Lockheed Martin loops Hercules at Farnborough 2018
As with the C-130 loop, if he kept a steady +1G all the way round, should cause no extra stress on the frame., surely? .... The Pax (if any) might not like it, much, though !
Join Date: Mar 2014
Location: WA STATE
Age: 78
Posts: 0
Likes: 0
Received 0 Likes
on
0 Posts
And tex johnson did a 1 g barrel roll in a 707 according to one of the test engineers who was knelling on the floor and taking pics out the window.
Of course one may not be able to do it in your microsoft simulator...
Bob Hoover routinely did 1 g loops in a rockwell aerocommander
He would have used "a bit more than 1g".
Hence my earlier query regarding how to do a 1g loop - it's not possible.
I wouldn't be throwing around "microsoft simulator" comments with the bunkum you just sprouted...
cheers
If you could perform a loop without pulling any more G than you experience when flying straight and level, then the world's air forces could save themselves a fortune by not bothering with all those expensive, cumbersome G-suits.
"Mildly" Eccentric Stardriver
CONSO. You do not have any idea of the (significant) difference between a loop and a barrel roll. I have tremendous respect for Bob Hoover, and what he could do with that aeroplane. However, his barrel rolls, like those of Tex in the 707, were not 1g, and no loop can ever be.
Helicopters get into the act as well.
Join Date: Mar 2014
Location: WA STATE
Age: 78
Posts: 0
Likes: 0
Received 0 Likes
on
0 Posts
Join Date: Oct 2017
Location: UK
Posts: 5
Likes: 0
Received 0 Likes
on
0 Posts
Let me tell you something CONSO.
In my flying career I have won three aerobatic competitions, one of them a major international.
Let me tell you something else - you exude the same substances from both ends of your body.
Perhaps this thread can now be closed.
In my flying career I have won three aerobatic competitions, one of them a major international.
Let me tell you something else - you exude the same substances from both ends of your body.
Perhaps this thread can now be closed.
I have followed this thread with interest regarding large aeroplanes flying aerobatic manoeuvres but also with some incredulity over the arguments regarding the 'g' used. If I may feed in some food for thought ...
To a pilot, 'g' is load factor = Lift/Weight. Therefore, if an aeroplane is flying at '1g' Lift=Weight. However, to a physicist, an object subject to 1g has an acceleration in a specified direction of 9.81m/sec2. Therefore, he/she would view an aircraft in straight and level, constant speed flight as being at 0g along all three axes. An accelerometer fitted to an aircraft to measure 'g' should, to a physicist, be calibrated to indicate 0g when level and static but, by convention and to be useful to a pilot as an indicator of load factor, it is calibrated to read +1.
Based on the above, I pose a question to those who have referred to a '1g' loop; what is your frame of reference? If it is 0g when straight and level then you could, if you had enough thrust, theoretically fly a loop with +1g indicated throughout.
For aircraft that have indications of longitudinal or lateral g, note that when at constant speed and under conditions of zero sideforce then the longitudinal and lateral 'g' respectively are both zero and when accelerating/decelerating and sideslipping these are true linear acceleration values.
To a pilot, 'g' is load factor = Lift/Weight. Therefore, if an aeroplane is flying at '1g' Lift=Weight. However, to a physicist, an object subject to 1g has an acceleration in a specified direction of 9.81m/sec2. Therefore, he/she would view an aircraft in straight and level, constant speed flight as being at 0g along all three axes. An accelerometer fitted to an aircraft to measure 'g' should, to a physicist, be calibrated to indicate 0g when level and static but, by convention and to be useful to a pilot as an indicator of load factor, it is calibrated to read +1.
Based on the above, I pose a question to those who have referred to a '1g' loop; what is your frame of reference? If it is 0g when straight and level then you could, if you had enough thrust, theoretically fly a loop with +1g indicated throughout.
For aircraft that have indications of longitudinal or lateral g, note that when at constant speed and under conditions of zero sideforce then the longitudinal and lateral 'g' respectively are both zero and when accelerating/decelerating and sideslipping these are true linear acceleration values.
Join Date: Mar 2014
Location: WA STATE
Age: 78
Posts: 0
Likes: 0
Received 0 Likes
on
0 Posts
What we have here is a failure to clearly communicate as to frame o reference My comments did NOT adequately explain that MY reference to 1 G was relative to the airplane, not the ' center ' of the earth or universe. Thus my mention of lack of or NO visual cues re the passenger ( or improbably- the pilot ) to determine the attitude of the airplane reference to the earth. 2nd miss-or incomplete communication has to do with how perfect is perfect re 1 G , ie 1.0 G or 1.00000 G. IOW using the airplane as a reference frame, 1 G downwards from head to butt ( in level flight ) is also technically 1 G aligned with the ' center ' of earth disregarding perfect arcs relative to earth surface, large deposits of iron ore, etc. With appropriate radius climb / turn at a specific speed- the 1 G vector airplane reference can still be from head to toe, etc. BUT relative to the universe or earth frame it is other than 1 g. , etc. which is why I used the term NET G as affects the passenger or pilot in a previous post..
BY the way Tex in his own words called his manuver a chandelle- not a ' barrel ' roll- and in some his not widely pubished comments indirectly referred to doin the same manuver in virtuall/y every plane h had been involved wih as a test pilot, eg B52, etc. And for those who want to believe that over lake washington was the first time he had done that, I have this neat floating bridge bridge for sale currently in cold storage in Hood Canal . Tex lived a long time by NOT being stupid or taking dumb risks with a new designed airplane NOT designed or stressed for aerobatics.
WE return you now to the game of ' whats YOUR reference frame ' for gravity ..
Last edited by CONSO; 21st Jul 2018 at 17:20. Reason: just noted preceeding post
Join Date: Oct 2017
Location: UK
Posts: 5
Likes: 0
Received 0 Likes
on
0 Posts
To a pilot, 'g' is load factor = Lift/Weight. Therefore, if an aeroplane is flying at '1g' Lift=Weight. However, to a physicist, an object subject to 1g has an acceleration in a specified direction of 9.81m/sec2. Therefore, he/she would view an aircraft in straight and level, constant speed flight as being at 0g along all three axes. An accelerometer fitted to an aircraft to measure 'g' should, to a physicist, be calibrated to indicate 0g when level and static but, by convention and to be useful to a pilot as an indicator of load factor, it is calibrated to read +1.
To a pilot, 'g' is load factor = Lift/Weight. Therefore, if an aeroplane is flying at '1g' Lift=Weight. However, to a physicist, an object subject to 1g has an acceleration in a specified direction of 9.81m/sec2. Therefore, he/she would view an aircraft in straight and level, constant speed flight as being at 0g along all three axes. An accelerometer fitted to an aircraft to measure 'g' should, to a physicist, be calibrated to indicate 0g when level and static but, by convention and to be useful to a pilot as an indicator of load factor, it is calibrated to read +1.
To a pilot, 'g' is load factor = Lift/Weight. Therefore, if an aeroplane is flying at '1g' Lift=Weight. However, to a physicist, an object subject to 1g has an acceleration in a specified direction of 9.81m/sec2. Therefore, he/she would view an aircraft in straight and level, constant speed flight as being at 0g along all three axes. An accelerometer fitted to an aircraft to measure 'g' should, to a physicist, be calibrated to indicate 0g when level and static but, by convention and to be useful to a pilot as an indicator of load factor, it is calibrated to read +1.
F = force
M = mass (quantity of matter per unit volume.
a = acceleration (rate of change of velocity)
Thus: for a given mass and g being gravitational acceleration the derived force is called weight.
Therefore if g were zero the aeroplane would be weightless.