is THESIS NL 3 1293 00991 0773 PLACE IN RETURN BOX to remove this checkout from your record. TO AVOID FINES return on or before date due. hfe ed » eee ts A STULY OF TH DESIGN OF AN aAUTOMOSILS MOOR. & Report yubmtted to the Yaoulty of the Miohigan agricultural College by iy Forest ki Mo¥arlmd oe Candidate for the Degree of Bachelor of solenoe. June, 1921. THES.” GENERaL OONSTHU CTION. The motor designed is intended to be used in a car of approximately one hundred sixteen inch wheel base and weighing about twenty four hundred pounds. In general the motor is of the four cylinder type of three and ons half inch bore by five inah stroke. The main ideas to be worked out in the design of the motor are, a orankshaft of the three bearing counter- weight type, an overhead camshaft operating two inlet and two exhaust valves to a cylinder through rocker arms, oonneoction between the orankshaft and camshaft by helical gears on a vertical shaft, generator and water pumps to be driven by a horizontal shaft at right angles to the crankshaft and oonneocted with it through helical gears, and pressure lubrication. The design of the parts will be given in the order it was taken up. It is believed this method will present the oomputations most olearly. 102395 oy > ORAMKSHAP? all) BEARINGS. Crankshaft. The crankshaft is to be of the three bearing counterweight type, dropforged to rough dimensions and them machined all over. Extra material is left in the counterweights as as to bring an unbalanced couple st these points to be balanced out by a (Carwin static Dynamic) balancing mohine. In case a Varwin machines is not available or it would not be good polioy to buy one ,the weights should be then computed so that the couple formed by the counterweights just balancesthe couple formed by the orank pins and the lower half of the conneoting rods. Let p = piston displacement, 4 = Diem: of orank pin ani min orank journal. ly lp eri lg = Length of min crank journals front to rear. we width of crank am. — t, = thiokmess of short orank arn, t,. = thickness of long crank arms. 1 Pp = i 3.6" (5) = 48.1 qm. ine From Heldt Vole 1 - 189, 190, i191. anlP alwin eo 1.733. a +f sre? It is desired to drill a 3/4" hole through the pins so let ws equate the section modulus of the solid shaft to that of a hollow shaft 3/4" inside dian. and 2" outside diameter. md? . 5. yr a,8 e-° Ee - e+ 1.738° = 25 ~ 75° 5.18 = 8 +.421 = 7.579 We are safe in the dimensions assumed. From Heldt. L = 1.25 d = 1.25 (1.733) = 2.168 = 2 3/16". Ly = L, = 1.26 L = 1.26 (2.168) = 2.71" Ly = £2.75" It is required that bearing one be as short as possible so let ws compute the area required and find the length required in this manner. Area required = 1.733 (2.71) = 4.70 sq. in. 4.70 Bde ine , 2" Length = = 2.35". Allowing 1/4" for lubrication grooves gives us 2 5/8". Lg = 1.75 L = 1.75 (2.168) = 3.79" 3.79 (1.733") Length required = ———y7—"— * 3.28". Use 3 5/8" to allow for 1/4" 011 groove, weaile4den= 1.4 (1.733) « 2.43" use 2.5" t= 6/9 6 [R27 8" 76S e876" use 1" . (ae ~ .9 pe 75 . ty -8)— 8 os # 1.17" use il 1/2 ta and ty are mde extra large to reduce bending of the shaft under lowto a minimum. | Steel for Crankshaft reoommonded by Heldt. Chemical Composition. Carbon coscccccncccccaccscovccccccece 0665 MANBANEOBE esccccccccccccccccccscscsece 060 SLLL0ON ceccccescscccccesese NOt Over 18 Sulphur cececcoscescsesscce NOt OVer 