Section I. SIMPLE HORIZONTAL CURVES TYPES OF CURVE
CHAPTER 3CURVESSection I. SIMPLE HORIZONTAL CURVESCURVE POINTSTYPES OFHORIZONTAL CURVESBy studying TM 5-232, the surveyor learns tolocate points using angles and distances. Inconstruction surveying, the surveyor mustoften establish the line of a curve for roadlayout or some other construction.A curve may be simple, compound, reverse, orspiral (figure 3-l). Compound and reversecurves are treated as a combination of two ormore simple curves, whereas the spiral curveis based on a varying radius.The surveyor can establish curves of shortradius, usually less than one tape length, byholding one end of the tape at the center of thecircle and swinging the tape in an arc,marking as many points as desired.SimpleThe simple curve is an arc of a circle. It is themost commonly used. The radius of the circledetermines the “sharpness” or “flatness” ofthe curve. The larger the radius, the “flatter”the curve.As the radius and length of curve increases,the tape becomes impractical, and thesurveyor must use other methods. Measuredangles and straight line distances are usuallypicked to locate selected points, known asstations, on the circumference of the arc.CompoundSurveyors often have to use a compoundcurve because of the terrain. This curve normally consists of two simple curves curvingin the same direction and joined together.3-1
FM 5-233ReverseA reverse curve consists of two simple curvesjoined together but curving in oppositedirections. For safety reasons, the surveyorshould not use this curve unless absolutelynecessary.SpiralThe spiral is a curve with varying radius usedon railroads and somemodern highways. Itprovides a transition from the tangent to asimple curve or between simple curves in acompound curve.STATIONINGOn route surveys, the surveyor numbers thestations forward from the beginning of theproject. For example, 0 00 indicates thebeginning of the project. The 15 52.96 wouldindicate a point 1,552,96 feet from thebeginning. A full station is 100 feet or 30meters, making 15 00 and 16 00 full stations.A plus station indicates a point between fullstations. (15 52.96 is a plus station.) Whenusing the metric system, the surveyor doesnot use the plus system of numbering stations.The station number simply becomes thedistance from the beginning of the project.ELEMENTS OF ASIMPLE CURVEFigure 3-2 shows the elements of a simplecurve. They are described as follows, andtheir abbreviations are given in parentheses.Point of Intersection (PI)The point of intersection marks the pointwhere the back and forward tangents3-2intersect. The surveyor indicates it one of thestations on the preliminary traverse.Intersecting Angle (I)The intersecting angle is the deflection angleat the PI. The surveyor either computes itsvalue from the preliminary traverse stationangles or measures it in the field.Radius (R)The radius is the radius of the circle of whichthe curve is an arc.Point of Curvature (PC)The point of curvature is the point where thecircular curve begins. The back tangent istangent to the curve at this point.Point of Tangency (PT)The point of tangency is the end of the curve.The forward tangent is tangent to the curveat this point.
FM 5-233Length of Curve (L)The length of curve is the distance from thePC to the PT measured along the curve.Long Chord (LC)The long chord is the chord from the PC to thePT.Tangent Distance (T)The tangent distance is the distance alongthe tangents from the PI to the PC or PT.These distances are equal on a simple curve.External Distance (E)The external distance is the distance from thePI to the midpoint of the curve. The externaldistance bisects the interior angle at the PI.Central AngleThe central angle is the angle formed by tworadii drawn from the center of the circle (0) tothe PC and PT. The central angle is equal invalue to the I angle.Middle Ordinate (M)The middle ordinate is the distance from themidpoint of the curve to the midpoint of thelong chord. The extension of the middleordinate bisects the central angle.3-3
