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Ok the imperial unit for the specific fuel consumption for a jet aircraft is shown below. Its the amount of fuel (in lbs) per hour required to produce 1 pound of thrust.

$$\frac{\frac{lbs}{h}}{lbs}$$

The metric unit is.

$$\frac{kg}{Ns}$$

I tried this.

$$\frac{\frac{lbs}{h}}{lbs}=\frac{lbs}{lbs}\cdot\frac{1}{h}=\frac{2.2\cdot kg}{2.2\cdot9.81\cdot N}\cdot\frac{1}{3600 \cdot s}=\frac{1}{9.81\cdot3600} \left[ \frac{kg}{Ns} \right]$$

But apparently it is. $$\frac{1}{2.2\cdot9.81\cdot3600} \left[ \frac{kg}{Ns} \right]$$

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    $\begingroup$ You need to be clear about the difference between pounds-force and pounds-mass. The two quantities you list have different dimensions: $lbs/h/lbs = 1/h = [T]^{-1}$, whereas $kg/N\cdot s = kg /(kg m/s) = s/m = [T] [L]^{-1}$. $\endgroup$ – Michael Brown Mar 8 '13 at 15:24
  • $\begingroup$ Also, a kg is 2.2 lbs, not the other way around. :) $\endgroup$ – Michael Brown Mar 8 '13 at 15:26
  • $\begingroup$ @MichaelBrown first the reason the lbs falls out is because the use of the same unit for mass and force in the imperial system. If you use kg-force instead of newton it will also fall out and leave only the second. Do you really think you could express the SFC in hertz. This is why the metric system is much clearer, it prevents these confusions. If you want just split the lbs-force in mass and acceleration. You will come to the same. $\endgroup$ – Aaron de Windt Mar 8 '13 at 22:38
  • $\begingroup$ @MichaelBrown I know a kg is 2.2 lbs. That's why there is a 2.2 in the numerator for a kg in the denominator which is afterwards converted to newton and a 2.2 in the denominator for the kg in the numerator. $1lbs=1/2.2kg$ $\endgroup$ – Aaron de Windt Mar 9 '13 at 1:22
  • $\begingroup$ "use of the same unit for mass and force in the imperial system" no this is wrong. Mass and force have different dimensions, they can't be measured in the same units in any system. You've written the SFC in Hertz. (NB: I grew up in the U.S. so I am familiar with imperial.) You're right about the 2.2 thing - I wrote that too late last night. $\endgroup$ – Michael Brown Mar 9 '13 at 3:21
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I think you are correct, and I think whatever source you've found quoting the factor of 2.2 is wrong. In fact a quick Google found this book on aircraft design that gives the conversion factor as 0.283 $\times$ 10$^{-4}$, which is just 1/(9.81*3600) without any rogue factors of 2.2.

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