04 PHOSPOLUsB ceocnevacccesesese NOt OVer 04 Physical Properties Tensile strength ecccceseee 90,000 lbs. per sq. in. Elastio Limit ecosccesesese 70,000 lds. per Bq. in. Wlongation in 2 inches ..ce 15% Contraction Of area wecccee 30% Crankshaft Bearings. Use die cast buehings. Heldt vol. 1 - 347. & = Bearing diameter «= 2" Be Outside diam. bushing = 1.284 « 1.2 (2) » 2.4" use 2/2” Thickness of flange = .1A = .1(2) = .2 use 1/4". Diameter of flange = B+ .24 = £2.44 .2(2) ¢ 2.8" use 2 7/6" Use 4 1/2" bolts for each bearing. FLYWHEEL Let flywheel] diameter equal 14" and thickness )}" “- | / — “ Ce OOD NT Welgnt flywheel « ae ( Sab, taizaxel = 8.4, Sab (5) | 22 Sw, Width (approx) = * a 1 « 28 TTAL aM (thickies:) Re ee Ieat Uciekosag of wab equal .f inches ahiah Js soverslly used. Diameter of VPieonsa « Lencth of streke = 5* Heldat 1 = °0% Width of Plance * 1/18 ( ‘nel. cf hore} (e vcore reot of number of oylincers) == 1/36 €- .5) (°) = 4-357" 3ccf efi” Dianster ef helts = "hiekren: of Plence = 6" 5A COMBUSTION CHAMBER. Asmme height of chamber » 1 ingh. Clearance volume = Depth (Area rectanghe - area rounded corner) - volume of piston abvre cylinder proper - 1 |4 (4.625) -] 2.262 ~ gl (2.26)? -[ = (3.6%) =/e]- 12-81 ou. in. fotal vol. = 12.81 cue ine + 48.1 aue in = 60.91 ou. ine 60.9 Compression ratio e« a 4.75 Assume an inlet pressure of 13 pounds per square inoh. Compreseion pressure = inlet pressure (ratio) 1.3 o 13(4.75)2°5 = 96.S#/s8q. in. absolute » 85.6 lbs. per 8qe in. gnuge. This pressure is satisfactory for a high speed engine so it will not be necessary to oarry the work further. CYLINDER ‘ND CRARKCASE. DIMENSIONS. Heldt VYol. 1 - 83. bore | Thickness = “35 Cit 1/8 in. = ~s + 1/8 = 2416 in. use 3/8 in. oylinderwa//. Water space = ]/£" Jacket wall = 3/16" on the sides, 3/4 to 11/2 in. on the top of Gylinder. Crank case thickness = J/:% in. Planges =|. (thickness) Loh/B) =» 3/3 in. Use 1/4 in. bolts with a 4 in. span. CONNECTING ROD AND PINS. Let ratio connecting rod to stroke equal 2.2 Length oonneoting rod 2.2(5) = 11 in. Use nioxel steel oe = 22000 lb. per sa. in. Heldt 1 - 143. Explosion force = gompression pressure (4) = 98.3 (4) = 393.2 Total force on piston = + 3.5% 393.2 = 8780 lbs. Heldt 1 - 220 gohart III. f » .169 use 5/32". Height of Yeotion = .159(5.7) = .906" use 15/16". Width of seotion = .159(3.8) = .604" use 5/8" The rod is of this apgssotion for entire length. Enlarging the rod at the crank pin end has no advantage as the greatest stress coms at the middle of the rod due to whipping. If any cOggection should be made larger this should be. Heldt 44 Equa. 24. ¥. » foroe due to reciprocating parts, we wt. of reoiproosting parts, eaasume 4.65 lb. L ¢ Length of stroke in inches, He RPM n =Fratio connecting rod "center to center" to stroke. Fg = .0000142 win®loove + 1/2 n cor 2 0) 1d, « 1 20000142 (4.5) 5 (3000)"(1 + —r- 3640 lbe. Héldt 60 Kqua. 31 ¥ = Uentrifugal force, we Weight of revolving parte, N= Rev. per sec. «= 3000/60 r= radius in feet, F = 1.226 WN” yr lb. Assume lower half of conneoting weighs 2 lbs. a:d r= 8".= Fr = 1.826 (2) 3000" 60 ye-~e 1620 lbs. fotal force = 3540 1lbde + 1530 Lbe s 6070 Lb. Using two bolts stress per bolt = 2036 lb. Assume a stress of 22,000 lb. per square inoh. 25 Area each volt =~ B28e Be B8aqe ine e2009 7 *215# 8a Use a 7/16 in. SeAe. bolt, a ew 12.7 Sqe ine Length bolts = { 1/2". 