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popsjumper

World Record - Speed

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Excuse my ignorance, i've been reading contradicting posts in various forums, have you guys finished CP nationals now? Or are they still coming up? If so, what were the results, who is on the team coming to SA later this year?

Thanks :)


Advertisio Rodriguez / Sky

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The tight shorts cut off circulation to your brain. Combined with Alzheimers that comes with someone your age, that amnesia is to be expected :)
In all seriousness, this was the toughest comp I've ever done with the level of competitors at an all time high IMO.

Nice job to everyone!

Blues,
Ian

Performance Designs Factory Team

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In all seriousness, this was the toughest comp I've ever done with the level of competitors at an all time high IMO.



Hey, I was just jazzed to be on the plane with all of my "swoop heros," it was really a lot of fun!:)
--"When I die, may I be surrounded by scattered chrome and burning gasoline."

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Average speed is one thing, the speed at the entry gate another. We had speed measurements of up to 94mph in the past, so I guess we are closing in now on 100mph!



It is possible to roughly estimate the gate entry speed from the average speed and time.

When flying horizontally, you decelerate at the fraction of g that is equal to the reciprocal of your momental L/D:

a = g/(L/D)

For the rough estimate, one can assume constant L/D (which is not true - as you slow down, your canopy must produce the same lift - equal to your weight - at slower speed, therefore, it must change angle of attack and as a result, L/D is constantly changing. But for an estimate, the changes in L/D are relatively small if one exits the course with high speed still.)

Assuming L/D ~ 2.5, we have a ~ 0.4g.

If V0 is the gate entry speed and V1 is the course exit speed, average speed Vavg = (V0 + V1)/2 (for constant deceleration), and V1 = V0 - a*t, where t is time. So, Vavg = (V0 + V0 - a*t)/2 = V0 - a*t/2, and

V0 = Vavg + a*t/2 = Vavg + g*t/2/(L/D)

In this particular case, V0 = 28.6m/s + 9.8m/s^2 * 2.45s/2/2.5 = 33.5m/s = 120.6km/h = 75mph.


altough i never understand shit, i always love to read your posts that involve mathematics.. :)
congrats to everyone!
“Some may never live, but the crazy never die.”
-Hunter S. Thompson
"No. Try not. Do... or do not. There is no try."
-Yoda

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Maybe a stupid quiestion but whats up with Edson Pacheco doing 2.1 seconds on the CPC-speed run ? Isnt that quite a bit faster than 2.44 ?

Also where were Jay Moledzki and Jeffro ? Didnt they compete at all ?

It seems confusing to have several different competitions at the same time ? (CPC, PST, nationals)

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Awesome, thanks Stu. And congrats to all you guys on your results :)
edited to add: so will the selection process be top eight at Nats plus two reserves?



No, only the top 8 go. If any one of the top 8 can't go then usually the next person on the list according to rankings is called and so on.

I've put in a bid to take on the Team Manager role, but I'll be there regardless :)
Blues,
Ian
Performance Designs Factory Team

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Does anyone know who the lady was that took the high-res pictures of my crash? I had given her my e-mail address so she could send me the pictures, but I wanted to contact her. Schwartz might know her, but I don't have Uncle Pete's contact info.
--"When I die, may I be surrounded by scattered chrome and burning gasoline."

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Average speed is one thing, the speed at the entry gate another. We had speed measurements of up to 94mph in the past, so I guess we are closing in now on 100mph!



It is possible to roughly estimate the gate entry speed from the average speed and time.

When flying horizontally, you decelerate at the fraction of g that is equal to the reciprocal of your momental L/D:

a = g/(L/D)

For the rough estimate, one can assume constant L/D (which is not true - as you slow down, your canopy must produce the same lift - equal to your weight - at slower speed, therefore, it must change angle of attack and as a result, L/D is constantly changing. But for an estimate, the changes in L/D are relatively small if one exits the course with high speed still.)

Assuming L/D ~ 2.5, we have a ~ 0.4g.

If V0 is the gate entry speed and V1 is the course exit speed, average speed Vavg = (V0 + V1)/2 (for constant deceleration), and V1 = V0 - a*t, where t is time. So, Vavg = (V0 + V0 - a*t)/2 = V0 - a*t/2, and

V0 = Vavg + a*t/2 = Vavg + g*t/2/(L/D)

In this particular case, V0 = 28.6m/s + 9.8m/s^2 * 2.45s/2/2.5 = 33.5m/s = 120.6km/h = 75mph.


altough i never understand shit, i always love to read your posts that involve mathematics.. :)
congrats to everyone!


Me too! :ph34r:

Smarties are sexy...:)

(75 mph!)

Action expresses priority. - Mahatma Ghandi

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Just to clear up what Dave is looking for.

He's not looking for the picture (yes, I only got one) that I took of his crash. He already as that one, plus he knows me and how to contact me;)

May your trails be crooked, winding, lonesome, dangerous, leading to the most amazing view. May your mountains rise into and above the clouds. - Edward Abbey

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I've put in a bid to take on the Team Manager role, but I'll be there regardless :)



Make sure you've got your new passport or they might not let you leave. :P
SCR #14809

"our attitude is the thing most capable of keeping us safe"
(look, grab, look, grab, peel, punch, punch, arch)

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I've put in a bid to take on the Team Manager role, but I'll be there regardless :)



Make sure you've got your new passport or they might not let you leave. :P


Hey hey, he's YOUR country's problem now, there's no way we want him back :P

Advertisio Rodriguez / Sky

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