I’ve made a very rough estimate (based on the speed of sound) of how high planes are as they cross the Lake Michigan shoreline on their final approach to O’Hare. I think that it varies by type of aircraft and weather conditions, and, of course, where along the shore they’re crossing. I’m thinking about an ENE approach, but I’m thinking in the 1500 - 2500 foot range. I’m just wondering how accurate my very rough estimate is. Can anyone tell me? Thanks, Dopers. xo, C.
Someone with an instrument rating will probably be through here to verify this, but according to this ILS approach platefor runway 270 @ Ohare, the height of the airplane at about 4.5 NM from the airport is 2200 msl, or about 1600 feet above the ground.
The map shows the shoreline, and looks to be around 3500’ or so…
IIRC the glide slope for an aircraft on final approach is something like 3°. I’m not sure how far the shore is from the runways, but with that distance and a bit of trig you should be able to estimate how high the airplanes are.
Check if my math is correct…
I converted silk1976’s example of distance of 4.5 nautical miles to 5.17 statute miles.
Multiplying 5.17 × 5280 feet and multiply that by tangent of 3 degrees into Wolfram gives me 1430 feet:
So it looks like the answer is in the ballpark: 1430 ≈ 1600
I don’t know how many miles (feet) away Lake Michigan is from the runway but someone else can recalculate with the adjusted distance.
Here’s a photo taken from a plane close to the Lake Michigan shoreline as it was coming in to land at O’Hare. The tallest building visible is about 1,450 feet high.
You can use Google Earth for live ORD inbound flight tracking. http://fboweb.com/flighttracking/
Just click on a flight’s icon and a popup will appear with its altitude and other info. For example, at this moment, flight RJA263 shows just over the lake shore at 3800 ft., ETA in 3 minutes.
It’d be nice if you could have 2 GE windows open at once, from slightly different locations. That way you could see them in live 3-D! Of course, you can just take 2 screen shots for a static 3-D view.
For some reason I feel like I’m not reading the question correctly, and granted it has been a long time since I looked at an approach plate, but at the BONZO intersection, just inside the shore, the altitude of an airplane (whether 747 or Cessna 150) would be 4,000 feet above sea level, or roughly 3,400 feet about ground level. That’s for an instrument approach.
Looking at the Chicago Terminal Area Chart, the floor of the O’Hare class B airspace at the shoreline to the E and NE of O’Hare is 3000’ MSL, or about 2400’ above the ground. Jet powered aircraft going into O’Hare are prohibited from operating below the floor of this airspace, so the answer must be greater than that.
It looks like about 11 NM or so to the runway for aircraft landing to the west, and a little farther for aircraft landing to the southwest. I’d say they’re probably around 3000’ or so above the ground at that distance.
Rule of thumb for an aircraft height on approach is 300 x distance from the runway. This approximates the 3 degree glide slope that is most common though 320 x distance is more accurate. So at 11 miles from the runway you’d expect the aircraft to be a bit over 3300 feet above the ground. in fact, we use that 3 degree decent profile from all the way up at cruising altitude if we aren’t given other restrictions. To work out a top of decent point we just multiply height to lose by 3. This gives another approximation of a 3 degree slope, though it is a bit steeper than the 300 x distance one.
It is commonly believed that pilots need to have a brain for maths, in reality they just need to know their three times table ;).
**How high is a plane that’s landing at O’Hare? **
Or is this a trick question, with the answer being zero feet AGL?
I found this rather confusing until I figured out that it should be taken to mean “Height in feet is 300 times the distance in miles.”
320 x 11 = 3520 
Sorry about that. I don’t even think of it as 300 times distance, I think of it as 3 times distance so it could have been even more confusing. It’s really the height above the runway as well, so you need to add the runway elevation to get the aircraft altitude. And the distance should actually be nautical track-miles from touchdown, if you are on downwind and doing a constant descent angle approach, which is where you attempt to fly a constant descent angle from cruising altitude to touchdown with no level segments*, you might be 3 nm from the runway but still have 10 track-miles to go as you fly the legs of the circuit and therefore be about 3000 above the runway.
So just to clear it all up, when descending on a 3 degree angle,
Aircraft altitude (feet) = 320 x track-miles (nm) to touchdown + runway elevation (feet).
That’s just simplified in my brain to 3 x distance, then I add zeros as appropriate to give a reasonable answer. It is easiest to just punch 3 degrees into the flight management system and let the autopilot sort it out, but a safety conscious pilot will be doing the calculation themselves every now and then to ensure the aeroplane is doing what they think it is doing.
- Level segments require an increase in power which results in higher noise levels for those living under the flight path and higher fuel burn.