformly loaded circular area. Basically for each aircraft and weight, pavement designs are done for CBR 3, 6, 10, and 15 subgrades to find the required depth of pavement in each case, and the deflection at top of subgrade level. Then a single load is applied and varied until it causes the same deflection at subgrade level, and that weight is the ESWL, and ACN is twice the ESWL. That is getting well into pavement engineering and is confusing for most pavement engineers.

The load influence from a wheel spreads downwards and out in a 45 degree cone shape. The drawing below shows that for a thin pavement (which is all that is needed for a strong subgrade), there is no stress overlap from adjoining tyres and the ESWL is the wheel load. As the subgrade gets weaker, the pavement gets thicker, the stresses overlap and they add together.

The load on the main gear is calculated at maximum aft CG at the maximum ramp weight. For your 777-300 baseline, this is 94.84% of the aircraft weight. With two main gear legs, this is 94.84/2 = 47.4% load on one gear.

Turning to your quote, ACN is numerically equal 2 times the equivalent load in tons on a single wheel with a standard tire inflation pressure of 12.5 bar (181.3 PSI) that would require the same pavement thickness as required by the aircraft for 10,000 coverages. To get this concept to work for multi-wheel gears, ACN is based on equivalent single wheel load (ESWL), which is intended to be the single wheel load which will cause pavement damage equal to that caused by a multi-wheel gear. In practice the ESWL is defined as the load on a single tyre which will produce the same maximum deflection at subgrade level as the multi-wheel load. The ESWL is a function of pavement depth. The current method for calculating the ESWL uses the single-layer Boussinesq expression for deflection beneath, and at offsets from, a uniformly loaded circular area. Basically for each aircraft and weight, pavement designs are done for CBR 3, 6, 10, and 15 subgrades to find the required depth of pavement in each case, and the deflection at top of subgrade level. Then a single load is applied and varied until it causes the same deflection at subgrade level, and that weight is the ESWL, and ACN is twice the ESWL. That is getting well into pavement engineering and is confusing for most pavement engineers.

The load influence from a wheel spreads downwards and out in a 45 degree cone shape. The drawing below shows that for a thin pavement (which is all that is needed for a strong subgrade), there is no stress overlap from adjoining tyres and the ESWL is the wheel load. As the subgrade gets weaker, the pavement gets thicker, the stresses overlap and they add together.

*Last edited by OverRun; 2nd Oct 2015 at 11:45.*

Join Date: Jun 2013

Location: Or-E-Gun, USA

Posts: 328

**@Haroon...Why?**

@Haroon, Sir: If your interest here is purely academic, go for it. OTOH, if the interest is an attempt to push the limits of your airplane or RWY-x, please reconsider. When dealing with flying machines, pushing performance to the absolute published limits or beyond them is a great way to kill yourself, your SLCs and/or seriously wrinkle an otherwise fine airplane. Please be very careful how you apply whatever you learn. This stuff can bite back.

Join Date: Jan 2000

Location: FL410

Posts: 842

Thread Starter

Join Date: Nov 2009

Location: PK

Posts: 187

Official Documents (Jeppesen in this case) gives the mass:

For 777-200 ER:

Empty 138346 - Max T/O 298464 (Jeppesen)

Empty 129200 - Max T/O 274000 (My Fleet)

For 777-200 LR:

Empty 145150 - Max T/O 348359 (Jeppesen)

Empty 140800 - Max T/O 340200 (My Fleet)

For 777-300 ER:

Empty 167829 - Max T/O 352441 (Jeppesen)

Empty 150500 - Max T/O 340200 (My Fleet)

The example in Jeppesen for 777-300 gives empty 157850 whereas the empty in the table is 160481. Instead if using ACN 25 they have used 24 (as also explained and confirmed by OverRun through performance software tool). Weight differential in my case is much more, so the need

For 777-200 ER:

Empty 138346 - Max T/O 298464 (Jeppesen)

Empty 129200 - Max T/O 274000 (My Fleet)

For 777-200 LR:

Empty 145150 - Max T/O 348359 (Jeppesen)

Empty 140800 - Max T/O 340200 (My Fleet)

For 777-300 ER:

Empty 167829 - Max T/O 352441 (Jeppesen)

Empty 150500 - Max T/O 340200 (My Fleet)

The example in Jeppesen for 777-300 gives empty 157850 whereas the empty in the table is 160481. Instead if using ACN 25 they have used 24 (as also explained and confirmed by OverRun through performance software tool). Weight differential in my case is much more, so the need