FM 5-233Degree of Curve (D)The degree of curve defines the “sharpness”or “flatness” of the curve (figure 3-3). Thereare two definitions commonly in use fordegree of curve, the arc definition and thechord definition.As the degree of curve increases, the radiusdecreases. It should be noted that for a givenintersecting angle or central angle, whenusing the arc definition, all the elements ofthe curve are inversely proportioned to thedegree of curve. This definition is primarilyused by civilian engineers in highwayconstruction.0English system. Substituting D 1 andlength of arc 100 feet, we obtain—Therefore,R 36,000 divided by6.283185308R 5,729.58 ftMetric system. In the metric system, using a30.48-meter length of arc and substituting D 1 , we obtain—Arc definition. The arc definition statesthat the degree of curve (D) is the angleformed by two radii drawn from the center ofthe circle (point O, figure 3-3) to the ends of anarc 100 feet or 30.48 meters long. In thisdefinition, the degree of curve and radius areinversely proportional using the followingformula:3-4Therefore,R 10,972.8 divided by6.283185308R 1,746.38 mChord definition. The chord definitionstates that the degree of curve is the angleformed by two radii drawn from the center ofthe circle (point O, figure 3-3) to the ends of achord 100 feet or 30.48 meters long. Theradius is computed by the following formula:
FM 5-233The radius and the degree of curve are notinversely proportional even though, as in thearc definition, the larger the degree of curvethe “sharper” the curve and the shorter theradius. The chord definition is used primarilyon railroads in civilian practice and for bothroads and railroads by the military.0Substituting D 1 and given Sin ½ 1 0.0087265335, solve for R as follows:0English system. Substituting D 1 andgiven Sin ½ 1 0.0087265355.50R 50ft orSin ½ D0.0087265355R 5,729.65 ftMetric system. Using a chord 30.48 meterslong, the surveyor computes R by the formulaR 15.24 m0.0087265355ChordsOn curves with long radii, it is impractical tostake the curve by locating the center of thecircle and swinging the arc with a tape. Thesurveyor lays these curves out by staking theends of a series of chords (figure 3-4). Sincethe ends of the chords lie on the circumferenceof the curve, the surveyor defines the arc inthe field. The length of the chords varies withthe degree of curve. To reduce the discrepancybetween the arc distance and chord distance,the surveyor uses the following chord lengths:3-5
FM 5-233SIMPLE CURVE FORMULASThe following formulas are used in thecomputation of a simple curve. All of theformulas, except those noted, apply to boththe arc and chord definitions.M R (l-COs ½ I)LC 2 R (Sin½ I)In the following formulas, C equals the chordlength and d equals the deflection angle. Allthe formulas are exact for the arc definitionand approximate for the chord definition.This formula gives an answer in degrees.,3048in the metric system. The answer will be inminutes.SOLUTION OF ASIMPLE CURVEL is the distance around the arc for the arcdefinition, or the distance along the chordsfor the chord definition.To solve a simple curve, the surveyor mustknow three elements. The first two are the PIstation value and the I angle. The third is thedegree of curve, which is given in the projectspecifications or computed using one of theelements limited by the terrain (see sectionII). The surveyor normally determines the PIand I angle on the preliminary traverse forthe road. This may also be done by triangulation when the PI is inaccessible.Chord DefinitionThe six-place natural trigonometric functionsfrom table A-1 were used in the example.When a calculator is used to obtain thetrigonometric functions, the results may varyslightly. Assume that the following is known:PI 18 00, I 45, and D 15 .3-6
FM 5-233Chord Definition (Feet)arc definitions are not exact for the chorddefinition. However, when a one-minuteinstrument is used to stake the curve, thesurveyor may use them for either definition.The deflection angles are—d 0.3’CDd std 0.3 x 25 x 15 112.5’ or 1 52.5’d1 0.3X 8.67X 15 0 39.015’d2 0.3 x 16.33 x 15 73.485’ or 1 13.485’The number of full chords is computed bysubtracting the first plus station divisible bythe chord length from the last plus stationdivisible by the chord length and dividing thedifference by the standard (std) chord length.Thus, we have (19 25 - 16 50)-25 equals 11full chords. Since there are 11 chords of 25feet, the sum of the deflection angles for 25foot chords is 11 x 1 52.5’ 20 37.5’.Chords. Since the degree of curve is 15degrees, the chord length is 25 feet. Thesurveyor customarily places the first stakeafter the PC at a plus station divisible by thechord length. The surveyor stakes thecenterline of the road at intervals of 10,25,50or 100 feet between curves. Thus, the levelparty is not confused when profile levels arerun on the centerline. The first stake after thePC for this curve will be at station 16 50.Therefore, the first chord length or subchordis 8.67 feet. Similarly, there will be a subchordat the end of the curve from station 19 25 tothe PT. This subchord will be 16,33 feet. Thesurveyor designates the subchord at thebeginning, C1 , an d at the end, C2 (figure 3-2).Deflection Angles. After the subchordshave been determined, the surveyor computesthe deflection angles using the formulas onpage 3-6. Technically, the formulas for theThe sum of d1, d2, and the deflections for thefull chords is—d1 0 39.015’d2 1 13.485’d std 20 37.500’Total 22 30.000’The surveyor should note that the total of thedeflection angles is equal to one half of the Iangle. If the total deflection does not equalone half of I, a mistake has been made in thecalculations. After the total deflection hasbeen decided, the surveyor determines theangles for each station on the curve. In thisstep, they are rounded off to the smallestreading of the instrument to be used in thefield. For this problem, the surveyor mustassume that a one-minute instrument is to beused. The curve station deflection angles arelisted on page 3-8.3-7
FM 5-233Special Cases. The curve that was justsolved had an I angle and degree of curvewhose values were whole degrees. When the Iangle and degree of curve consist of degreesand minutes, the procedure in solving thecurve does not change, but the surveyor musttake care in substituting these values into theformulas for length and deflection angles.For example, if I 42 15’ and D 5 37’, the3-8surveyor must change the minutes in eachangle to a decimal part of a degree, or D 42.25000 , I 5.61667 . To obtain the requiredaccuracy, the surveyor should convert valuesto five decimal places.An alternate method for computing the lengthis to convert the I angle and degree of curve to
FM 5-233minutes; thus, 42 15’ 2,535 minutes and 5 37’ 337 minutes. Substituting into the lengthformula givescorrection of 0.83 feet is obtained for thetangent distance and for the externaldistance, 0.29 feet.L 2,535 x 100 752.23 feet.337The surveyor adds the corrections to thetangent distance and external distanceobtained from table A-5. This gives a tangentdistance of 293.94 feet and an externaldistance of 99.79 feet for the chord definition.This method gives an exact result. If thesurveyor converts the minutes to a decimalpart of a degree to the nearest five places, thesame result is obtained.Since the total of the deflection angles shouldbe one half of the I angle, a problem ariseswhen the I angle contains an odd number ofminutes and the instrument used is a oneminute instrument. Since the surveyornormally stakes the PT prior to running thecurve, the total deflection will be a check onthe PT. Therefore, the surveyor shouldcompute to the nearest 0.5 degree. If the totaldeflection checks to the nearest minute in thefield, it can be considered correct.Curve TablesThe surveyor can simplify the computationof simple curves by using tables. Table A-5lists long chords, middle ordinates, externals,and tangents for a l-degree curve with aradius of 5,730 feet for various angles ofintersection. Table A-6 lists the tangent,external distance corrections (chord definition) for various angles of intersection anddegrees of curve.Arc Definition. Since the degree of curve byarc definition is inversely proportional to theother functions of the curve, the values for aone-degree curve are divided by the degree ofcurve to obtain the element desired. Forexample, table A-5 lists the tangent distanceand external distance for an I angle of 75degrees to be 4,396.7 feet and 1,492,5 feet,respectively. Dividing by 15 degrees, thedegree of curve, the surveyor obtains atangent distance of 293.11 feet and anexternal distance of 99.50 feet.Chord Definition. To convert these valuesto the chord definition, the surveyor uses thevalues in table A-5. From table A-6, aAfter the tangent and external distances areextracted from the tables, the surveyorcomputes the remainder of the curve.COMPARISON OF ARCAND CHORD DEFINITIONSMisunderstandings occur between surveyorsin the field concerning the arc and chorddefinitions. It must be remembered that onedefinition is no better than the other.Different ElementsTwo different circles are involved incomparing two curves with the same degreeof curve. The difference is that one is computed by the arc definition and the other bythe chord definition. Since the two curveshave different radii, the other elements arealso different.5,730-Foot DefinitionSome engineers prefer to use a value of 5,730feet for the radius of a l-degree curve, and thearc definition formulas. When compared withthe pure arc method using 5,729.58, the 5,730method produces discrepancies of less thanone part in 10,000 parts. This is much betterthan the accuracy of the measurements madein the field and is acceptable in all but themost extreme cases. Table A-5 is based onthis definition.CURVE LAYOUTThe following is the procedure to lay out acurve using a one-minute instrument with ahorizontal circle that reads to the right. Thevalues are the same as those used todemonstrate the solution of a simple curve(pages 3-6 through 3-8).3-9
