![]() |
If this is your first visit, be sure to check out the FAQ by clicking the link above. You may have to register before you can post: click the register link above to proceed. To start viewing messages, select the forum that you want to visit from the selection below. |
|
|
Thread Tools | Display Modes |
#4
|
|||
|
|||
![]()
Ok Going out on a limb here.....
The ballasted ship is heavier. The ballasted ship has a flatter glide angle. So an I correct to assume that the unballasted ship must pull MORE G to get the same climb angle as the ballasted ship. If the unballasted ship needs to put more +g on the airplane to achieve the same climb angle, then it will bleed energy thru drag a bunch faster than the less g loaded ballasted ship. In fact, the unballasted ship must pull up a little bit to momentarily match the flatter glide angle of the ballasted ship. Scott "szd41a" wrote in message ... HMMMM this is getting puzzling!?1 Let's get back to the basics 1. Total energy at the start MUST equal total energy at the end + drag(defined here as interplay between inertial and viscous forces due to the nature of the fluid in which we play (a mix of nitrogen, oxygen and carbon dioxyde). No matter how lift, weigth or drag acts,at what rate, more at the beginning less at the end , the above will always be true 2. Let us forget about the difference of stall speed, and drag, the above statement yields 130 metres gain of height for my Jantar in laboratary vacuum and let us attribute it to the balalsted glider. 3 Now the pro ballasted say that there is a detectable difference between ballasted and dry.Let us say 25 metres, that 75 feet is readable on the clock. So the dry glider only went to 105 metres. 4 The difference is equivalent to the work necessary to raise 100 kgs (weight of ballast) on 25 metres which is (100 * 9.81 * 25) that is 24 500 Joules. 5. The force doing the work to hold the dry glider has to be drag, no???. So someone has to come with a demonstration. Because No 1 has to apply no matter what, and is easy to evaluate P1+ K1=P2+K2. "Derrick Steed" a écrit dans le message de ... Hmmm... maybe, but not. All you have demonstrated, in the same way that Galileo did, is the equivalence between gravitational mass and inertial mass (physics 101), and you should note it only applies in a vacuum where the effects of drag can be neglected, and as you rightly state it has nothing to do with kinetic or potential energy. However, as a later post explains there are other factors at work due to the aerodynamic effects which must apply to any real world glider, namely the interplay between the inertial and viscous forces due to the nature of the fluid in which we play. Rgds, Derrick. |
Thread Tools | |
Display Modes | |
|
|
![]() |
||||
Thread | Thread Starter | Forum | Replies | Last Post |
VOR/DME Approach Question | Chip Jones | Instrument Flight Rules | 47 | August 29th 04 05:03 AM |
"Cleared Straight-In Runway X; Report Y Miles Final" | Jim Cummiskey | Piloting | 86 | August 16th 04 06:23 PM |
For Keith Willshaw... | robert arndt | Military Aviation | 253 | July 6th 04 05:18 AM |
Air Force Museum Working Group to release final report | Otis Willie | Military Aviation | 0 | November 18th 03 12:28 AM |
Special Flight Setup Question (COF) | Dudley Henriques | Simulators | 4 | October 11th 03 12:14 AM |