Nealius Posted June 13, 2022 Share Posted June 13, 2022 (edited) I've been trying to run the engine according to parameters on Chuck's Guide, namely: Max cont. 82% RPM at Baro+95 and Nominal 70% RPM at Baro+75 On an 11C day with QNH 762, that would be 857mmHg and 837mmHg respectively. However I am unable to get any higher than 800, maybe 810 when flying between 500~1000m. Am I doing something wrong, or is the manifold pressure drop-off supposed to be that early at such low altitudes? Or is the cooler weather to blame? Edited June 13, 2022 by Nealius Link to comment Share on other sites More sharing options...
Toriy Posted June 14, 2022 Share Posted June 14, 2022 (edited) At 1000 м: 762-84=678, at 500m 762 - 48 = 714. When the height changes by 10.5 meters, the atmospheric pressure changes by 1 mm Hg. The supercharger helps a bit. The first nominal value must be maintained +/- 15 excess to a height of 1000 meters at normal conditions. The second nominal value must be maintained +/- 15 to a height of 1800 meters. Edited June 14, 2022 by Toriy 1 Link to comment Share on other sites More sharing options...
Nealius Posted June 15, 2022 Author Share Posted June 15, 2022 Ah, so it's pressure at current altitude plus x-mmhg. The manuals omit that but it makes sense now that I think about it. Math isn't my strong suit, so bare with me for a moment: How did you get -84 at 1000m? I got -95mmhg. 500m checks out (47.6 rounded up). So at 500m max continuous QNH 762 I would need (762-48)+75 = 789 which is definitely doable. 18 hours ago, Toriy said: The first nominal value must be maintained +/- 15 excess to a height of 1000 meters at normal conditions. The second nominal value must be maintained +/- 15 to a height of 1800 meters. Can you go a little deeper on this? I don't follow the first/second nominals. I see they're also notated on Chuck's Guide (max cont./nominal 1, then nominal 2) but they aren't explained. I can't find them in the official DCS manual either. Link to comment Share on other sites More sharing options...
Toriy Posted June 15, 2022 Share Posted June 15, 2022 (edited) Atm pressure drops by 1 mmHg with a height of 10.5 m (approximately) (https://www.yaklass.ru/p/geografiya/6-klass/atmosfera-gazovaia-obolochka-zemli-5987236/atmosfernoe-davlenie-6007530/re-fce1085c-4895-4ee5-aae7-c6322b4d6823) sorry, its in Russian. 1. Let us find mmHg change at 500 m if QNH 762. a. 500 m : 10.5 m = 47.6 mm = 48 mm Hg. b. Determine the atm. pressure. at a height. 762 - 48 = 714 mmHg. 2. At 1000m. a. 1000 : 10.5=95,2 b. 762-95=667. However, some books state that according to the pressure change formula, it drops by about 1 mm every 12 meters. In this case, we receive those -84 (actually 83.3333)at 1000m. ( https://fb.ru/article/472406/kak-s-vyisotoy-izmenyaetsya-atmosfernoe-davlenie-formula-grafik ) Which statement is correct? I have no idea:) The correct formula is here ( https://ru.frwiki.wiki/wiki/Variation_de_la_pression_atmosphérique_avec_l'altitude ) Also, if air temperature drops, pressure increases - with every 1C for 0.28 mm Hg As for nominals, it is not like we as pilots have to maintain them, it is just about an aircraft and its engine capabilities. The engine must be able to keep the first nominal value +/- 15 excess to a height of 1000 meters at normal conditions. The second nominal value +/- 15 to a height of 1800 meters. Nominals are used when climbing and in horizontal flight with increased speed in difficult meteorological conditions. Edited June 15, 2022 by Toriy 1 1 Link to comment Share on other sites More sharing options...
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