Question

ть KŽ Uc F 1 Forces on the wheel and suspension F, = K, (x– x,,) F, = K (x, -x)+C(, x;) mw кр F, Equations of motion muž, = F

Please explain and describe in details by step step how the transfer function between the road surface displacement and the acceleration of the car body is derived and how the transfer function between the road surface displacement and the relative movement between the car body and the tyre is derived.

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Answer #1

Forces on wheel a suspension Frakplag tw) Feks (aw-lo) + Cgliw-in) Equations of motion No 5 = Fe - Fe mbah = Fs Apply loplacesubstitute equ @ into ③ (mustus 5+ ks) deb (3) · (kercsa) (kpXgcs) + Cassiks) Xb(s) ) Emosies, epoks) Xhirsisember cssiks) -Denominator Can t . *acs) be further solved ть му ѕч. mb es s*.4 mhkos, htc, s., те се с 3 stessotsests + maks 3² est55+ ks k2 f (sobedya [Casque 004 saylan 55254 +v](5298- 536x] Cos you (5194-(57mx)335254 - Csobraze (1998-1999X datmu 2 mbs? 2 (539xt (rus)-ть:)) (tes tes) (~ься -to) mbs (mws²4 kp) Ерхү04) [uts) - Yь (3) s)7». mbs*(no s“-ko) (kg-s) (~ь з аmоѕчко); kP (mbsRelative movement a - ده хь consider Vals) - Vb(s)2 s (xw(s)-xb(s) Vw (s)-Vb(s) z lcp (mb mw sht mbkps) [mb miste mb kopsht (

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