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On Sat, 22 Jul 2006 11:35:27 -0500, Big John
wrote in :: Flight load factor Flaps up = +4.4 G's and -1.76 G's So, in inverted flight a C-172 has only a 76% of a G margin to carry additional G force. That isn't much. Thanks. |
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Larry Dighera wrote:
On Sat, 22 Jul 2006 11:35:27 -0500, Big John wrote in :: Flight load factor Flaps up = +4.4 G's and -1.76 G's So, in inverted flight a C-172 has only a 76% of a G margin to carry additional G force. That isn't much. Thanks. No, 176% of a G. Matt |
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On Sun, 23 Jul 2006 01:54:35 GMT, Matt Whiting
wrote in :: Flight load factor Flaps up = +4.4 G's and -1.76 G's So, in inverted flight a C-172 has only a 76% of a G margin to carry additional G force. That isn't much. Thanks. No, 176% of a G. No. A _margin_ of only a 76% of a G to carry G forces in addition to the one G natural force. |
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Larry Dighera wrote:
On Sun, 23 Jul 2006 01:54:35 GMT, Matt Whiting wrote in :: Flight load factor Flaps up = +4.4 G's and -1.76 G's So, in inverted flight a C-172 has only a 76% of a G margin to carry additional G force. That isn't much. Thanks. No, 176% of a G. No. A _margin_ of only a 76% of a G to carry G forces in addition to the one G natural force. When you are pulling negative G, there is no one natural G force. It takes -1 G of acceleration to counter gravity and get you to 0 G. You can then add -1.76 G of additional acceleration and still be within load limits. The negative G load factor is referenced to 0 G, not 1 G straight and level. Matt |
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In article ,
Matt Whiting wrote: When you are pulling negative G, there is no one natural G force. It takes -1 G of acceleration to counter gravity and get you to 0 G. You can then add -1.76 G of additional acceleration and still be within load limits. The negative G load factor is referenced to 0 G, not 1 G straight and level. Is negative G an up force or a down force? |
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![]() "john smith" wrote in message ... In article , Matt Whiting wrote: When you are pulling negative G, there is no one natural G force. It takes -1 G of acceleration to counter gravity and get you to 0 G. You can then add -1.76 G of additional acceleration and still be within load limits. The negative G load factor is referenced to 0 G, not 1 G straight and level. Is negative G an up force or a down force? I'll disagree, here. Straight level flight, right side up is 1 G Free-fall, so you are not touching the seat or seatbelt is O G Straight level flight upside down, is -1G If the plane is upside down, and pushes up elevator, until a100KG weight weighs 176KG, that is a -1.76 G factor. Not much extra for upside down flight, is it. -- Jim in NC |
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john smith wrote:
In article , Matt Whiting wrote: When you are pulling negative G, there is no one natural G force. It takes -1 G of acceleration to counter gravity and get you to 0 G. You can then add -1.76 G of additional acceleration and still be within load limits. The negative G load factor is referenced to 0 G, not 1 G straight and level. Is negative G an up force or a down force? It is an acceleration opposite the normal acceleration due to gravity, so it would be downward with respect to the airframe. Matt |
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On Sun, 23 Jul 2006 14:00:37 GMT, Matt Whiting
wrote in :: Larry Dighera wrote: On Sun, 23 Jul 2006 01:54:35 GMT, Matt Whiting wrote in :: Flight load factor Flaps up = +4.4 G's and -1.76 G's So, in inverted flight a C-172 has only a 76% of a G margin to carry additional G force. That isn't much. Thanks. No, 176% of a G. No. A _margin_ of only a 76% of a G to carry G forces in addition to the one G natural force. When you are pulling negative G, there is no one natural G force. While you are in the vicinity of the Earth, your aircraft is being acted upon by the Earth's one G gravitational force. If you are flying straight and level while inverted, the airframe is experiencing -1G, not 0 G. That provides the remaining 0.76 (76%) of a G of the C-172's negative load factor specification of -1.76 to carry the load of any acceleration that may subsequently occur. It takes -1 G of acceleration to counter gravity and get you to 0 G. Agreed. You can then add -1.76 G of additional acceleration and still be within load limits. That analysis presumes the aircraft is not inverted. The negative G load factor is referenced to 0 G, not 1 G straight and level. Are not both the positive and negative load factors referenced to 0 G? You don't set your G-meter to 0 when you are on the taxiway; you set it to read 1 G, right? We were discussing the negative load that might be encountered during the inverted recovery from a barrel roll, so the earth's gravity would add to any accelerative force while the aircraft is inverted, right? |
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![]() Larry Dighera wrote: We were discussing the negative load that might be encountered during the inverted recovery from a barrel roll If the barrel roll is performed correctly it is a +G maneuver throughout, even while inverted. |
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Larry Dighera wrote:
On Sun, 23 Jul 2006 14:00:37 GMT, Matt Whiting wrote in :: Larry Dighera wrote: On Sun, 23 Jul 2006 01:54:35 GMT, Matt Whiting wrote in :: Flight load factor Flaps up = +4.4 G's and -1.76 G's So, in inverted flight a C-172 has only a 76% of a G margin to carry additional G force. That isn't much. Thanks. No, 176% of a G. No. A _margin_ of only a 76% of a G to carry G forces in addition to the one G natural force. When you are pulling negative G, there is no one natural G force. While you are in the vicinity of the Earth, your aircraft is being acted upon by the Earth's one G gravitational force. If you are flying straight and level while inverted, the airframe is experiencing -1G, not 0 G. That provides the remaining 0.76 (76%) of a G of the C-172's negative load factor specification of -1.76 to carry the load of any acceleration that may subsequently occur. If you had said 76% of a negative 1 G, I'd have agreed with you. :-) Just saying of a G, implies positive 1 G by most conventions. Matt |
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