FM 5-233Setting PC and PTWith the instrument at the PI, the instrumentman sights on the preceding PI andkeeps the head tapeman on line while thetangent distance is measured. A stake is seton line and marked to show the PC and itsstation value.The instrumentman now points the instrument on the forward PI, and the tangentdistance is measured to set and mark a stakefor the PT.Laying Out Curve from PCThe procedure for laying out a curve from thePC is described as follows. Note that theprocedure varies depending on whether theroad curves to the left or to the right.Road Curves to Right. The instrument isset up at the PC with the horizontal circle at0 00’ on the PI.(1) The angle to the PT is measured if the PTcan be seen. This angle will equal one halfof the I angle if the PC and PT are locatedproperly.(2)Without touching the lower motion, thefirst deflection angle, d1 (0 39’), is set onthe horizontal circle. The instrumentmankeeps the head tapeman on line while thefirst subchord distance, C1 (8.67 feet), ismeasured from the PC to set and markstation 16 50.(3) The instrumentman now sets the seconddeflection angle, d1 dstd (2 32’), on thehorizontal circle. The tapemen measurethe standard chord (25 feet) from thepreviously set station (16 50) while theinstrument man keeps the head tapemanon line to set station 16 75.(4) The succeeding stations are staked out inthe same manner. If the work is donecorrectly, the last deflection angle willpoint on the PT, and the last distance willbe the subchord length, C2 (16.33 feet), tothe PT.3-10Road Curves to Left. As in the proceduresnoted, the instrument occupies the PC and isset at 0 00’ pointing on the PI.(1) The angle is measured to the PT, ifpossible, and subtracted from 360 degrees.The result will equal one half the I angle ifthe PC and PT are positioned properly.(2) The first deflection, dl (0 39’), issubtracted from 360 degrees, and theremainder is set on the horizontal circle.The first subchord, Cl (8.67 feet), ismeasured from the PC, and a stake is seton line and marked for station 16 50.(3)The remaining stations are set bycontinuing to subtract their deflectionangles from 360 degrees and setting theresults on the horizontal circles. The chorddistances are measured from the previously set station.(4)The last station set before the PT shouldbe C2 (16.33 feet from the PT), and itsdeflection should equal the angle measured in (1) above plus the last deflection,d2 (1 14’).Laying Out Curve fromIntermediate SetupWhen it is impossible to stake the entire curvefrom the PC, the surveyor must use anadaptation of the above procedure.(1) Stake out as many stations from the PCas possible.(2) Move the instrument forward to anystation on the curve.(3) Pick another station already in place, andset the deflection angle for that station onthe horizontal circle. Sight that stationwith the instruments telescope in thereverse position.(4)Plunge the telescope, and set the remaining stations as if the instrument wasset over the PC.
FM 5-233Laying Out Curve from PTIf a setup on the curve has been made and it isstill impossible to set all the remainingstations due to some obstruction, the surveyorcan “back in” the remainder of the curvefrom the PT. Although this procedure hasbeen set up as a method to avoid obstructions,it is widely used for laying out curves. Whenusing the “backing in method,” the surveyorsets approximately one half the curve stationsfrom the PC and the remainder from the PT.With this method, any error in the curve is inits center where it is less noticeable.Road Curves to Right. Occupy the PT, andsight the PI with one half of the I angle on thehorizontal circle. The instrument is noworiented so that if the PC is sighted, theinstrument will read 0 00’.The remaining stations can be set by usingtheir deflections and chord distances fromthe PC or in their reverse order from the PT.Road Curves to Left. Occupy the PT andsight the PI with 360 degrees minus one halfof the I angle on the horizontal circle. Theinstrument should read 0 00’ if the PC issighted.Set the remaining stations by using theirdeflections and chord distances as ifcomputed from the PC or by computing thedeflections in reverse order from the PT.CHORD CORRECTIONSFrequently, the surveyor must lay out curvesmore precisely than is possible by using thechord lengths previously described.To eliminate the discrepancy between