10 PISTON AND OISTON PIN. Heldt 144 - 146. Average piston prossure = 2500 lb. per eqs ine Area Of besring «a eane =» 1.51 square inch. Assume 7/8" outside diameter. i 1.51 " Length = 4/6 = 1.73 say 1.75". Let dj = inside dinneter, d =» outside diameter, b « piston bore, p = Kxplosion pressure, s «= Stress = 20,000 lb. per sqe in. (3.6 niokel steel ) 4 a, = [at - avdp * forse - 876(3.6)° 3760 28 2 (20,000) -681 use 5/8" PISTON DIMENSION. Average practice as given by Heldt at 1800 ReP.M. an average side thrust of 110 lba. was found. Allowing 9 lb. per sq. in. bearing pressure on the piston we have a1 Length piston ye = 5.5" bearing length. Diameter piston at top = .998 (3.5) = 8.493 in. Diameter piston at bottom = .9995 (3.5) = 3,498 in. Thickness of head » .032 (3.5) + .060 = .182 use 2 Depth of ring grooves = .04( d:). J.) ) 604(8.5) = e140 Say 9/64" thiokness of piston wall upper rod » e062 (3.5) ¢ e190 & .317 use 5/16. Thickness wall lower end ¢ .02(3.65) + .05 = 12 use 1/8 in. Use & 1/4 in. set sorews to nold the pin in position. A 6004 grade of cast iron to be used. 12 VALVES AHD KOCKER ARMS An effprt has been made to keep the weight of Valves and rooker arm to as low a figure as possible to inéurs quiet running. Maximum epeed = 3000 R.?P.Me Maximum piston speed = 3000. 10/18 » £2500 Feet per minute. Pavary 64 - 66. Let p = piston diam. «= 3.5" Let 8 = piston speed = 2800 ft. per min. Let G@ = gas velocity = 8180 ft. per min. Let A @ Area of one valve (two intake and two exhaust per oylinder) A - 2. 7854 (p)® 9 | 2 (7854) 8.5" p600) | G z 8180 147 Bde in. Diame of valve . 1.47 « 1.37" use il 3/8 in. | oe 7854 Left of valve ? <7. + (a «1 1/4) 4. . 52 $2 wi 29 2. 20 re go (1 3/8 ~ 11/4) 1/8 =» a7 way ey or ge 13 Valve Dimension: Outside diam. » 1.18 (1.375) = 1.58 say 1 9/16" Thiekness of head = .15 (1.375) = .200 say 3/16" Valve stem diameter = .15 (1.375) + .15 = .36860 say 6/18" Radius of valve = .25 (1.376) = .344 Say 6/16" 14 VALVE GPRINGS Weight of valve, disc, Spring and rooker erm. Wte Val ve 228 08 . ( C2 (.3125)2 + .1876 _- (1.375)2/ « 28(.680) = 174 1b. Wt. spring cap = .28 ([ (28) .0625 = .28 (.196) ¢ .056 1b. affective weight of spring = 1/2 total weight. assume a spring of 10 ooile, 21.n diam and .134 in. wire diam. | Wt. spring = «26 10 f27) 2 (1342) « .248 1b. Effective weight = 1/2 (.248) lb. =.124 ld. Consider rocker arm as two bare 1/2 in wide tapering to 1/4 inch, 5/16 in. thick, 1 1/4 in. long and revolving through an aro of 5/16 in with one end fixed. Wt. rooker arm = .28(2) (1.25) (.376) (.3125) = O82 lb. The ir weight with the motion spoken of it spproximately equivalent to half their weight reciprocating through a Aistance of 6A16 in. Therefore, the approximte weight « e041 lbs. Total weight = 174 + .055 + 1B + .041 = .594 lb. Let us use a mashroom oam with a top oirole of 1/10 in. Distance center top cirole to center of camshaft «= 15 3 = .875 which will be arrived at in the canshaft computation. Heldt 1 ~ 2665. Let ¥. = spring DProe, we wt. reciprocating parts = .394 lb. Mo e= kKevolution of camshaft = 1600 Ow Value Given above = 22.08 lb. 2 2 pao WM _D_ . 2394 (1500)” 8750 . 