|07-20-2004, 08:42 AM||#11|
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Hate to rain on your parade, but you require a slight mathematical correction.
Since most manufactures(save SimJet and Artes) require 5% by volume of turbine oil in the fuel, the correct calculation is as follows:
9.46L / .95 = 9.958L Kerosene/turbine oil mixture
9.958L - 9.46L = .498L or 498 ml of turbine oil
What you have calculated above is actually 4.76% turbine oil/kerosene mixture. So half a quart(473ml) of oil per jug is actually on the light side of 5%.
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|07-20-2004, 11:19 AM||#12|
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I am: Andrew Coholic
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Picky Picky.. I forgot that we want 5% of the total volume (oil included). Yup, its 496ml per big CT jug of kero.
I know the wren 54 will work as low as 2.5% - I imagine that much of a difference in the kj66's wont be a problem either (that is, having 0.0025 less oil than 5%)
I bet there is more varience than that in the restriction that feeds the bearings from the fuel intake, VS the temp and viscosity of the oil/kero, etc. There has to be some kind of allowed variation for this?
Andrew Coholic -MAAC #26287L
1/2A to giant scale, IMAC, SAM, R/C sport, turbine jets, Heli's...
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