chordand arc lengths, the chords must be correctedusing the values taken from the nomographyin table A-11. This gives the corrections to beapplied if the curve was computed by the arcdefinition.Table A-10 gives the corrections to be appliedif the curve was computed by the chorddefinition. The surveyor should recall thatthe length of a curve computed by the chorddefinition was the length along the chords.Figure 3-5 illustrates the example given intable A-9. The chord distance from station18 00 to station 19 00 is 100 feet. The nominallength of the subchords is 50 feet.INTERMEDIATE STAKEIf the surveyor desires to place a stake atstation 18 50, a correction must be applied tothe chords, since the distance from 18 00through 18 50 to 19 00 is greater than thechord from 18 00 to 19 00. Therefore, acorrection must be applied to the subchordsto keep station 19 00 100 feet from 18 00. Infigure 3-5, if the chord length is nominally 50feet, then the correction is 0.19 feet. The chorddistance from 18 00 to 18 50 and 18 50 to19 00 would be 50.19.3-11
FM 5-233Section II. OBSTACLES TO CURVE LOCATIONTERRAIN RESTRICTIONSTo solve a simple curve, the surveyor mustknow three parts. Normally, these will be thePI, I angle, and degree of curve. Sometimes,however, the terrain features limit the size ofvarious elements of the curve. If this happens,the surveyor must determine the degree ofcurve from the limiting factor.Inaccessible PIUnder certain conditions, it may be impossible or impractical to occupy the PI. Inthis case, the surveyor locates the curveelements by using the following steps (figure3-6).3-12(1) Mark two intervisible points A and B, oneon each of the tangents, so that line AB (arandom line connecting the tangents)will clear the obstruction.(2) Measure angles a and b by setting up atboth A and B.(3) Measure the distance AB.(4) Compute inaccessible distances AV andBV as follows:I a b
FM 5-233(5) Determine the tangent distance from thePI to the PC on the basis of the degree ofcurve or other given limiting factor.(6) Locate the PC at a distance T minus AVfrom the point A and the PT at distance Tminus BV from point B.(7) Proceed with the curve computation andlayout.Inaccessible PCWhen the PC is inaccessible, as illustrated infigure 3-7, and both the PI and PT are set andreadily accessible, the surveyor mustestablish the location of an offset station atthe PC.(1)Place the instrument on the PT and backthe curve in as far as possible.(2) Select one of the stations (for example,“P”) on the curve, so that a line PQ,parallel to the tangent line AV, will clearthe obstacle at the PC.(3) Compute and record the length of line PWso that point W is on the tangent line AVand line PW is perpendicular to thetangent. The length of line PW R (l - Cosdp), where dp is that portion of the centralangle subtended by AP and equal to twotimes the deflection angle of P.(4) Establish point W on the tangent line bysetting the instrument at the PI andlaying off angle V (V 180 - I). Thissights the instrument along the tangent3-13
FM 5-233AV. Swing a tape using the computedlength of line PW and the line of sight toset point W.(5) Measure and record the length of line VWalong the tangent.(6) Place the instrument at point P. Backsightpoint W and lay off a 90-degree angle tosight along line PQ, parallel to AV.(7) Measure along this line of sight to a pointQ beyond the obstacle. Set point Q, andrecord the distance PQ.(8) Place the instrument at point Q, backsightP, and lay off a 90-degree angle to sightalong line QS. Measure, along this line ofsight, a distance QS equals PW, and setpoint S. Note that the station number ofpoint S PI - (line VW line PQ).3-14(9) Set an offset PC at point Y by measuringfrom point Q toward point P a distanceequal to the station of the PC minusstation S. To set the PC after the obstaclehas been removed, place the instrumentat point Y, backsight point Q, lay off a90-degree angle and a distance from Y tothe PC equal to line PW and QS. Carefullyset reference points for points Q, S, Y, andW to insure points are available to set thePC after clearing and construction havebegun.Inaccessible PTWhen the PT is inaccessible, as illustrated infigure 3-8, and both the PI and PC are readilyaccessible, the surveyor must establish an
