Running :: essays research papers

  Ã‚  Ã‚  Ã‚  Ã‚  In this study, I investigate the affects that running has on reducing the risk of some health problems. I am doing this because I run about 40 to 60 miles per week, and my family has a history of health problems. For instance, my grandfather suffered a heart attack, and he also had cancer when he was about the age of 50. Furthermore, my grandfather, on my dad’s side of the family, has also had triple bi-pass heart surgery from a heart attack he has had recently.   Ã‚  Ã‚  Ã‚  Ã‚  Here, I present information from some sources that talk about the affects that running has on reducing health risks. My sources agree that running, and some other aerobic exercises, reduce the risk of: Diabetes, diverticular disease, heart decease, several types of cancer, and even common sicknesses like a cold.   Ã‚  Ã‚  Ã‚  Ã‚  One source agrees that running reduces the risk of diabetes. Jim Harmon writes, in Sports Illustrated, about Bruce Leonard, a marathoner with a masters degree in public health. Bruce Leonard went to study the Zuni Indian tribe. This tribe has had a bad history of diabetes until they started to run. Leonard said, After the Zuni tribe started running, â€Å"many Zuni were able to reduce or eliminate their diabetes medication.†(5)   Ã‚  Ã‚  Ã‚  Ã‚  My research also reveled that diverticular disease can be reduced in men that run. For instance, Marty Munson and Teresa Yeykal writes in the article â€Å"Outrun trouble† which says, â€Å"guys who racked up the most ours doing vigorous exercise reduced their risks of diverticular decease by a third.†(38) They also say â€Å"it’s good advice to make your lifestyle to consume high fiber.†(38) I found that running can also reduce the risk of heart disease. The article â€Å"Run for your life† talks about Mitchell H. Whaley, the director of adult physical fitness program at Ball State University. He analyzed data collected from a group of men, which included runners. Mitchell Whaley found that â€Å"individuals with low aerobic capacity have a higher risk of developing premature coronary artery decease that those who were more fit.†(47) The article also talks about Martha L. Slattery, Ph.D., from the University of Utah Medical School. Maria Slattery also found that running is a good preventive measure for heart disease. She says, â€Å"The greatest increase in protection was between those men who were sedentary and those who had some activity.†(48)   Ã‚  Ã‚  Ã‚  Ã‚  Another source I found says that running can help prevent many types of cancer. In the article â€Å"Running for your life†, Doctor Leonard Cohen talks about how running helps reduce colon,

Steel Design

STEEL BEAM DESIGN Laterally Unrestrained Beam Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 1 Non-dimensional slenderness Beam behaviour analogous to yielding/buckling of columns. M Wyfy Material yielding (in-plane bending) MEd MEd Elastic member buckling Mcr Lcr 1. 0 Dr. A Aziz Saim 2010 EC3 Non-dimensional slenderness Unrestrained Beam ? LT 2 Lateral torsional buckling Lateral torsional buckling Lateral torsional buckling is the member buckling mode associated with slender beams loaded about their major axis, without continuous lateral restraint.If continuous lateral restraint is provided to the beam, then lateral torsional buckling will be prevented and failure will occur in another mode, generally in-plane bending (and/or shear). Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 3 Eurocode 3 Eurocode 3 states, as with BS 5950, that both crosssectional and member bending resistance must be verified: MEd ? Mc ,Rd Cross-section check (In-plane bending) MEd ? Mb,Rd Dr. A Aziz Saim 2010 EC3 Unr estrained Beam Member buckling check 4 Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 5 Laterally Unrestrained BeamThe design of beam in this Lecture 3 is considering beams in which either no lateral restraint or only intermittent lateral restraint is provided to the compression flange Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 6 Lateral Torsional Buckling Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 7 Lateral Torsional Buckling Figure 3-1 shows an unrestrained beam subjected to load increment. The compression flange unrestrained and beam is not stiff enough. There is a tendency for the beam to deform sideways and twist about the longitudinal axis. The failure mode which may occur to the beam is called lateral torsional buckling.Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 8 ?Involves both deflection and twisting rotation ?Out-of plane buckling. Bending Resistance M c, Rd ? M pl ? W pl f y ?M0 Due to the effect of LTB, the bending resistance of cross section become less. Failure may occurs earlier then expected Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 9 Examples of Laterally Unrestrained Beam Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 10 Restrained Beam Comparsion Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 11 Intermittent Lateral Restrained Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 12Torsional restraint Usually both flanges are held in their relative positions by external members during bending. May be provided by load bearing stiffeners or provision of adequate end connection details. See Figure 3-4. Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 13 Beam without torsional restraint Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 14 Can be discounted when: †¢ Minor axis bending †¢ CHS, SHS, circular or square bar †¢ Fully laterally restrained beams †¢ ? LT< 0. 2 (or 0. 4 in some cases) – Unrestrained length Cross-sectional shape End restrained condition The moment along the beam Loading – tension or compression Unrestrained Beam 16Dr. A Azi z Saim 2010 EC3 Lateral torsional buckling resistance Checks should be carried out on all unrestrained segments of beams (between the points where lateral restraint exists). Lateral restraint Lateral restraint Lcr = 1. 0 L Lateral restraint Beam on plan Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 17 Three methods to check LTB in EC3: †¢ The primary method adopts the lateral torsional buckling curves given by equations 6. 56 and 6. 57, and is set out in clause 6. 3. 2. 2 (general case) and clause 6. 3. 2. 3 (for rolled sections and equivalent welded sections). The second is a simplified assessment method for beams with restraints in buildings, and is set out in clause 6. 3. 2. 4. †¢ The third is a general method for lateral and lateral torsional buckling of structural components, given in clause 6. 3. 4. Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 18 Eurocode 3 states, as with BS 5950, that both cross-sectional and member bending resistance must be verified: MEd ? Mc ,Rd Cros s-section check (In-plane bending) MEd ? Mb,Rd Dr. A Aziz Saim 2010 EC3 Unrestrained Beam Member buckling check 19 Lateral-torsional buckling Eurocode 3 design approach for lateral torsional buckling is analogous to the olumn buckling treatment. The design buckling resistance Mb,Rd of a laterally unrestrained beam (or segment of beam) should be taken as: Mb,Rd ? ?LT Wy fy ? M1 Reduction factor for LTB Lateral torsional buckling resistance: Mb,Rd = ?LT Wy fy ? M1 Equation (6. 55) Wy will be Wpl,y or Wel,y ?LT Dr. A Aziz Saim 2010 EC3 is the reduction factor for lateral torsional buckling Unrestrained Beam 21 Buckling curves – general case (Cl 6. 3. 2. 2) Lateral torsional buckling curves for the general case are given below : (as in Eq (6. 56)) ?LT ? 1 2 ? LT ? ?LT ? ?2 LT but ? LT ? 1. 0 ?LT ? 0. 5 [ 1 ? ?LT (? LT ? 0. ) ? ?2 ] LT Plateau length Imperfection factor from Table 6. 3 Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 22 Imperfection factor ? LT Imperfection factors ? LT for 4 buckling curves: (refer Table 6. 3) Buckling curve Imperfection factor ? LT a 0. 21 b 0. 34 c 0. 49 d 0. 76 Buckling curve selection For the general case, refer to Table 6. 4: Cross-section Rolled I-sections Welded Isections Limits h/b ? 2 h/b > 2 h/b ? 2 h/b > 2 – Buckling curve a b c d d Other crosssections Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 24 LTB curves 4 buckling curves for LTB (a, b, c and d) 1. 2 Reduction factor ? LT . 0 0. 8 0. 6 0. 4 0. 2 0. 0 0 0. 5 1 1. 5 Curve a Curve b Curve c Curve d 2 2. 5 0. 