4 ms·
Inch is diagonal so a 17” 16:9 would be to wide. Unless I’m thinking wrong we’d need something at (sqrt(17)*(16^2+9^2)/(16^2+10^2))^2, which should put us at
by hultner 5y ago
Inch is diagonal so a 17” 16:9 would be to wide.
Unless I’m thinking wrong we’d need something at (sqrt(17)*(16^2+9^2)/(16^2+10^2))^2, which should put us at 15.23”, I know 15,4” exists so that might be doable depending on how much room there is to spare, 5mm is a bit but it could maybe work.
- hultner 5y agoSorry I did this wrong yesterday, wrote it late in bed and lost an exponent a long the way, should be 16,54” With the following calculation, 1.6 is for 16/10 sqrt((1.617/sqrt(1.6^2+1))^2+((9/16)(1.6*17/sqrt(1.6^2+1)))^2)