FM 5-233offset station at the PT using the method forinaccessible PC with the following exceptions.(1) Letter the curve so that point A is at thePT instead of the PC (see figure 3-8).(2)Lay the curve in as far as possible fromthe PC instead of the PT.(3) Angle dp is the angle at the center of thecurve between point P and the PT, whichis equal to two times the differencebetween the deflection at P and one halfof I. Follow the steps for inaccessible PCto set lines PQ and QS. Note that thestation at point S equals the computedstation value of PT plus YQ.(4)Use station S to number the stations ofthe alignment ahead.Obstacle on CurveSome curves have obstacles large enough tointerfere with the line of sight and taping.Normally, only a few stations are affected.The surveyor should not waste too much timeon preliminary work. Figure 3-9 illustrates amethod of bypassing an obstacle on a curve.(1) Set the instrument over the PC with the0horizontal circle at 0 00’, and sight on thePI.Check I/2 from the PI to the PT, ifpossible.(2)Set as many stations on the curve aspossible before the obstacle, point b.(3) Set the instrument over the PT with theplates at the value of I/2. Sight on the PI.3-15
FM 5-233(4)Back in as many stations as possiblebeyond the obstacle, point e.(2) Measure line y, the distance from the PI tothe fixed point.(5) After the obstacle is removed, theobstructed stations c and d can be set.(3) Compute angles c, b, and a in triangleCOP.c 90 - (d I/2)CURVE THROUGHFIXED POINTBecause of topographic features or otherobstacles, the surveyor may find it necessaryto determine the radius of a curve which willpass through or avoid a fixed point andconnect two given tangents. This may beaccomplished as follows (figure 3-10):(1)Given the PI and the I angle from thepreliminary traverse, place the instrument on the PI and measure angle d,so that angle d is the angle between thefixed point and the tangent line that lieson the same side of the curve as the fixedpoint.3-16To find angle b, first solve for angle eSin e Sin cCos I/2Angle b 180 - angle ea 180 - (b c)(4)Compute the radius of the desired curveusing the formula
FM 5-233(5) Compute the degree of curve to fivedecimal places, using the followingformulas:(arc method) D 5,729.58 ft/RD 1,746.385 meters/R(chord method) Sin D 2 (50 feet/R)Sin D 2 (15.24 meters/R)(6) Compute the remaining elements of thecurve and the deflection angles, and stakethe curve.LIMITING FACTORSIn some cases, the surveyor may have to useelements other than the radius as the limitingfactor in determining the size of the curve.These are usually the tangent T, external E,or middle ordinate M. When any limitingfactor is given, it will usually be presented inthe form of T equals some value x,x. In any case, the first step is to determinethe radius using one of the followingformulas:Given: Tangent; then R T/(Tan ½I)External; then R E/[(l/Cos ½I) - 1]Middle Ordinate; then R M/(l - Cos ½I)The surveyor next determines D. If thelimiting factor is presented in the form Tequals some value x, the surveyor mustcompute D, hold to five decimal places, andcompute the remainder of the curve. If thethen D islimiting factor is presented asrounded down to the nearest ½ degree. Forexample, if E 50 feet, the surveyor wouldround down to the nearest ½ degree, recompute E, and compute the rest of the curvedata using the rounded value of D, The newvalue of E will be equal to or greater than 50feet.If the limiting factor is the D is rounded isto the nearest ½ degree. For example, if M45 feet, then D would be rounded up to thenearest ½ degree, M would be recomputed,and the rest of the curve data computed usingthe rounded value of D. The new value of Mwill be equal to or less than 45 feet.The surveyor may also use the values fromtable B-5 to compute the value of D. This isdone by dividing the tabulated value oftangent, external, or middle ordinate for al-degree curve by the given value of thelimiting factor. For example, given a limitingtangent T 45 feet and I 20 20’, the T for al-degree curve from table B-5 is 1,027.6 and D 1,027.6/45.00 22.836 . Rounded up to thenearest half degree, D 23 . Use this roundedvalue to recompute D, T and the rest of thecurve data.Section III.COMPOUND AND REVERSE CURVESCOMPOUND CURVESA compound curve is two or more simplecurves which have different centers, bend inthe same direction, lie on the same side oftheir common tangent, and connect to form acontinuous arc. The point where the twocurves connect (namely, the point at whichthe PT of the first curve equals the PC of thesecond curve) is referred to as the point ofcompound curvature (PCC).Since their tangent lengths vary, compoundcurves fit the topography much better thansimple curves. These curves easily adapt tomountainous terrain or areas cut by large,winding rivers. However, since compoundcurves are more hazardous than simplecurves, they should never be used where asimple curve will do.3-17
