about top gun...

actually ive read that all the air manuevers(sp?) done in that movie were actual maneuvers performed by f14''s and perfectly capable

Just thought of another thing that is partly on topic, and partly the old fog-of-war argument.
In real war these days there are spy satellites. The combatants tend to more or less know when an enemy satellite is going to pass overhead, and can hide or set up decoys/feints during those periods.
It could be pretty cool for a game to incorporate a periodic satellite flyover which shows all the enemy''s units for just a few minutes, but can be easily fooled by decoys, etc. The mini-map display could show a large circle moving across the map at regular intervals to indicate when the enemy was known to be watching. Each side gets the benefit of overhead surveillance, but also get the opportunity to completely fool the opponent about their readiness and deployment, again adding another layer of strategic options.

quote: Original post by InnocuousFox
That is, if a signal is strength x at 100'', it is sqrt(x) at 200''.

I''d say x/4, not sqrt(x). It''s not exponential, it''s inverse quadratic (the total power is spread over an ever-growing sphere).

Actually, it occurs to me that my first post didn''t imply a linear relationship at all, in fact it doesn''t matter what the formula is. In order for the target to be detected, the signal must travel exactly twice the distance between the antenna and the target, right? And at it''s maximum effective range, the signal, after the round trip, will have just barely enough energy to be detected. So, in my example with the target at the max range of 100km, the transmitted signal is just strong enough to travel 200km, whether it bounces off the target or not. Granted, the signal will lose more energy on the return trip than on the way out, but that doesn''t make any difference whatsoever; if the reflected signal can be detected after travelling a total of 200km, then a direct unreflected signal can be too.

I''m all for...

• Real radar simulation including atmospheric diffraction. What''s the matter, can''t you afford an extra few floating point calculations in a radar sweep?
  
• Spy satellites, and using or evading them.
  
• Optical instruments that open up a huge playing field in detail if you use them to scan the field. Imagine spying and identifying an enemy tank on a ridge silhouetted against the morning light through a magnifying scope. Suddenly the terrain plays a very significant role in how you position yourself.
  

___________________________________

quote: Original post by Plasmadog
Actually, it occurs to me that my first post didn''t imply a linear relationship at all, in fact it doesn''t matter what the formula is. In order for the target to be detected, the signal must travel exactly twice the distance between the antenna and the target, right?

You are correct... we are both correct. You didn''t imply, but the inverse square law is the way to define it. My correction was correct but ill-advised.



Dave Mark
Intrinsic Algorithm Development

quote: Original post by Fruny
I''d say x/4, not sqrt(x). It''s not exponential, it''s inverse quadratic (the total power is spread over an ever-growing sphere).

If you said that, you would be wrong. It''s straight up physics, bro.



Dave Mark
Intrinsic Algorithm Development

quote: Original post by Fruny
I'd say x/4, not sqrt(x). It's not exponential, it's inverse quadratic (the total power is spread over an ever-growing sphere).

quote: Original post by InnocuousFox
If you said that, you would be wrong. It's straight up physics, bro.

He'd be right, actually.

Inverse square law:

x = g(z) * 1/a^2

where z is independent of a

Hence for b = 2*a,
f(b) = g(z) * 1/b^2
f(b) = g(z) * 1/ (2*a)^2
f(b) = g(z) * 1/ 4*(a^2)
f(b) = (g(z) * 1/a^2) * 1/4
f(b) = x/4

As to the original point, Terra Nova apparently accounted for this as well as atmospheric diffraction of laser weapons & so on. Having played the game over and over, I still cannot notice the effect. It's not worth the cycles unless it's an integral part of the design (HGII may do it better; I haven't played it), so you'd have to present a design where this effect mattered for me to feel your pain

ld

Edited for clarity

Edited by - liquiddark on January 17, 2002 10:18:14 PM

... anyone here read: The Bear and the Dragon by Tom Clancy?

Zaptruder

quote: Original post by liquiddark
He'd be right, actually.

Narf!

Sorry...

Dave Mark
Intrinsic Algorithm Development

Edited by - InnocuousFox on January 18, 2002 8:28:50 AM