In the equations above, we are assuming that the adhesion coefficient (peak value) of the road surface is minimum

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Luke Ray
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1 ME 466 SPRING 017 PERFORMANCE OF ROAD VEHICLES HOMEWORK #4 SOLUTIONS Solution 1: a) F p P i A mc l p η p Pa π( m) N (Ans. ) d η i (μ i ) utilized (μ i ) utilized F xi F zi F xf K f P f F xr K r P r π( ) 4 F xi K i P i n ia wci η ci C R i P i r w π( ) 4 So the fixed bfdf is 5300 / (5300+3) 0, N N 0. 3 d unloaded d loaded d (F x) total W 5300 N + 3 N 1179 kg m s 0. 65g 5300 N + 3 N 1587 kg m s 0. 48g In the equations above, we are assuming that the adhesion coefficient (peak value) of the road surface is minimum 1 P a g e

2 id 0,705*0,65 f 0,58 b h 559 d 0,65 0,65 l l i d 0,95* 0,65 r 0,91 a h 559 d 0,35 0,65 l l 616 0,91 for the unloaded case, and id 0,705*0,48 f 0,5 b h 50 d 0,55 0,48 l l i d 0,95*0,48 r 0,40 a h 50 d 0,45 0,48 l l 616 0,56 for the loaded case, therefore we are assuming that the adhesion coefficient of the road surface is minimum 0,91. F zf W ( b l + h l d) F zf,unloaded 1179 kg m 559 ( ) 914 N s 616 F zf,loaded 1587 kg m 50 ( ) N s 616 F zr W ( a l h l d) F zr,unloaded 1179 kg m 559 ( ) 44 N s 616 F zr,loaded 1587 kg m 50 ( ) 550 N s 616 (μ f,unloaded ) utilized F xf 5300 N F zf,unloaded 914 N P a g e

3 (μ f,loaded ) utilized (μ r,unloaded ) utilized (μ r,loaded ) utilized F xf 5300 N 0. 5 F zf,loaded N F xr 3 N F zr,unloaded 44 N F xr 3 N 0. 4 F zr,loaded 550 N η f,unloaded η r,unloaded (Ans. ) The small one is chosen which is 0.71 η f,loaded (Ans. ) 0. 5 η f,loaded The small one is chosen which is 0.9 b) i b l + h l d i d unloaded d unloaded 0. 57g i d loaded d loaded g 3 P a g e

4 For the unloaded condition, front wheels lock first for decelerations up to 0.57g. Above 0.57g, rear wheels lock first. For the loaded condition, front wheels lock first for decelerations up to 0.780g. Above 0.780g, rear wheels lock first. While the loaded condition can be accepted, the unloaded condition cannot be accepted for this brake system because European standards require that the front wheels must lock-up first for decelerations up to around 0.80g. It can be concluded that this brake system was designed considering the loaded condition. Solution : a) d f μ (b l ) i μ ( h l ) μ ( a l d r ) (1 i) + μ ( h ) l The Matlab script for the solution is as the following: close all clear i0.:0.01:1; l.865; % wheelbase [m] a_yuksuz1.333; a_yuklu1.458; b_yuksuzl-a_yuksuz; b_yuklul-a_yuklu; mu0.85; h_yuksuz0.63; h_yuklu0.508; df_yuksuzmu*b_yuksuz/l./(i-mu*h_yuksuz/l); df_yuklumu*b_yuklu/l./(i-mu*h_yuklu/l); dr_yuksuzmu*a_yuksuz/l./(1-i+mu*h_yuksuz/l); dr_yuklumu*a_yuklu/l./(1-i+mu*h_yuklu/l); figure plot(i*100,[df_yuksuz;df_yuklu;dr_yuksuz;dr_yuklu],'linewidth',) grid 4 P a g e

5 legend('d_f unloaded','d_f loaded','d_r unloaded','d_r loaded') ylim([0.5 1]) xlim([30 100]) xlabel('brake force distribution factor') ylabel('max decel before front - rear wheel lock [g]') b) The BFDF corresponding to the intersection of the curves for the front wheels of the loaded vehicle and the rear wheels of the unloaded vehicle can be seen as 0.68 on the following figure: 5 P a g e

