Physics for the Laymen: Dives and Gs(Help!)

[HoHun]

Hi Badboy,

>Hope that helps.

Hm, actually it confuses me. What's the physical background of your curves? Using the calculation I outlined initially, I get a pull-out radius of 2088 ft for a 500 mph/8 G recovery, which should be equal to the pull-out altitude in a 90° dive. Your chart indicates 1000 ft more, though.

I'm not surprised you don't agree with my simplistic calculation, but I'd have expected the accurate value to be lower than my estimate, not higher :-)

(The reason would be the energy loss due to pulling Gs, as pointed out by Bozon.)

Regards,

Henning (HoHun)

[Badboy]

Originally posted by HoHun
Hi Badboy,

I'm not surprised you don't agree with my simplistic calculation, but I'd have expected the accurate value to be lower than my estimate, not higher :-)

Yep, that's because I posted the chart without checking it. The calculations were actually in knots and because the chart was originally produced for the recovery from a dive bomb delivery it also included a frag pattern allowance.

In this chart I have removed the safe frag clearance altitude and corrected for mph. For a 90° dive at 500mph, your figure is correct. Well spotted.


 


Sorry for any confusion!

Badboy

[Kweassa]

Terms and terms!

 What's the differece between knots and mph? How do they convert?

 Also, what does "frag clearance" mean??

[Badboy]

Originally posted by Kweassa
Terms and terms!

 What's the differece between knots and mph? How do they convert?

 Also, what does "frag clearance" mean??

The Knot in correct usage denotes a speed of one nautical mile per hour. Nautical miles are defined as the average length of one minute of latitude on the earths surface. There are two values for the knot, one is exactly 6080ft and the other is 1852m or about  6076ft. This is the international value, and the one I was using. The statute mile is of course 5280ft so the conversion factor is just 6076/5280, or approximately 1.151 statute miles to the nautical mile.

The frag distance is illustrated in this diagram, along with the equation for the radius. If you have any questions about how that equation was derived, just ask. However, the curves in my diagram, take into account factors that aren't included in that equation, for example, the reduction in g when the aircraft becomes g limited.    


 

Hope that helps.

Leon "Badboy" Smith