20.08 ip, $5, B00 56, 200 allowing 20% for friction of gHrdes, eto., we have P = 22.08 (1.80) ad 26.496 ibe ) ¢ os 26.6 lbe Heldt 582. Let us use No. 11 wire diam. .12 in. and 1 1/4" diam. spring. ‘taking ¥ =» 10,000,000 the ellowable Load = 27.1 1b. and the deflaction per ooil per 100g = 7538". Defleution for 27.l# = .7538 Bree « 2 in. Vhen the spring is fully defleoted the losd should not inorease by mre than 2y vince the load in proportional to the deflection the allowable deflection per oof] = .20 (.204) =» .O#dsin. No ooilea = total deflection + allowable deflection 05125 per coil. = ap" 7.6% Uae 10 ocollis, 8 active. Height of spring when valve is closed,alloring 3/8" between the ooils when valve is opened wide = 10(.12) + 3/8 + .5125 = 1.8875 say 1 7/8 “. Height of spring when free should be 1] 7/&" + 211 (8) = 3.66". 16 17 OaMsHar?, KUOKEH AM DHA ANRD SKAKINGS. Heldt 1 ~- 276. Use 3 bearings, Length between bearings = I = 11.0" X= ratio of distance of cam from bearing to total distance = 8/1) = .727 a= .069 Exhaust Valves. Load on camshaft = Pressure of valve spring + Pressure due to inertia of reciprocating parts + pressure on exhaust valve. : Pressure Of valve spring » 27.1 lb. Pressure due to inertia of reoiprocating parte = 26.5 lb. “ressure on exhaust valve =» 650 + (1.875)? » 73.5 1d. Total pressure » 127.1 lb. Total pressapre two valves = 2 (127.1) » 254.8 lb. Consider that the two valves are both located at the point 6" from the farther bearing. Their oondition my be safely essumed as it will cause a greater deflection than will the two followerswhere they actually are. “au ~ 4 P 1a " 3800 ,00 01.008) 18 where @ = diam. camshaft required P= load exerted by follower, ae factor depending on ratio given above e002 = allowable deflection. -4 4 ( g A = ] 256 25) 969 ‘= 1.078 say 1 1/16 in. TUBE ON WHIOH RUUKEx ARM TURNS. Boyd's strength of Materials, page 176, equation 6 solved, gives for a beam fixed at the ends with concentrated 10ad at the middle. 2 3 £ Bly = —- + OX where Mo = =: Solving thia equa. fur the case where a= b= Xx —_ } and substituting the moment of inertia 4 —z—0f a s0lid oiroular section for I and solving for a = 4\ 64 PL Fe (1T Ey) where 4 = diam shaft 4 we get P » load | L% = length between bearings = 4.375" Zw Modulus of elasticity = 30,000,000 y = Allos deflection = .002 in. Our greatest load on the rocker arm tube will be when the cam starts to raise the valve. 19 Prom soaling off the drawing we find we have a load of 254 1b. at 280 to horizontal ( cam on follower), and a load of practioally 254 lbs. acting vertically up- ward. Solving to get the resultant load on the rooker arm shaft we have R= (254 - £54s¢m 2ef)2 + (254 cog 280)" = 262 lbs. 4 | 4.5760 80 d = | _. &» 6444 in. 192 4 ) 30,000,000 (.002) However it is desired to use gs hollow tube. Let us see if a 1/2" tube with a 1/4" hole Aas a seotion modulus equal to that of the .444 in. solid tube. ma . _j7p8 § pr a8 $2 62 32 aF wp. ad 04445 = 65 » .260 20875 = .125 ~ .25 = .1094 the hollow tube is the stronger and will give even less deflection than the solid tube. The best timing of the valves for a motor of this speed (up to 8000 R.P.M.) to compromise between high erfleienoy and power at high speeds and smooth Operation at low speeds seomsto be - Let inlet oam open 16° past top deaé center, Let inlet cam alose 40° past lower dead oenter, Let exhaust cam open 48° before lower dead center, Let exhaust cam close 10° past top dead center. £0 Inlet can: - Heldt 1 ~ 264. HK, = radius of oan at base = .626" Yr = radius of top circle = .0625 L = left = .312 B= Center top circle to center of shaft « R, + L- Fe .6250 + 3125 - .0625 = «87650- g = Angle of Oopming on camshaft = 180 + 40° ~ 157 . 