I finally gave in and ordered a Benq HT2150ST. Maybe to watch movies and play computer games. I am also thinking to do the heretical thing of using it for JAMMA games. Anyone has any experience running 15KHZ games on projectors??
- Thread starter suverman
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Ok I will check it with my OSSC.15khz works perfect on a CRT projector
On a modern projector I would use a OSSC, but no first hand experience (yet)
I've been (console) gaming on projectors exclusively since 2002.
My old DLP did really well with 15KHz because it had an excellent scaler chip designed specifically for 15KHz content. my current projector one does not though so I use an OSSC.
If I had the space I'd dedicated a little corner with a couch and a nice big Sony Trinitron for 15KHz stuff but the projector is still fun for those old 4-player split screen games.
projectors are no different than any other type of display, some are great, most are trash.
My old DLP did really well with 15KHz because it had an excellent scaler chip designed specifically for 15KHz content. my current projector one does not though so I use an OSSC.
If I had the space I'd dedicated a little corner with a couch and a nice big Sony Trinitron for 15KHz stuff but the projector is still fun for those old 4-player split screen games.
projectors are no different than any other type of display, some are great, most are trash.
Can you recommend some 4 player games? I and my mates are hooked onto Super Bomberman 2 Battle mode on Wii SNES emulator - its surprisingly fun with the wiimote xD They arent skilled enough to play SF2 sadly and beatemups are fun if they need to pay - since they dont they get bored.
on NES: https://en.wikipedia.org/wiki/NES_Four_Score
on SNES: http://nintendo.wikia.com/wiki/Super_Multitap
on Mega Drive: https://segaretro.org/4_Way_Play
on N64: http://www.nintendo64ever.com/Nintendo-64-4-players-Games.html
on Saturn: http://www.racketboy.com/retro/saturn-multiplayer-games-best-multi
on PS1: https://en.wikipedia.org/wiki/PlayStation_Multitap
on Dreamcast: http://www.racketboy.com/retro/the-best-dreamcast-multiplayer-games
Some good lists here: https://en.wikipedia.org/wiki/List_of_local_multiplayer_video_games_by_system#GameCube
and here: https://www.co-optimus.com/
the Dreamcast and the Nintendo Consoles from N64 forward are probably the best for this since they had 4 controller ports a lot of games supported 4 players.
There are some Gems on the Original Xbox, Saturn and PS1 though.
on SNES: http://nintendo.wikia.com/wiki/Super_Multitap
on Mega Drive: https://segaretro.org/4_Way_Play
on N64: http://www.nintendo64ever.com/Nintendo-64-4-players-Games.html
on Saturn: http://www.racketboy.com/retro/saturn-multiplayer-games-best-multi
on PS1: https://en.wikipedia.org/wiki/PlayStation_Multitap
on Dreamcast: http://www.racketboy.com/retro/the-best-dreamcast-multiplayer-games
Some good lists here: https://en.wikipedia.org/wiki/List_of_local_multiplayer_video_games_by_system#GameCube
and here: https://www.co-optimus.com/
the Dreamcast and the Nintendo Consoles from N64 forward are probably the best for this since they had 4 controller ports a lot of games supported 4 players.
There are some Gems on the Original Xbox, Saturn and PS1 though.
@twistedsymphony Thanks for the lists I will check and sort out the ones which is still playable now :] Also never knew of Co-Optimus, looks like the website I am looking for.
@nem That game reminds me of Scorched Earth, multiplayer would be fantastic :}
@nem That game reminds me of Scorched Earth, multiplayer would be fantastic :}
Big fan of couch multiplayer. Reason why I have a 37" 4:3 CRT.
10 player Saturn Bomberman
6 player Death Tank (we couldn't get the 7th controller working even though it should support it)
are both top tier on the Saturn. We also tried Fire Pro, Street Racer and Vatlva. They weren't that fun. I completely forgot Space Jam supports six player as well, that could have been good.
I'm hoping to try 8 player Megadrive Micro Machines 2 and 5 player PC Engine Moto Roader at some point.
10 player Saturn Bomberman
6 player Death Tank (we couldn't get the 7th controller working even though it should support it)
are both top tier on the Saturn. We also tried Fire Pro, Street Racer and Vatlva. They weren't that fun. I completely forgot Space Jam supports six player as well, that could have been good.
I'm hoping to try 8 player Megadrive Micro Machines 2 and 5 player PC Engine Moto Roader at some point.
I have two Saturns both with dead laser, hopefully the Satiator is released soon. Till then I will miss out on the Saturn Bomberman. I have Pete Sampras tennis on Megadrive which allows 4 players but not sure how many people will be even able to play it.
I will give Wii games another go as well, digging out the fun ones.
I will give Wii games another go as well, digging out the fun ones.
Is there a rule of thumb somewhere for getting 4:3 content size from the projector's 16:9 image size?
Like this projector says min throw distance is 3.6ft and minimum screen size is 59.8". How would I figure out the minimum 4:3 screen size?
Like this projector says min throw distance is 3.6ft and minimum screen size is 59.8". How would I figure out the minimum 4:3 screen size?
math.
