Glide ratio beyond the curvature of the earth.

[OP]

I guess you would be in orbit at that point, right?

I did the math first, then checked wikipedia. We had almost same answer: beyond a 3,750:1 glide.

If a top notch 31m glider has a 70:1 glide. That is still 50x short.

[chadness]

Hey George,

I was gonna post the same response but you beat me to the punch. And did a more thorough job than I would have.

But also a fun thought to me, is to consider a glide ratio of infinity to one. Basically a perfect wing that had a way to avoid all energy losses from drag, resistance, etc (not possible to achieve but possible to approach).

Perhaps consider this perfect L/D would be at a particular airspeed. Now when you set out on a glide, all you would be thinking of it what line to pick and how fast to fly. Your handy Flytech vario guestimeter would always have to say you could make the glide to anywhere at the same alt you are now. Flying close to the ground (say 5 feet) would get real interesting because it is hard to imagine too much sink (or lift) when so close to the ground.

Cheers,
Chad

[stebbins]

I guess you would be in orbit at that point, right?

I did the math first, then checked wikipedia. We had almost same answer: beyond a 3,750:1 glide.

If a top notch 31m glider has a 70:1 glide. That is still 50x short.

Unfortunately, this isn't true.

Why? Because as one moves forward, the gravity vector (against which glide is defined) changes direction. Your answer assumes it remains unchanged. For any reasonable glide ratio, the change is in the noise, and can be ignored. But if you are talking about large glide ratios things don't work that way. This is espectially true if you are going "level" (orbit-like) and not dropping at all (relative to the earth's center, or MSL). Then you are talking glides long enough for the gravity vector change to matter.

For those more visually inclined, picture a "level" glide all around the earth. It is a circle, yes? That means that 1/2 way around the earth, the gravity vector has changed direction by 180 degrees. Clearly you can't use a straight line to approximate a circle except for a short distance compared to the radius of the circle.

For those who like to use energy arguments, consider this: If you remain on the circle, you don't require input of energy (friction asside.) If you deviate from that circle, you are rising or falling. Thus you require additional energy to rise (deviate upward from the circle). You also gain additional energy (as speed) when you deviate downward from the circle, as we normally do.

Still, a great photo, and an interesting idea that a circle the size of the earth deviates from a straight line by 1 part in 3750 over short distances. I didn't do the math before reading, and would have naively expected the glide ratio deviation to be larger.

Thanks for the mental exercise!

[stebbins]

......Flying close to the ground (say 5 feet) would get real interesting because it is hard to imagine too much sink (or lift) when so close to the ground.

Cheers,
Chad

But those irregularities will bite you big time, and instruments are blind...

One time, coming back from the Owens Valley, I saw a sailplane breaking down in a large turnout on the side of the road. (Really!) He had clearly not made the landing zone he really wanted. Upon reflection, I realized why. He was a few miles south of Inyokern, where the ground rises up before dropping down past Red Rock and then to California City. I assume his instrument said he had the glide (or close to it) to some safe landing on the other side of the rise. But, that pesky rising ground got in the way...

It took me some time to think this out, because I was just so flabbergasted to see a high-end sailplane that had clearly landed on the turnout along highway 14.... I was thinking WTF, more than "Why'd that happen."

[OP]

Ok, I get the energy argument. I made a picture:


If the red arrow is the perfect glide slope infinity:1, then it is receiving free energy. As E = mgh then it starts at lower h and ends up with a higher h.




So now what about the slope of 3750:1. Constant height above the earth.

[chadness]

Hey Orion,

that is what I was talking about in my previous post (which was a response to George but has somehow appeared before his response).

Glide ratios should be thought of with respect to a perpendicular angle with respect to gravity. So on an infinitely large non rotating perfectly round planet, infinity-to-one would be infinity-to-one with respect to Cartesian coordinates, but on earth it would be roughly equal to the curvature of the earth (3750:1 as you informed us).

Trying to get there. Comp wing, racing harness. But seems it is gonna require alot more invention (and breaking some laws of physics). Or, adding some energy from, say a mosquito. Yeah, maybe I should try that some day:)

Cheers,
Chad

[dhmartens]

If the earth were smaller, like the diameter of the moon, it might work. But if you reduce the mass of the the earth, the gravity would reduce and the atmosphere would blow off into space. An asteroid made of lead could have a dense enough atmosphere to continuously glide around it but you might actually be able to glide out in to orbit and would need a pressure suit.

I like those solar powered drones that recharge while flying during the day and try to stay up on batteries at night. I was envisioning a cross country 3,000 mile hang glider race that use only a solar panel, Lithium Ion battery, and electric motor. You recharge in the morning, use the battery to get airborne, fly as far as you can until dark, land and camp until the next day and repeat until the goal is met.