From: billphil@ix.netcom.com (Badwater Bill) Newsgroups: rec.aviation.homebuilt Subject: Re: How Many G's is your Prop Tip Under? Date: Sun, 21 Feb 1999 19:15:58 GMT Thank you all for participating. The answer is: 6500 to 7500 g's depending on the prop you have and the rpm. On a prop of 2 meter diameter spinning at 2400 rpm it's about 6500 g's If you work it out in the English system with a 72 inch prop at 2700 rpm (John Ampmeter did this one) it's about 7500 g's So, if you put one gram of paint on one side of your prop at the tip and spin it up, that results in about 7000 grams of weight differential or 7 kilograms which is about 15 pounds. I find that really amazing. My guess when first asked this question was about 100 g's. That sounds like a lot to me since I hurt when pulling 4 g's. But...7000 g's??? I'd have never guessed it. Here's the calculation: A=V^2 / R G's = A/9.8 m/s-s at 2400 rpm or 40 rps the tip travels 2 pi meters or 6.28 meters in one revolution, or 251 meters in one second. Square that and get 63000, divide by one meter radius (R) and you still have 63000. Divide that by 9.8 meters/s-s for one g on Earth and you get 6430 about. So, a prop whose radius is one meter spinning at 40 rps has a force on the tip of 6430 g's. BWB

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