2 Dr. A Aziz Saim 2010 EC3 Non-dimensional slenderness Unrestrained Beam ?LT 25 Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 26 lateral torsional buckling slenderness ? LT Mcr ? Wy f y Mcr Elastic critical buckling moment Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 27 Non-dimensional slenderness †¢ Calculate lateral torsional buckling slenderness: ? LT ? Wy f y Mcr †¢ Buckling curves as for compression (except curve a0) †¢ Wy depends on section classification †¢ Mcr is the elastic critical LTB moment Dr. A Aziz Saim 2010 EC3Unrestrained Beam 28 BS EN 1993-1-1 does not give a method for determining the elastic critical moment for lateraltorsional buckling Mcr !!!!!!!! May use ‘LTBeam’ software (can be downloaded from CTICM website) Or may use method presented by L. Gardner †¦Ã¢â‚¬ ¦. Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 29 Mcr under uniform moment For typical end conditions, and under uniform moment the elastic critical lateral torsional buckling moment Mcr is: Mcr ,0 G IT Iw Iz Lcr ? EIz ? 2 Lcr 2 ? Iw Lcr GIT ? ? ? 2 ? ? EIz ? ? Iz 2 0. 5 is the shear modulus is the torsion constant is the warping constant is the inor axis second moment of area is the buckling length of the beam Unrestrained Beam 30 Dr. A Aziz Saim 2010 EC3 Mcr under non-uniform moment Numerical solutions have been calculated for a number of other loading conditions. For uniform doubly-symmetric cross-sections, loaded through the shear centre at the level of the centroidal axis, and with the standard conditions of restraint described, Mcr may be calculated by: ? EIz Mcr ? C1 2 Lcr 2 Dr. A Aziz Saim 2010 EC3 Unrestrained Beam ? Iw Lcr GIT ? ? ? 2 ? ? EIz ? ? Iz 2 0. 5 31 C1 factor – end momentsFor end moment loading C1 may be approximated by the equation below, though other approximations also exist. C1= 1. 88 – 1. 40y + 0. 52y2 but C1 ? 2. 70 where y is the ratio of the end moments (defined in the following table). Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 32 C1 factor – transverse loading Loading and support conditions Bending moment diagram Value of C1 1. 132 1. 285 1. 365 1. 565 1. 046 Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 33 Design procedure for LTB Design procedure for LTB: 1. Determine BMD and SFD from design loads 2. Select section and determine geometry 3. Classify cross-section (Class 1, 2, 3 or 4) 4.Determine effective (buckling) length Lcr – depends on bounda ry conditions and load level 5. Calculate Mcr and Wyfy Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 34 Design procedure for LTB 6. Non-dimensional slenderness ? LT ? Wy fy Mcr 7. Determine imperfection factor ? LT 8. Calculate buckling reduction factor ? LT 9. Design buckling resistance 10. Check Mb,Rd ? ?LT Wy fy ? M1 MEd ? 1. 0 Mb,Rd for each unrestrained portion Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 35 LTB Example General arrangement Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 36 LTB Example Design loading is as follows: 425. 1 kN A B C 319. 6 kN D 2. 5 m 3. 2 m 5. 1 mLoading Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 37 LTB Example 267. 1 kN A B D 52. 5 kN SF C 477. 6 kN Shear force diagram B A C D BM 1194 kNm 1362 kNm Bending moment diagram Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 38 LTB Example For the purposes of this example, lateral torsional buckling curves for the general case will be utilised. Lateral torsional buckling checks to be carried out on segments BC and CD. By inspection, segment AB is not critical. Try 762? 267? 173 UB in grade S 275 steel. Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 39 LTB Example b z tw h d y y r z tf h = 762. 2 mm b = 266. 7 mm tw = 14. 3 mm tf = 21. 6 mm r = 16. mm A = 22000 mm2 Wy,pl = 6198? 103 mm3 Iz = 68. 50? 106 mm4 It = 2670? 103 mm4 Iw = 9390? 109 mm6 Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 40 LTB Example For a nominal material thickness (tf = 21. 6 mm and tw = 14. 3 mm) of between 16 mm and 40 mm the nominal values of yield strength fy for grade S 275 steel (to EN 10025-2) is 265 N/mm2. From clause 3. 2. 6: N/mm2. E = 210000 N/mm2 and G ? 81000 Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 41 LTB Example Cross-section classification (clause 5. 5. 2): e ? 235 / fy ? 235 / 265 ? 0. 94 Outstand flanges (Table 5. 2, sheet 2) cf = (b – tw – 2r) / 2 = 109. 7 mm cf / tf = 109. 7 / 21. 6 = 5. 8 Limit for Class 1 flange = 9e = 8. 48 > 5. 08 ? Flange is Class 1 Dr. A Aziz Saim 2010 EC3 Unrestrained Bea m 42 LTB Example Web – internal part in bending (Table 5. 