Could not resist having a little fun with this. Given that the measured relative length of an object is directly proportionate to its distance from the viewer one can determine what it's relative size will be at any given distance without having to actually place it at that distance, in this case 500 feet. Using two known measures, in my case a 76" dresser (better yet, use your plane's wingspan) and a 12" ruler, hold the ruler at arm's length, close your non-dominant eye and back away from the object until the length of the ruler equals the length of the object. Measure the distance from you to the object, in my case 86". Solve the equation X = 12(distance in inches of the object at arm's length-could be any length you choose)*86(measured distance in inches from object when covered by 12" held at arm's length)/500(distance in question in feet)*12(converting feet to inches). In this case the 76" dresser would be the same length as 0.172" held at arms length. Wider and it is below 500 feet, wingspan shorter it is above 500 feet. This technique will vary somewhat for each individual depending on the length of their arm and the circumference of their eye, but will give you a very quick and fairly accurate estimation of altitude. What amazed me is just how small (equivalent to looking at a .172 inch wingspan plane at arm's length) a 6+ foot wing span plane appears at 500" Verify it for yourself, if I doubled the distance I was standing from the dresser to 172" it was indeed covered by 6" on the ruler.