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<br />...... <br />I <br />(.1l <br /> <br />,aDo NAT <br />HAT QlI' <br />IECNO <br />Q <br />TIM! <br />IILOP! <br /> <br />The solution to the Method 4 example by HEC-2 is shown below. A dis.charge of <br />1,000 cfs was used. For comparison purposes: <br /> <br />Q1 = O.OlK <br /> <br />NATQ1 = Conveyance for natural water surface elevation in the first line <br />of Statement 2800 and conveyance for the higher water surface <br />elevation in the second line. <br /> <br />RATIOS LOB, CH, ROB. The ratios of conveyance at higher water surface <br />elevations. <br /> <br />TARGET · Ratio of required conveyance reduction. <br /> <br />QJ' \0'01l.1i WSIL-. 6!.oq INC <br />115". ~ATIOs LOB,CH,Roe- .10112 <br />D!PTH CWSEL CRIW! WII!L~ <br />GLOB GCM GAOB ALOe <br />yLOB YCH Y~OB XNL. <br />XLOBL XLCH XLOBR ITRTAL <br /> <br />\:1;, <br />.00 <br />'U, <br />.t' <br />o. <br /> <br />.. -,,~ <br />\117.11 <br />$!l.oq <br />~~1. <br />.0110 <br />o <br /> <br />> <br />." <br />." <br />.,., <br />:z <br />c <br />...... <br />>< <br /> <br />).,0 !NCROACHMENT STATIONS- <br />1.00 Ut~OO ...00 <br />tooo. ~1: .'l~ <br />.00 .1~ .3~ <br />.000001 O. o. <br /> <br />...... <br /> <br />Qt- 10'0'~" W'~L~ <br />..". .2!" WI!LI <br />!~ HY HL <br />ACH AR08 VQL <br />XN~H XNR _ wTN_ <br />IDC ICONT COAAR <br /> <br />.000' <br /> <br />...qo RATIO- <br />....00 <br />OLe!!! <br />T",A <br />eLMIN <br />TO~WID <br /> <br />TVP.U <br />...OQ <br />IQ'O. <br />,no <br />o <br /> <br />ii, TAA~n.. <br />.OQ .oq <br />1119'. .. O. <br />.050 .~oo <br />I .00 <br /> <br />. <br />.05' <br />.00 <br />.o~ <br />,,2~,OO <br />lU.TI <br /> <br />IU~~ ELn <br />LIPT I".GHT . <br />uu <br />rNon <br /> <br />50;00 <br />,q~OO <br />,! "71 <br />1111.42 <br />