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I just read David McCollough's "The Wright Brothers," and was surprised to find several references to the exact altitude attained by various Wright airplanes. For example, on page 235:

The next day came news from Potsdam, Germany, that Orville had flown to an altitude of 984 feet, higher than anyone had yet flown in an airplane.

The flight in question occurred in 1909. But the altimeter would not be invented until 1928. How can anyone be certain that such a specific altitude was reached? I don't see any instrumentation in the photos of their aircraft. Was the altitude calculated from the ground somehow?

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    $\begingroup$ While the barometric altitude was invented on that year, the way how altitude could be measured using a baometer has already been known for centuries. $\endgroup$ Mar 12, 2023 at 16:18
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    $\begingroup$ Kollsman's invention was the mechanism to accurately calculate and display altitude from barometric pressure, adjust for ground-level pressure, and temperature-compensate the aneroid. Pressure altimetry was already written into regulations by 1919. $\endgroup$
    – user71659
    Mar 13, 2023 at 20:05

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In my opinion the measurement referred to in your question could have come from either an "aneroid barometer (or barograph)" and/or from a "triangulation calculation."

In this article , referring to a flight about a year later, the pilot (Walter Richard Brookins) was able to fly his Wright Flyer to an altitude of 6175 ft. This was recorded from an aneroid barometer and also an engineering triangulation calculation.

(highlighting/underlining in the image below is mine)

Excerpt from the article mentioned above: enter image description here

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  • $\begingroup$ How did they know what angle to point the sights so that the plane would fly through the intersection? Wouldn't this mean they knew the altitude ahead of time? $\endgroup$
    – Michael
    Mar 14, 2023 at 22:53
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By triangulation. Knowing where the airplane is above the ground and knowing what the angle is between horizontal and the sightline vector pointing towards the plane, you can solve for the height of the resulting triangle and hence the altitude of the plane.

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    $\begingroup$ This is a classic translation problem. 984 feet is 299.9 meters. I'm willing to bet the original report was 300 meters. Which got translated into 984 feet. As 300 meters is a nice round number, that could have been a proxy for "299 - 301", "295 - 305", etc. or even "275 meters to 325 meters" - i.e., a range easily determined by triangulation with the tools of the time. In fact, if the measurement had been in the US, and they thought it was "984 feet plus or minus a bit" they would have rounded to a nice big 1,000 feet and then the foreign press would have reported 304 or 305 meters. $\endgroup$ Mar 12, 2023 at 3:43
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    $\begingroup$ @nuggethead you use two measurement points simultaneously, with a surveyed relation to each other. $\endgroup$
    – fectin
    Mar 12, 2023 at 15:53
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    $\begingroup$ @manassehkatz-Moving2Codidact There's a similar issue with people referring to "normal" body temperature as being 98.6. It was originally only two significant figures (37) in Celsius, but it turned into three significant figured when converted to Fahrenheit (and even 37 is overly precise, as "normal" body temperature varies from person to person, by time of day, etc. etc.). $\endgroup$ Mar 13, 2023 at 3:33
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    $\begingroup$ @Acccumulation: Even worse, Fahrenheit degrees are smaller (closer together) than Celsius degrees, amplifying the problem of introducing excess precision. So it's about 18x more precise, not just 10x. 37+-1 C is like 98.6 +- 1.8 F. (Some would argue that you should round 98.6 to 99, moving the center of your uncertainty range due to rounding error as well as the original uncertainty / variability, although presumably 37 wasn't right in the middle of typical range either.) Anyway yeah, naivety about sig figs is widespread, but multiple roundings can accumulate more error. $\endgroup$ Mar 13, 2023 at 12:53
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    $\begingroup$ @Accumulation: it only requires each observer to have an assistant with a reasonably accurate watch, which are synchronized before and after the experiment. They track the plane through their theodolite sights and read out numbers every few seconds like "Now: three-zero-six. Now: three-one-four.", while the assistant looks at their watch and writes down the time of each "now", and the number. When the timestamps of the two observers differ, you interpolate between them. I've been the assistant in somewhat similar measurements. $\endgroup$
    – Dronir
    Mar 13, 2023 at 13:51

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