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Hi Prof.In your calculation for the effective cross sectional area, I believe you accounted twice for the corner areas when subtracting from the initial area. I get WHo - 2Wd* - 2(Ho-2d*)d*.
You are correct but I ignored the corner effects since delta* is so small compared to W or H, so it is insignificant.
@@johncimbala Noted. Also, I thank you sincerely for this well crafted video series. An unbelievably helpful exam-prep resource.
@@thomas_hall Thank you. I appreciate your comments.
Hi Prof.
In your calculation for the effective cross sectional area, I believe you accounted twice for the corner areas when subtracting from the initial area. I get WHo - 2Wd* - 2(Ho-2d*)d*.
You are correct but I ignored the corner effects since delta* is so small compared to W or H, so it is insignificant.
@@johncimbala Noted. Also, I thank you sincerely for this well crafted video series. An unbelievably helpful exam-prep resource.
@@thomas_hall Thank you. I appreciate your comments.