102.50. 2 Kk = Headius of flank oirole « p? . r® R28» Bt) Bon 2 (Ry - r - Boom G- | (.8750)" - ,06257 + .625" - 2(.625) 875 ($84) . 16.26 in. 2 (2625 - .0625 - (.6750)..626) Exhaust aam:- g - 180° + 46° - 107 = 218° 2 f= 54.5° ums-2 = .68 *« 109° on camshaft. Ry, = «626 re .0625 P = «87650 28750" - .06252 « .628% ~ 2/.625) . 875 (e580 Ke ( 660) 4.66" 2 (.626 - .0625 - (.875).58) Heldt 1 ~- 276. Exhaust oam width = 1/3 equivalent diameter of valve equal in area to the two valves.=« 1/3 (1.94) = .667 in. Allowing 1/16" for oil hole in follower gives .647 + 20625 = .7095 say 3/4". Inlet oam width =» 1/3 (1.575) + 1/16" for o11 hole od e458 > 20625 . = .5205 say 9/16" CAMSHAFT BRAKING. Practice is to mke the length of the oam shaft bearings from 2 to S times camsheft diameter or from 11/6 (2 to 3) or 21/4" to 3 3/8". Let us make front bearing 3 in. long Let us make middle bearing 3 in. long Let us make rear bearing 2 1/2 in. long GEARS DRIVING CAMSHAFT, WATER PUMY , GENERATOK, and BEARINGS. Horse Power to Drive Uanshaft. Consider the work done in one revolution. Load to opm one exhaust valve = 127.1 Load to open eight exhaust valves « 8(127.1) = 1016.8 Load to open one inlet valve = 60.46 Load to open eight inlet valves = 8(50.45) = 403.60 Total load = 1466.40. Work per rev. = total load (Lert) = 1420.40 .5/16 1/12 = 370 ft. lb. HP, = ROrk per reve (KePeMe) _ 311500) . 4 6g2 33,000 33,000 Allowing 20% for friction loss HP = 1.682 (1-20) = 2.01840 Torque «65,000 . (HP) ) 3022000 (2) » lb. RePeMe is,000 «SH Ame 1 ¥ront End Gears. Aseume diam. of 3" for gear on crankshaft Asaune diam. of 2" for lower gear on vertical shaft Assume diam. 1 1/4" for upper gear on vertical sheft assume diam. 3 3/4" for gear on camshaft. Marks 729 - 30 } Compider gear on oranxshaft and the one on vertical shaft meshing with it. Lot My = Xo2.. of crankshaft = 3000 Hg = RPM. Of vertioal shaft = 4500 Kp = Kadius of crankshaft gear = 1.5 in. Hg = Radius of vertiosl shaft gear «= 1 in. Xq = Z of true pitoh line to axis of gear D Xp = & of true pitoh line to axis of gear D og 1:5 coe Xp 4 oor X, COB Xq = CoB Ky but CoB Xqg = Bin. (90 - X,) = sin. Xq Sin Xg = oof X, Xq = 458 Xy = 90° - 45° = 45° Let us now consider the camshaft gear B and the vertical shaft gear A. Notation as before Bp Ry CoB Xa 4500 1.876 ocosX, 1500 ~ .625 ooxsX, 1 = Cox Xp Cos Xa X= te ® 458 Let us check the gears for strength. Marks 752. 25 Load on gear 3 a S84 ine 1be 1 gag an. 1.875 in. Speed of pitch line (apparent) = 1600 _ (1.875) o 12 334 ft. per min. f= 12,000 (tadle 21 cast steel gears) X = pitoh line angle = 45? b «= width of face = ra ine P » circular pita Let us try diametral pitch of & ps gy Unwin's formila for helical gears, Load = = .0833 b p f ces. *& = 0833 (re—) 3 (12000) .7072 os 06147 Oversize gears are to be used to make the radial loads and thrust loads smaller and to obtain a fairly large number of teeth per gear in order that the gears may run sil@mtly. Load on teeth of upper gear on the vertical Bhaft = 44.8 lb. Barring the thrust load due to friction which would be emall,44.8 1b. is the thrust to be taken oare Of in the vertical shaft and camshaft. This thrust load is greater than the thrust loads on the lower gear of the vertical shaft or the thrust load on the pump and generator shaft. This shaft runs at 11/2 orankshaft speed and meshes with the crankshaft by a gear the same size as that on the lower end of the vertioal shaft, and will ordinarily take less than 2 horse power. Because of this a oombined radial the thrust ball bearing suitable to take the loads here can als0 he used in the other places. Let us take a new departure comined radial and thrust bearing 20 mm. bore, (.787") a » 1.027" in. inside shoulder, B. =» 1.596 in. outside shoulder, diam ; 2.05" width = .