> .0-3 bjbj}+}+ "(AAl\\\\hi ?N#
Z|ЗxR\
90iPreamble:
The Vector Diagram (pp. 132) in Jeff Goins definitive treatise The Powered Paragliding Bible got me wondering if I could computerize it. No big deal coming up with a spreadsheet, until. . . what, exactly, does Glide Ratio imply? Yours Truly was soon abashed: (try expounding on any subject, to an imaginary audience, and see if/when you run out of steam).
In retrospect I would have used a different style and tone in my original Explanation - it might be misconstrued as condescending or sarcastic That was not my intent, just an unprofessional attempt at trying to sound conversational.
Arbitrarily chosen limits are:
Airspeed: 10-50 miles per hour; 16-80 kilometers per hour
Wind: 0-50 miles per hour; 0-80 kilometers per hour
X/C distance: 200 miles; 300 kilometers
Glide Ratio: 5-20 to one
This version is intended to combine the In/Out-bound realities of an X/C flight.
Explanation:
Sometimes it's fun to imagine "what if? Writing computer applications can facilitate the process. For the sake of argument may we assume that, legally and/or otherwise, there is an "absolute altitude": 18,000 feet (AGL/ASL?) - something like Einstein's "c" (186,000 Miles/Second). Well, if I, me, my PPG, and, if absolutely necessary, "Bobby McGee" (acknowledgements to Kris Kristofferson) are at FL 180, what happens when the "fan" decides to take a "union coffee break"? Bobby can use her "dirty red bandana" to mop her brow; I'll go into my best Chuck Yeager-style trivialize-the-problem lingo: "Ah, Bobby, ummm, we're over Salinas right now and, ah, heh-heh, you can start searchin' for that 'home', uh, soon's we land this mother. But will you watch where you point that damned harpoon?!"
Sorry. Where was I? Oh, right, eighteen thou'. At a 6:1 glide ratio I can cover ((6 x 18000) feet ==> 20.5 miles). Simple, you say? Simple, say I. Simple I am. Isn't there something about "thinner air" upstairs? Ipso facto the glide will be less; plus what about Groundspeed? However, this is "let's pretend", so:
Is it OK if, on a flight from "A" to "B" the Air- and Wind- speeds remain constant? Further, given a wing's Airspeed (the glide ratio doesn't come into play just yet) and the Wind's contribution to Groundspeed, there is a point (say "X") between "A" and "B" where the time to "Return to Base" equals the time to "Press On". (In still air or a 90-degree crosswind "X" is midway.)
Let's take a simple case: the distance from "A" to "B" is 22 (Miles/Kilometers, it doesn't matter); Airspeed is 22; Windspeed is 10 and directly from "A" to "B".
Starting with the basics: "Distance = Time x Velocity" (D = T x V). Transposing gives "Time = Distance / Velocity" (T = D / V). If point "X" is the unknown distance from "A", then the distance from "X" to "B" is the distance from "A" to "B" minus "X": (22 - "X"); D1 and D2, respectively. The outbound Groundspeed is, in this example, the Airspeed plus the TAILwind: (22 + 10) = 32. In the event of having to turn back before reaching "B", the inbound Groundspeed is the Airspeed minus the (now) HEADwind: (22 - 10) = 12.
Before deriving a magic formula for "X", consider the obvious. An outbound flight with a HEADwind implies a speedier return from any point on the course. Conversely, setting out with a wind from the nether regions (TAIL) should temper a "Pollyanna Glad Game" outlook of how beautiful PPG life CAN be: it's going to be a longer, and possibly a LOT longer journey home. Wind, for some reason, affects Groundspeed. The MINIMUM G/S (12) must be used to determine point "X".
To re-state the problem, it will take as long (Time 1, "T1") to return to "A" from "X" as it will ("T2") to continue from "X" to "B"; i.e. "T1 = T2". The minimum G/S is 12 (let's call it Velocity 1, "V1"). The maximum G/S is 32 ("V2"). Remembering that "T = D / V", "T1 = D1 / V1", and "T2 = D2 / V2". Someone who (to my knowledge, anyway) never had any affiliations with "The Flat Earth Society" once posited, "Things equal to the same thing are equal to each other. So if "Time" is the "thing" (or "of the essence", or whatever), and "T1 = Time" just like "T2 = Time", then "D1 / V1" = "D2 / V2". We're now about to enter the shadowy world of "simultaneous equations". Fortunately there're only two and they involve only one unknown, "X".