Right. But what's the equation?math.
since I can't help myself you'd use the Pythagorean Theorem: https://en.wikipedia.org/wiki/Pythagorean_theorem
a2[/sup] + b[sup]2[/sup] = c[sup]2
So lets say a = height, b = width and c= diagonal
we know that on a 4:3 screen the width = (4 * height)/3 and that on a 16:9 screen the width = (16 * height)/9
lets also sub-script these variables with f for full-screen and w for wide-screen to keep them apart so:
af[/sub]2[/sup] + ((4*a[sub]f[/sub])/3)[sup]2[/sup] = c[sub]f[sup]2
and
aw[/sub]2[/sup] + ((4*a[sub]w[/sub])/3)[sup]2[/sup] = c[sub]w[sup]2
and we know that [the height of a 4:3 projected image] = [the height of a 16:9 projected image] so:
aw[/sub] = a[sub]f
That means we can drop out the sub-scripts for a:
a2[/sup] + ((4*a)/3)[sup]2[/sup] = cf[sup]2
and
a2[/sup] + ((16*a)/9)[sup]2[/sup] = cw[sup]2
what we want is to solve for the full screen diagonal (cf[/sub]) given the only input as the wide-sceen diagonal (c[sub]w).
so that means we need to solve for cf on the first equation which is pretty easy:
√(a2[/sup] + ((4*a)/3)[sup]2) = cf
simplified:
cf = (5/3) * a
on the other equation we need to solve for the height (a) which is a little more complicated but we end up with:
√(a2[/sup] + ((16*a)/9)[sup]2) = cw
simplified:
(√337/9) * a = cw
rearranged:
a = (9/√337) * cw
now we can insert the second equation in place of a on the first equation:
cf[/sub] = (5/3) * (9/√337) * c[sub]w
simplified:
cf[/sub] = ((15 * √337)/337) * c[sub]w
or if you don't care about whole numbers:
cf[/sub] = 0.8171 * c[sub]w
That means if you have 60in (16:9) wide-screen and you want to know the (4:3) full-screen size you just sub in 60in for cw and solve:
cf = 0.8171 * 60in = ~49in
a2[/sup] + b[sup]2[/sup] = c[sup]2
So lets say a = height, b = width and c= diagonal
we know that on a 4:3 screen the width = (4 * height)/3 and that on a 16:9 screen the width = (16 * height)/9
lets also sub-script these variables with f for full-screen and w for wide-screen to keep them apart so:
af[/sub]2[/sup] + ((4*a[sub]f[/sub])/3)[sup]2[/sup] = c[sub]f[sup]2
and
aw[/sub]2[/sup] + ((4*a[sub]w[/sub])/3)[sup]2[/sup] = c[sub]w[sup]2
and we know that [the height of a 4:3 projected image] = [the height of a 16:9 projected image] so:
aw[/sub] = a[sub]f
That means we can drop out the sub-scripts for a:
a2[/sup] + ((4*a)/3)[sup]2[/sup] = cf[sup]2
and
a2[/sup] + ((16*a)/9)[sup]2[/sup] = cw[sup]2
what we want is to solve for the full screen diagonal (cf[/sub]) given the only input as the wide-sceen diagonal (c[sub]w).
so that means we need to solve for cf on the first equation which is pretty easy:
√(a2[/sup] + ((4*a)/3)[sup]2) = cf
simplified:
cf = (5/3) * a
on the other equation we need to solve for the height (a) which is a little more complicated but we end up with:
√(a2[/sup] + ((16*a)/9)[sup]2) = cw
simplified:
(√337/9) * a = cw
rearranged:
a = (9/√337) * cw
now we can insert the second equation in place of a on the first equation:
cf[/sub] = (5/3) * (9/√337) * c[sub]w
simplified:
cf[/sub] = ((15 * √337)/337) * c[sub]w
or if you don't care about whole numbers:
cf[/sub] = 0.8171 * c[sub]w
That means if you have 60in (16:9) wide-screen and you want to know the (4:3) full-screen size you just sub in 60in for cw and solve:
cf = 0.8171 * 60in = ~49in
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You can also use that equation to work the other way.
lets say you have a 4:3 CRT and want to know what sized wide-screen display would offer the same size 4:3 image
cf[/sub] = 0.8171 * c[sub]w
rearranged to:
cw[/sub] = c[sub]f / 0.8171
so for a 29in 4:3...
cw = 29in / 0.8171 = ~35.5in
lets say you have a 4:3 CRT and want to know what sized wide-screen display would offer the same size 4:3 image
cf[/sub] = 0.8171 * c[sub]w
rearranged to:
cw[/sub] = c[sub]f / 0.8171
so for a 29in 4:3...
cw = 29in / 0.8171 = ~35.5in
I'll do just that. I need to find a backup option for the Megalo if the parts of two original projectors I have don't make a whole working projector in the end.
you're talking a really big screen... Megalos are what 50in? that's a ~61in widescreen
Like T says, math, but if you feeling lazy
Something like this?
http://www.comparetvsizes.com/?tv1=59.8&a1=169&tv2=48.8&a2=43
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