§91". This bearing will take oare of 125 lb. at 3000 R.P.M. so is of ample sise. On the vertical shaft usc 0 modified Square — gs al Yorns, Bloc Ty pe of Uriver; couple it with the shafts carrying the gears. ad ow INTAKS AND RXHAUST MANIFOLDS. Intake Manifold. The design of the intake mnifold is of oonsiderable importance in the performance in an engine. The one factor is the smoothness in the operation of an engine, is that each aylinder receive a charge identical with that received by esaoh of the other oylinders. In the design, at attempt has bem mde to give each cylinder a charge of the same quality and the sane volume at the same temperature or presmre, as that of each of the other ay lindere. Sharp corners have been avoided in order that the friction loss be as emall as possiblo. Fh Exhaust Manifold. As moh radiation surface as possible has been given the exhaust manifold s0 that the volume of the gas will be reduced,thereby reducing the pressure at the exhaust valvos. £5 WATER PUMP AND COOLING SYSTEM. A single bearing centrifugal pump is to be used. Heldt 1 ~ 612 - 513. Dy » Impeller diameter, V = Cue in. piston displacement, N= Hatio of pump speed to orankshaft speed, C = Vane width at impeller circumference, Cy = Vane width at inlet cirmmference, 4 V 192.4 Dy = 1.1 Hee aa To = 3.9 say 4 in. Cs 12D = .12 (4) = .48 = .48 say 1/2" Cpe ofS De ote (4) = .7oeay 2 3/4" Water jacket espace = 1/2". An attempt has been made to provide for equal cooling for the cylinders by making the jackets as symmetrical as posaibile. FAN CALOQULATIONS. Sise and Capacity. Heldt 499 | The 3.T.U.e. t be disposed of per minute approximately equals 68 (maxinmwm Brake Horse Power). This figure is oniy a fairly close approximtion as it is ordinarily used with an L or T head mtor. aBeuming the maximum brake horse power at 48, the B.7T.U. per minute to be disposed of = 68(48) = 3260. assuming maximum speed of 70 miles per hour, general practice has shown that .6 cue ft. of air is required per 3.T.U. entering radiator. ° ous ft. of air per minute maximum speed = 3260 (.6) = 1956.0 Let us design a four blade face of the helical blade type. Assume outside diameter = 16" " ins ide " =» 4" " width of 2" n pitoh of 20" " maximgm speed of 4000 R.P.M. " a 3" diam. for fan pulley, " ea 4" diem. for driving pulley, Heldt 1 - 492. 27 VaNW Ts — (K® - R,2) ) 2 xP 2 xp R's £2 3 y Zo 2 — 1 4 8xFe pe (ko - RY) QBs. pe ge - 72 2104 600 0683 2.642 oa we» 2.0902 e628 0866 a 5° Se » 0764 e660 0605 g . al = .0625 686 0429 2 2 a = .0486 706 0348 ge 2 | osas 713 “laa 7 @ eo e 29248 Total oe 2714 Ved —s— 3.142 (4000) .2770°/ min. The capaoity im emmough greater to take care of slip in the belt. POWER TO DRIVEL FAN. Heldt 1 ~- 446. Nn 377 w pe pe 4] . 2 =o re oo 4 2 HF 49, 660,000,000 R TT 1°, Where N = no bladeg, n= RPM. =» width of blades in feet, » pitoh in feet, = Radius to large oiroumferenoe, 4 ww we 4 Radius to small ciroumferaqoe, : 2. 