FM 5-233Compound Curve DataThe computation of compound curves presents two basic problems. The first is wherethe compound curve is to be laid out betweentwo successive PIs on the preliminarytraverse. The second is where the curve is tobe laid in between two successive tangents onthe preliminary traverse. (See figure 3-11.)Compound Curve between SuccessivePIs. The calculations and procedure forlaying out a compound curve betweensuccessive PIs are outlined in the followingsteps. This procedure is illustrated in figure3-11a.(1) Determine the PI of the first curve atpoint A from field data or previouscomputations.the tangent for the second curve must beheld exact, the value of D2 must be carriedto five decimal places.(9) Compare D1 and D2. They should notdiffer by more than 3 degrees, If they varyby more than 3 degrees, the surveyorshould consider changing the configuration of the curve.(l0) If the two Ds are acceptable, then computethe remaining data and deflection anglesfor the first curve.(11) Compute the PI of the second curve. Sincethe PCC is at the same station as the PTof the first curve, then PI2 PT1 T2.(2) Obtain I1, I2, and distance AB from thefield data.(12) Compute the remaining data and deflection angles for the second curve, andlay in the curves.(3) Determine the value of D1 , the D for thefirst curve. This may be computed from alimiting factor based on a scaled valuefrom the road plan or furnished by theproject engineer.Compound Curve between SuccessiveTangents. The following steps explain thelaying out of a compound curve betweensuccessive tangents. This procedure isillustrated in figure 3-llb.(4) Compute R1, the radius of the first curveas shown on pages 3-6 through 3-8.(1) Determine the PI and I angle from thefield data and/or previous computations.(5) Compute T1, the tangent of the first curve.(2) Determine the value of I1 and distanceAB. The surveyor may do this by fieldmeasurements or by scaling the distanceand angle from the plan and profile sheet.T1 R1 (Tan ½ I)(6) Compute T2, the tangent of the secondcur
FM 5-233 Length of Curve (L) Long Chord (LC) The length of curve is the distance from the The long chord is the chord
Texts of Wow Rosh Hashana II 5780 - Congregation Shearith Israel, Atlanta Georgia Wow ׳ג ׳א:׳א תישארב (א) ׃ץרֶָֽאָּהָּ תאֵֵ֥וְּ םִימִַׁ֖שַָּה תאֵֵ֥ םיקִִ֑לֹאֱ ארָָּ֣ Îָּ תישִִׁ֖ארֵ Îְּ(ב) חַורְָּ֣ו ם
SOLIDWORKS creates surface bodies by forming a mesh of curves called U-V curves. The U curves run along a 4-sided surface and are shown in the magenta color. The curves that run perpendicular to the U curves are called the V Curves, and they are shown in the green color. Certain commands may offer the preview mesh where these
Figure 10. Centrifugal Pump Performance Curves 37 Figure 11. Family of Pump Performance Curves 38 Figure 12. Performance Curves for Different Impeller Sizes 38 Figure 13. Performance Curves for a 4x1.5-6 Pump Used for Water Service 39 Figure 14. Multiple Pump Operation 44 Figure 15. Multiple-Speed Pump Performance Curves 45 Figure 16.
applications. Smooth degree-3 curves, known as elliptic curves, were used in Andrew Wiles’s proof of Fermat’s Last Theorem [11]. The points on elliptic curves form a group with a nice geometric description. Hendrick Lenstra [5] exploited this group structure to show that elliptic curves can be used to factor large numbers with a relatively .
Sebastian Montiel, Antonio Ros, \Curves and surfaces", American Mathematical Society 1998 Alfred Gray, \Modern di erential geometry of curves and surfaces", CRC Press 1993 5. 6 CHAPTER 1. CURVES . Contents: This course is about the analysis of curves and surfaces in 2-and 3-space using the tools of calculus and linear algebra. Emphasis will
Hermite Curves Bezier Curves and Surfaces [Angel 10.1-10.6] Parametric Representations Cubic Polynomial Forms Hermite Curves Bezier Curves and Surfaces . Hermite geometry matrix M H satisfying. 02/11/2003 15-462 Graphics I 25 Blending FunctionsBlending Functions Explicit Hermite geometry matrix
improve current MAC curve generation is given. The next section will explain and compare in detail the existing approaches to the generation of MAC curves, namely expert-based and model-derived cost curves. Section 3 then turns towards climate policy instruments and demonstrates the degree to which MAC curves can
on curves with moderate and sharp curvature, and upgrading specific roadside improvements. Expected costs should be com pared with estimated accident reductions to determine whether geometric improvements are warranted. Horizontal curves are a considerable safety problem on rural two-Jane highways. A 1981 study estimated that there are
