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#121
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MoGas Tips, Tricks, Concerns, How To
It is close, as TEL is only about 2/3 lead by weight. The molecular
weight of TEL is 323 according to my chemistry book, and the tetraethyl part is about 116. The net numbers work out as you indicated. The spcific density of TEL is 1.659 gm/ml. |
#122
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MoGas Tips, Tricks, Concerns, How To
That's indeed quite a bit, huh? If I take a 1/3lb lead ball to a C-206 flyer and tell him that's how much lead there is in his full tanks, I bet he'll be surprised. nrp wrote: A couple of points: 1) My copy of ASTM spec D910 page 0539 states that the allowable TEL content of 100LL avgas is 2 milliliters per gallon NOT 2 grams per gallon. This means there will be over 1/3rd pound of lead in an 80 gallon tank of 100LL, plus about 1/5th pound of ethlylene dibromide scavenging agent. |
#123
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MoGas Tips, Tricks, Concerns, How To
"M" wrote in message ups.com... That's indeed quite a bit, huh? If I take a 1/3lb lead ball to a C-206 flyer and tell him that's how much lead there is in his full tanks, I bet he'll be surprised. How big do you think that ball would be? |
#124
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MoGas Tips, Tricks, Concerns, How To
Matt Barrow wrote:
"M" wrote in message ups.com... That's indeed quite a bit, huh? If I take a 1/3lb lead ball to a C-206 flyer and tell him that's how much lead there is in his full tanks, I bet he'll be surprised. How big do you think that ball would be? Roughly somewhere between the size of a candy jaw breaker and a big marble (I know there is a name for the big ones, but since I lost all my marbles, I can't remember what it is). -- Jim Pennino Remove .spam.sux to reply. |
#125
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MoGas Tips, Tricks, Concerns, How To
("Matt Barrow" wrote)
That's indeed quite a bit, huh? If I take a 1/3lb lead ball to a C-206 flyer and tell him that's how much lead there is in his full tanks, I bet he'll be surprised. How big do you think that ball would be? "I know, I know. Me. Me. Call on Me!" http://www.tesarta.com/www/resources...y/weights.html Lead: [Cubic foot = 1,728 cubic inches] 710 lbs / cubic foot 1 lb. = 2.43 cubic inchs of lead x 0.33 lbs. = 0.8 c.i. (cubic inches of lead) http://www.1728.com/diam.htm Volume of a sphere ..."I knew where to look it up!" 1.15 inches [Diameter] seems to be close for 0.8 cubic inches. A 25 cent [Quarter] measures 92% of an inch across. About 15/16 - not quite an inch. (I squinted because I needed 0.92! g) ..92 [Diameter] into the sphere equation = 0.4 c.i. (x 2) = 0.8 c.i. So, two spheres, each with a diameter of a 25 cent [Quarter] = 0.8 cubic inches ...which = 1/3 lb of lead. One lead ball for each side! Montblack Time for Gym. Heading outside for my 3 mile walk.... |
#126
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MoGas Tips, Tricks, Concerns, How To
"Montblack" wrote in message ... ("Matt Barrow" wrote) That's indeed quite a bit, huh? If I take a 1/3lb lead ball to a C-206 flyer and tell him that's how much lead there is in his full tanks, I bet he'll be surprised. How big do you think that ball would be? "I know, I know. Me. Me. Call on Me!" http://www.tesarta.com/www/resources...y/weights.html Lead: [Cubic foot = 1,728 cubic inches] 710 lbs / cubic foot 1 lb. = 2.43 cubic inchs of lead x 0.33 lbs. = 0.8 c.i. (cubic inches of lead) http://www.1728.com/diam.htm Volume of a sphere ..."I knew where to look it up!" 1.15 inches [Diameter] seems to be close for 0.8 cubic inches. A 25 cent [Quarter] measures 92% of an inch across. About 15/16 - not quite an inch. (I squinted because I needed 0.92! g) .92 [Diameter] into the sphere equation = 0.4 c.i. (x 2) = 0.8 c.i. So, two spheres, each with a diameter of a 25 cent [Quarter] = 0.8 cubic inches ...which = 1/3 lb of lead. One lead ball for each side! In the words of Don Rickles, "Very good; you win a cookie!" :~) |
#127
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MoGas Tips, Tricks, Concerns, How To
"I know, I know. Me. Me. Call on Me!"
http://www.tesarta.com/www/resources...y/weights.html Lead: [Cubic foot = 1,728 cubic inches] 710 lbs / cubic foot 1 lb. = 2.43 cubic inchs of lead x 0.33 lbs. = 0.8 c.i. (cubic inches of lead) http://www.1728.com/diam.htm Volume of a sphere ..."I knew where to look it up!" 1.15 inches [Diameter] seems to be close for 0.8 cubic inches. A 25 cent [Quarter] measures 92% of an inch across. About 15/16 - not quite an inch. (I squinted because I needed 0.92! g) .92 [Diameter] into the sphere equation = 0.4 c.i. (x 2) = 0.8 c.i. So, two spheres, each with a diameter of a 25 cent [Quarter] = 0.8 cubic inches ...which = 1/3 lb of lead. One lead ball for each side! So there you have it. We can now legitimately claim that using unleaded Mogas increases your useful load, too! :-) -- Jay Honeck Iowa City, IA Pathfinder N56993 www.AlexisParkInn.com "Your Aviation Destination" |
#128
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MoGas Tips, Tricks, Concerns, How To
("Matt Barrow" wrote)
So, two spheres, each with a diameter of a 25 cent [Quarter] = 0.8 cubic inches ...which = 1/3 lb of lead. One lead ball for each side! In the words of Don Rickles, "Very good; you win a cookie!" :~) How big's the cookie? Montblack Radius is all I need... g |
#129
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MoGas Tips, Tricks, Concerns, How To
Montblack wrote:
("Matt Barrow" wrote) So, two spheres, each with a diameter of a 25 cent [Quarter] = 0.8 cubic inches ...which = 1/3 lb of lead. One lead ball for each side! In the words of Don Rickles, "Very good; you win a cookie!" :~) How big's the cookie? ....and what's the lead content? |
#130
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MoGas Tips, Tricks, Concerns, How To
("Dave Butler" wrote)
How big's the cookie? ...and what's the lead content? Hmm? I thought, in cookies, Pb stood for peanut butter. Drat! Montblack "And we talked about some old times And we drank ourselves some beers Still crazy afler all these years Oh, still crazy after all these years" |
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