6 At this point d0.78, i0.68 and μ0.85 which is its maximum value. c) η r,unloaded η f,unloaded d μ d F xf F zf η f,loaded η r,loaded d μ d F xr F zr d d (Ans. ) μ max Wid W( b l +h l d) d ( b l + h l d) i d (Ans. ) μ max W(1 i)d W( a l h l d) ( a l h l d) 1 i The smaller ones are chosen which are both 0.9 The Matlab script for the solution is as the following: close all ( ) ( ) P a g e

7 clear i0.:0.01:1; l.865; % wheelbase [m] a_yuksuz1.333; a_yuklu1.458; b_yuksuzl-a_yuksuz; b_yuklul-a_yuklu; mu0.; h_yuksuz0.63; h_yuklu0.508; df_yuksuzmu*b_yuksuz/l./(i-mu*h_yuksuz/l); df_yuklumu*b_yuklu/l./(i-mu*h_yuklu/l); dr_yuksuzmu*a_yuksuz/l./(1-i+mu*h_yuksuz/l); dr_yuklumu*a_yuklu/l./(1-i+mu*h_yuklu/l); figure plot(i*100,[df_yuksuz;df_yuklu;dr_yuksuz;dr_yuklu],'linewidth',) grid legend('d_f unloaded','d_f loaded','d_r unloaded','d_r loaded') ylim([0.1 0.]) xlim([15 95]) xlabel('brake force distribution factor') ylabel('max decel before front - rear wheel lock [g]') 7 P a g e

8 At this point d0.19, i0.55 and μ0.0 which is its maximum value. η r,unloaded η f,unloaded d μ d F xf F zf η f,loaded η r,loaded d μ d F xr F zr d d (Ans. ) μ max 0. 0 Wid W( b l +h l d) d ( b l + h l d) i d (Ans. ) μ max 0. 0 W(1 i)d W( a l h l d) ( a l h l d) 1 i The smaller ones are chosen which are both 0.95 Solution 3: For power limited gradability: ( ) ( ) W f 1396 kg kg f W r 1396 kg kg f 400 a f ( ) ( ) a r ( ) ( ) a a fw f + a r W r W inches m r w [ m] m 1 inch P 157 Nm 3750 rpm Ω π rad 1 rev 1 min 60 s 3750 rpm rpm W 8 P a g e

9 A Ω ( P ) ( 1 ) (0. 65) ( P 1 Ω (1 Ω) ( ) B 1 A 1 1 Ω ) ( 1 (0. 684) ) C 1 Ω B A f m m m i t i d a 1 η t [ r w (1 S) ] i t i d a η t [ r w (1 S) ] P 1C n 1 3 KC DA f [ m ( ) ] W ( ) π rad (5750 rpm 1 m s ( 3. 6 kph ) P 1B n 1 bw rev [ m ( ) ] W π rad (5750 rpm kph 1396 kg m N s 1 rev kph a i t i d 3 η t r w (1 S) P 1A W(a + sinθ) n m ( ) W (0. 684) π rad 5750 rpm m N s 1 rev N kph 1 min 60 s )3 1 min 60 s ) 1 min 60 s 1 m s 3. 6 kph 1396 kg Θ sin 1 { 1 W [ (a ) + a 4a 3 ]} 1 sin { 1396 kg m [ ( ) ]} ( ) s 56% 9 P a g e

10 For slip limited gradabality: This is an indirect gearbox, so it can be FEFWD or RERWD vehicle. From the load distribution, it can be concluded that it is FEFWD vehicle. Θ tan 1 l ( μ f r 1 + h μ ) l b f r a + bv Velocity dependent rolling resistance coefficient, b, can be neglected. For dry, cement- concrete surface: Θ tan ( ) % Applying the same formula for all cases gives the table below: Gradability [%] Road Surface/Condition Dry Wet Greasy Frozen Cement-Concrete Asphalt Cobbled stone-large Cobbled stone-small It can be seen that all the values for slip limited gradability are smaller than that for the power limited gradabality 9.4. Those values in the table are the gradabality results for corresponding conditions. 10 P a g e

Vehicle Dynamics CEE 320. Winter 2006 CEE 320 Steve Muench

Vehicle Dynamics CEE 320. Winter 2006 CEE 320 Steve Muench Vehicle Dynamics Steve Muench Outline 1. Resistance a. Aerodynamic b. Rolling c. Grade. Tractive Effort 3. Acceleration 4. Braking Force 5. Stopping Sight Distance (SSD) Main Concepts Resistance Tractive