2, sheet 1) cw = h – 2tf – 2r = 686. 0 mm cw / tw= 686. 0 / 14. 3 = 48. 0 Limit for Class 1 web = 72 e = 67. 8 > 48. 0 ? Web is Class 1 Overall cross-section classification is therefore Class 1. Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 43 LTB Example Bending resistance of cross-section (clause 6. 2. 5): Mc ,y,Rd ? Wpl,y fy ? M0 for Class 1 and 2 sec tions 6198 ? 103 ? 265 ? ? 1642 ? 106 Nmm 1. 0 ? 1642 kNm ? 1362 kNm ? Cross-section resistance in bending is OK.Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 44 LTB Example Lateral torsional buckling check (clause 6. 3. 2. 2) – Segment BC: MEd ? 1362 kNm Mb ,Rd ? ? LT Wy fy ? M1 where Wy = Wpl,y for Class 1 and 2 sections Determine Mcr for segment BC (Lcr = 3200 mm) Dr. A Aziz Saim 2010 EC3 ? EIz Mcr ? C1 2 Lcr 2 ? Iw Lcr GIT ? ? ? 2 ? ? EIz ? ? Iz Unrestrained Beam 2 0. 5 45 LTB Example For end moment loading C1 may be approximated from: C1 = 1. 88 â⠂¬â€œ 1. 40y + 0. 52y2 but C1 ? 2. 70 1194 y is the ratio of the end moments ? ? 0. 88 1362 ? C1 ? 1. 05 ? 2 ? 210000 ? 68. 5 ? 106 Mcr ? 1. 05 ? 32002 ? 9390 ? 109 32002 ? 81000 ? 2670 ? 103 ? ? ? 68. 5 ? 106 ? 2 ? 210000 ? 68. 5 ? 106 ? ? 0. 5 = 5699Ãâ€"106 Nmm = 5699 kNm Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 46 LTB Example Non-dimensional lateral torsional slenderness for segment BC: ? LT ? Wy fy Mcr 6198 ? 103 ? 265 ? ? 0. 54 6 5699 ? 10 Select buckling curve and imperfection factor ? LT: From Table 6. 4: h/b = 762. 2/266. 7 = 2. 85 For a rolled I-section with h/b > 2, use buckling curve b Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 47 LTB Example From Table 6. 3 of EN 1993-1-1: For buckling curve b, ? LT = 0. 34 Calculate reduction factor for lateral torsional buckling, ? LT – Segment BC: ?LT ? 1 ? LT ? ? 2 LT LT but ? LT ? 1. 0 where ? LT ? 0. 5 [ 1 ? ?LT (? LT ? 0. 2) ? ?2 ] LT Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 48 LTB Example ?LT = 0. 5[1+0. 34(0. 54-0. 2) + 0. 542] = 0. 70 ? ? LT ? 1 0. 70 ? 0. 70 ? 0. 54 2 2 ? 0. 87 Lateral torsional buckling resistance Mb,Rd – Segment BC : Mb,Rd ? ? LT Wy fy ? M1 265 ? 0. 87 ? 6198 ? 10 ? 1 . 0 3 ? 1425 ? 106 Nmm ? 1425 kNm Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 49 LTB Example MEd 1362 ? ? 0. 96 ? 1. 0 ? Segment BC is OK Mb,Rd 1425 Lateral torsional buckling check (clause 6. 3. 2. 2) – Segment CD: MEd ? 1362 kNm Mb ,Rd ? ? LT Wy fy ? M1 where Wy = Wpl,y for Class 1 and 2 sectionsDetermine Mcr for segment CD (Lcr = 5100 mm) Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 50 LTB Example ? EIz Mcr ? C1 2 Lcr 2 ? Iw Lcr GIT ? ? ? 2 ? Iz ? EIz ? ? 2 0. 5 Determine y from Table: 0 y is the ratio of the end moments ? ?0 1362 ? C1 ? 1. 88 ? 2 ? 210000 ? 68. 5 ? 106 Mcr ? 1. 88 51002 ? 9390 ? 109 51002 ? 81000 ? 2670 ? 103 ? ? ? ? 68. 5 ? 106 ? 2 ? 210000 ? 68. 5 ? 106 ? ? 0. 5 = 4311? 106 Nmm = 4311 kNm Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 51 LTB Example Non-dimensional lateral torsio nal slenderness for segment CD: ? LT ? Wy fy Mcr 6198 ? 103 ? 265 ? ? 0. 62 6 4311? 10 The buckling curve and imperfection factor ?LT are as for segment BC. Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 52 LTB Example Calculate reduction factor for lateral torsional buckling, ? LT – Segment CD: ?LT ? 1 ? LT ? ? 2 LT 2 LT but ? LT ? 1. 0 where ? LT ? 0. 5 [ 1 ? ?LT (? LT ? 0. 2) ? ?2 ] LT = 0. 5[1+0. 34(0. 62-0. 2) + 0. 622] = 0. 76 ? ? LT Dr. A Aziz Saim 2010 EC3 ? 1 0. 76 ? 0. 76 ? 0. 62 2 Unrestrained Beam 2 ? 0. 83 53 LTB Example Lateral torsional buckling resistance Mb,Rd – Segment CD : Mb,Rd ? ?LT Wy fy ? M1 265 ? 0. 83 ? 6198 ? 10 ? 1. 0 3 ? 1360 ? 106 Nmm ? 1360 kNm MEd 1362 ? ? 1. 00 Mb,Rd 1360 Segment CD is critical and marginally fails LTB check.Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 54 Blank Page Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 55 Simplified assessment of ? LT For hot-rolled doubly symmetric I and H sections without destabilising loads,? may be conservatively simplified to: LT ? LT ? 1 0. 9 ? z ? C1 ?z 1 0. 9 ? 1 C1 E ? z ? L / iz ; ? 1 ? ? fy As a further simplification, C1 may also be conservatively taken = 1. 0. Simplified assessment of ? LT Substituting in numerical values for simplified expressions result. ? 