"Whazzat?!" Well, in the spirit of "a rose by any other name. . ." (whatever that implies): "D1" = "X", "V1" = 12, "D2" = (22 - "X"); "V2" = 32. Thus, ("X" / 12) = ((22 - "X") / 32). Still not clear? OK, how about "12X = 264 - 32X"? Oh... Then I bet you can cancel IFR if "44X = 264". Egg-zackly: "X" = 6! If your whole day is ruined at or before 6 (SIX, I say again, SIX - Miles/Km) from "A", turn back; otherwise boogie (22 - 6 = 16 Miles/Km) for "B".
"Sure you got that right?" "Affirmatative." Flame-out over "X"; outbound G/S is 32; inbound is 12; Time to fly from "X" back to "A" = (6 / 12) = 0.5 hours (30 minutes). Time to fly from "X" onwards to "B" = (16 / 32) = 0.5 hours (30 minutes). Q.E.D.
"Small question: Fly? Don't you mean glide? Like, I need altitude, don't I?" Ahem, yes. Glide ratio, please? Six-to-one you say. Distance? Also 6 (miles)? Airspeed is 22 (mph)?
From "X" to "A" is 6 miles, right? And the GROUNDspeed is 12 mph, am I not incorrect? AND, if I may continue, 12 mph = ((12 x 5280) / 60) = 1056 Feet/Minute; any argument there? How many feet in 6 miles: (6 x 5280) = 31,680 feet. Glide is, what again; ah yes, 6. That's with NO WIND, though, by the bye. "Whaddaya mean?" See "The PPG Bible: Section I, Chapter 5; Section IV, Chapter 22: Glide Ratio"; I bought my copy. Point is, that 10 mph headwind reduces your glide ratio from 6 to 3.3 ((6 x (22 - 10) / 22) = 3.27 to be exact). The altitude required to glide 31,680 feet, at a glide ratio of 3.272727... is 9680 feet.
"Well, OK; but... 10 grand lets me glide 6 miles from "X" back to "A". You said I could also glide from "X" to "B", and that's 16 miles. Don't make sense to me." We're not paying attention... What's the GROUNDspeed from "X" to "B"? Correct me if I'm wrong, but does 32 mph center your needle? Guess what that does to your glide ratio: ((6 x 32 / 22) = 8.7272). Sixteen miles is 84,480 feet. From an altitude of 9680, with a glide of 8.7272..., you'll cover, why bless my soul, 84,480 feet.
Have a good one!
OJQJ^JCJOJQJ^JaJ
z{gh">?
t
u
$%
7$8$H$`7$8$H$%]^7$8$H$ 1h/ =!"#$%
i8@8NormalCJ_HaJmH sH tH <A@<Default Paragraph Font(
z{gh">?t u
$%
]^000000000000000000000000000000000000000000% Glen Boyd4C:\Documents and Settings\glenboyd\Desktop\GCalc.doc Glen BoydaC:\Documents and Settings\glenboyd\Application Data\Microsoft\Word\AutoRecovery save of GCalc.asd Glen BoydaC:\Documents and Settings\glenboyd\Application Data\Microsoft\Word\AutoRecovery save of GCalc.asd Glen BoydaC:\Documents and Settings\glenboyd\Application Data\Microsoft\Word\AutoRecovery save of GCalc.asd Glen Boyd4C:\Documents and Settings\glenboyd\Desktop\GCalc.doc Glen Boyd4C:\Documents and Settings\glenboyd\Desktop\GCalc.doc Glen BoydaC:\Documents and Settings\glenboyd\Application Data\Microsoft\Word\AutoRecovery save of GCalc.asd Glen Boyd4C:\Documents and Settings\glenboyd\Desktop\GCalc.doc Glen Boyd4C:\Documents and Settings\glenboyd\Desktop\GCalc.doc Glen Boyd4C:\Documents and Settings\glenboyd\Desktop\GCalc.doc@@@UnknownGz Times New Roman5Symbol3&z Arial"h[\ r
+!r0d2 Glen Boyd Glen BoydOh+'0p
,8
DPX`hGlen BoydNormal.dotGlen Boyd9Microsoft Word 9.0@,@|.@HRr՜.+,0hp
NBDOT+
# Title
!"#$&'()*+,/Root Entry F9R11TableWordDocument"(SummaryInformation(DocumentSummaryInformation8%CompObjjObjectPool9R9R
FMicrosoft Word Document
MSWordDocWord.Document.89q