For pilots returning eastbound from EAA AirVenture in Oshkosh, the VFR choices include skirting Chicago to the west (adding a lot of distance and time to the trip); hugging a few miles off the coastline at 3,500 feet or lower; or climbing high and going direct across Lake Michigan. Assuming a tailwind (most of us suffered 25-knot headwinds on the way out), at what point would you turn back into the wind as opposed to continuing on with the tailwind to make the far coast? A formula posted on a Cessna Cardinal user group's website (courtesy of Flying contributing editor Tom Benenson) offers some guidance, and it works for any over-water trip. According to the theory, you multiply the over-water distance (Lake Michigan is about 100 nm wide at its widest point) by the 'return' groundspeed (best glide speed minus the wind, in the case of an engine failure); then divide that number by sum of the return groundspeed plus the value of the 'continue-on' groundspeed. So, if 90 knots true is your best glide speed, with a 20-knot tailwind from the west the point of no return would be: 100 x 70 kt divided by 70 + 110. That adds up to about 39 miles east of where you crossed over. Now, you can use your airplane's known glide ratio to determine if you'll make dry land from your cruising altitude. Either way, it will take you .54 hours to get to land. Do you have enough altitude to stay aloft that long? If not, start looking for boats -- and hope they have the makings of a stiff drink on board. You'll need it.