I am having difficulty and struggling in trying to find a effective method in determining figures for pressure altitude. It is easy if the number is 29.4, but much more difficult if the number is 29.38. Any help or guidance would be greatly appreciated.

Okay, do this then: A standard day is always 29.92, so it remains constant. Then take the current altimeter reading for the airport in question, let’s say it’s 29.75. Take the 29.92 minus the current altimeter, multiply by 1000 and then add field elevation. Here is an example for you: (29.92 - 29.75) x 1,000 + 690 = 860ft Does that aid in your study?

I think he is talking about Figure 8 from the Airman Knowledge Testing Supplement. 29.38 does not appear in pressure altitude correction factor the table, so you have to look at 29.3 and 29.4. Altimeter Setting | Pressure altitude conversion factor 29.3 | 579 feet 29.4 | 485 feet If the question wanted you to determine pressure altitude with an altimeter setting of 29.35, you could average those two, but instead, with an altimeter setting of 29.38, you have to interpolate. Since 29.38 is 80% of the difference between 29.3 and 29.4 579 - 485 = 94 feet 94 * 80% = 75 feet 579 - 75 = 504 feet

Thank you. I know that formula. Here is the question on the test. Determine the pressure altitude with an indicated altitude of 1380 msl with an altitude setting of 28.22 at standard temperature. So I have taken 29.92 minus 28.22 = 1.7 (1,700 feet). I then took 1,380 indicated altitude with 1,700 feet and have 3080 feet. Here is the kicker, the answer is 2991... I have attached a photo.

From quizlet: 1. Since the altimeter setting that is given is not shown in FAA Figure 8, interpolation is necessary. Locate the settings immediately above and below the given value of 28.22" Hg: Altimeter Setting Conversion Factor 28.2 1,630 feet 28.3 1,533 feet 2. Determine the difference in the two conversion factors: 1,630 - 1,533 = 97 feet 3. Determine the amount of the difference to be subtracted from the 28.20" Hg conversion factor (2/10 of 97). 97.0 x .2 = 19.4 4. Subtract the amount of difference from the amount shown for the 28.20" Hg conversion factor: 1,630.0 - 19.4 = 1,610.6 5. Add the correction factor to the indicated altitude to find the pressure altitude: 1,610.6 + 1,380.0 = 2,990.6 feet MSL (pressure altitude) There's a typo in the example you posted. The author referenced 1650 when they should have referenced 1630. In common terms linear interpolation is being used to extract a value from tabular data. It's a common method and sometimes understood better if it is done graphically.

Thank you for the post of the problem. I understand the formula to get pressure altitude but I'm not feeling the interpolate process. I wonder if there will be a test question that would require interpolation.

A graphical example can demonstrate the problem. Call X0 28.2 and Y0 1,630 then X1 is 28.3 and Y1 is 1,533. If drawn to scale you could read y = 1611 for the given x= 28.22

Yeah it's a simple thing when the picture is drawn. The word problems make it sound like some huge math problem. Of course we can work through the math problem now if you want: remember that the equation of a line is y=mx+b where m is the slope and b is the y intercept. The slope is defined as (y1-y2)/(x1-x2). Once the slope has been calculated b can be determined using b = y1-mx1. The equation of the line is now fully determined and the unknown y for the known x can be calculated directly by the diligent student. Of course the equation only yields a valid linear interpolation between x1 and x2. Remember to do your homework and assigned reading for the next class meeting....

The process takes a bit getting used to, but after a few q and a, it will come naturally for test time. Thanks again.

Yeah. I just deleted that post. Reading the OP again I saw he got that. It was just when things got carried out a few decimal places that it got harder

Yeah, typically FAA test stuff that will never be used in practice. Heck if pressure is 28.22 chances are most of us aren't flying much less worried about pressure altitude.

Don’t know about test questions, but almost every POH out there has nasty small printed graphs that you’ll be sliding around to get performance numbers from, that require you to guess how far you are between two lines. Linear interpolation. One line says 700 feet, the next line says 800 feet and your ruler/pencil mark falls about 60/40 between them. What’s your takeoff roll distance? Helps immensely if you take said crappy POH to a copy machine and blow the graph up with the zoom function about twice as big as it is printed in the manual, if you’re trying to be accurate. Of course, you probably don’t need to be that accurate unless you’re cutting it really close on a limitation of the aircraft. Normally you’d just estimate whichever direction gives you more safety margin. Add a little for “mom, the kids, and the dogs” as they say.

Question: Why then, when these numbers are plugged into the Sporty's E6B Flight Computer (Electronic): 1380 ft Indicated Alt, 28.22 InHg Barometric Pressure, 15 deg Celsius True Temperature (the written problem says 59 deg Fahrenheit) does the computer return a result of 3080 ft Pressure Altitude?

Don't waste any more brain cells on this. First, it is an unrealistic question; second, the odds of that question being one of the sixty you are given (if it exists at all) are extremely small. ASA 's 2018 Test Prep book does not include this question or anything similar to it. If the worst happens and you get such a question, guess at it and accept the resulting grade of 98. Bob Gardner

www.faa.gov/training_testing/testing/media/whats_new_general.pdf Note what it says about interpolations. Bob

I want to thank everyone here for their suggestions and guidance on this particular matter. I am getting 93's on all my practice tests, and now I am fine tuning my studies. Interpolation questions have popped up here and there, so I felt it was worth a post on here.