20 Neaé@ ne 4000 wes ft. P= TF x, 8 -.: no R = wr as 2 —, “Tz F° ss The Il. Pe solves out to 6.57. Torque = 63 000 H Le os 63 000 (6.57) = 10.36 in. 1b. 5 4,000 Net driving tension = 10.56 6.9 lb. 15 DESIGN OF FAN BHLT. Let uw design a V belt of built up oonstruction with sides at an angle of 28°. Machine Lesign ~- Leutwiler ~ 173. 2 TyEp = AL - — oo Sores -1 4 UO © sin.B Trt, » Gi fference in tension, % = Maximm tension, We wt. of belt per foot of lengt, V = Velocity of belt in ft. per sea. U = Coefficient of friotion, 9 = aro of contact in radipa; B= angle of sides of belt, T{Fp ad 6. 9 Let mean width of the belt be 1/2 in. and the thickness 1/2 in. assume blocks to weigh 1/4 of belt proper. Wt. belting = .035 lb. per ou. in. . » Wee per ft. of belt = 1.25 (1/201/2) 12 (.038) = # elS13 lb. V = 4000 ( Tg = 5142 ft. per min. = 52.36 ft. per sec. U from table = .50 2636 8 _ 0 (TT) 6.9 = T) - 01213 oe > Bs EAR ° 08 (TT) 0292 649 6.9 «= ty - Lled 650 6.91 se ty - llel Ty = 18.01 Lbde ioad . 18.0 fs “ area DEt/e » 72.04 lb. per sq. in. fg hs well above 400 1b. per sq. in. so the belt is of ample proportions. LUBRICATION Quantity of O11 Kequired. Allow 25 ou. ine of oil per min. per sq. in. of projected bearing area. Bearing area crankshaft = 2(4(2 3/16) + 2 6/8 + 2 3/4 + 3 58) # 2 (17.77) = 55564 9g. in. Bearing area camshaft = 1.0625 (2.75 + 3+ 2.5) = 8.76 Sq-In. Bearing aren vertical shaft and pup shaft = .75 (3) 1 « 2e25 Baq,q ine Total area = 35. Bet + 8.76 + 2.25 = 46.55 sq.e in. Cu. Ine Of o11 per min. for bearings = 46.55 (25) = 1163.75 ou. in. per min. are Phexe,12 - 1/16 in. holes in the rooker arms, lubricating the canah aft. Assume a 5 lb. per inch head. $= 0, a 2 gh (12) (60)(1728) = “61 (IL) ( tp) ( raz) [64-4 (5) 2.3)(18) 60 (1728) « 441 cu. in. per min. This figure is liberal enough to take oare of the rocker arm bearings also. Total quantity of oil per minute = 1163.75 + 441 = 1604.75 Gue ine per minute. DESIGN OF PUMP. Let us design a pump of the gear type. Heldt 1 ~ 338. Q = Quantity in cue in. per minute D = pitch diameter, # = Face of gear, He Ko: Hu. Of PUMP», P = diametral pitch. Q > Steel cue in. per min. or Fs FeQ QTD Assume a pitok diameter of 1.25" and a diametral pitch of 8 = g 600) 72 " = @ " tT 3 (1025) 4600 0565" = Bay 5 S52 MOUNTING OF KLECTRICAL UNITS. The motor is designed to take a two unit system with the distributor mounted on the end of the Generator. S.A.K- standard mountings provide for the starter md generator. A tube is provided for the protection of the ignition cables. SUS2ENSION OF MOTOR FROM FPHAME. The mtor is provided with three point suspension, two faces on the flywheel housing and one point in front. 335 BIBLIOGRAPHY. The Gasoline Automobile - Yol. 1 - P. M. Hel at, The Gasoline Automobile - Vol. 3 =- PP. Me Holdt, Motor Vehicle " eT . MA Af} a e ir qj i iG ik > oH > s —s~ a - —= ~s _ at _ => —e———_— 2 — N IV. LIBRARIES HLL 910773 a MM 312 a i . NS a mH “q EEEEE & Wy — TQ 5 eae . te ie, cee c 2 <. ae -_ £ Sit i 4 sa. = ——_ ~— 4 iM aE a } ad _—_— STATE a senate ULM 31293009910773 vi yi , ” iw me = a ao ao 7 4 RAR roe STATE UN TES ACA 31293009910773 V. LIB MHI 107 9 ve y = i. “™., ™ =~ 7 ee : - . la — —_ ~ “| a ie az a : = — = a . iin ii - at B, ° ; = — — a | 4 » sal tt) = es ms = Se one as . a: I cl leila irs atpgrcen teen — a \- TNL a yl &. - ~. . ~ a ~~. ——. + i. 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