1 , the following S235 ? LT ? 1 L / iz C1 104 S275 ? LT ? 1 L / iz C1 96 S355 ? LT ? 1 L / iz C1 85 C1 may be conservatively taken = 1. , though the level of conservatism increases the more the actual bending moment diagram differs from uniform moment. Simplified method (Cl. 6. 3. 2. 4) Simplified method for beams with restraints in buildings (Clause 6. 3. 2. 4) This method treats the compression flange of the beam and part of the web as a strut: b b Compression h Tension Compression flange + 1/3 of the compressed area of web Strut Dr. A Aziz Saim 2010 EC3 Beam Unrestrained Beam 58 General method (Cl. 6. 3. 4) General method for lateral and lateral torsional buckling of structural components †¢ May be applied to single members, plane frames etc. Requires determination of plastic and elastic (buckling) resistance of structure, which subsequently defines global slenderness †¢ Generally requires FE Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 59 Blank Page Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 60 Important Notes: (End Connections) When full torsional restraint exist: -both the compression and tension flanges are fully restrained against rotation on plan -both flanges are partially restrained against rotation on plan – both flanges are free to rotate on plan Unrestrained Beam 61 Dr. A Aziz Saim 2010 EC3 Connection DetailDr. A Aziz Saim 2010 EC3 Unrestrained Beam 62 Important Notes: (End Connections) Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 63 Important Notes: (End Connections) When both flanges are free to rotate on plan and the compression flange is unrestrained: i. torsional restraint is provided solely by connection of the tension flange to the supports, ii. torsional re straint is provided solely by dead bearing of the tension flange on support. Unrestrained Beam 64 Dr. A Aziz Saim 2010 EC3 Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 65 Dr. A Aziz Saim 2010 EC3 Unrestrained Beam 66

Understanding Dinosaur Combat

In Hollywood movies, dinosaur fights have clear winners and losers, carefully demarcated arenas (say, an open patch of scrubland or the cafeteria in Jurassic Park), and usually a bunch of scared-out-of-their-wits human spectators. In real life, though, dinosaur fights were more like confused, chaotic bar brawls than Ultimate Fighting matches, and rather than persisting for multiple rounds, they were usually over in the blink of a Jurassic eye. (See a list of the Deadliest Dinosaurs, as well as Prehistoric Battles featuring your favorite dinosaurs, reptiles, and mammals.) Its important at the outset to distinguish between the two main types of dinosaur combat. Predator/prey encounters (for example, between a hungry Tyrannosaurus Rex and alone, juvenile Triceratops) were quick and brutal, with no rules except kill or be killed. But intra-species clashes (say, two male Pachycephalosaurus head-butting each other for the right to mate with available females) had a more ritualistic aspect, and rarely resulted in a combatants death (though one presumes serious injuries were common). Of course, in order to fight successfully, you need to be equipped with suitable weapons. Dinosaurs didnt have access to firearms (or even blunt instruments), but they were endowed with naturally evolved adaptations that helped them either to hunt down their lunch, avoid being lunch or propagate the species in order to restock the global lunch menu. Offensive weapons (like sharp teeth and long claws) were almost exclusively the province of meat-eating dinosaurs, which preyed on one another or on gentler herbivores, while defensive weapons (like armor plating and tail clubs) were evolved by plant-eaters in order to fend off attacks by predators. A third type of weapon consisted of sexually selected adaptations (such as sharp horns and thickened skulls), wielded by the males of some dinosaur species in order to dominate the herd or compete for the attention of females. Offensive Dinosaur Weapons Teeth. Meat-eating dinosaurs like T. Rex and Allosaurus didnt evolve big, sharp teeth merely to eat their prey; like modern cheetahs and great white sharks, they used these choppers to deliver quick, powerful, and (if they were delivered in the right place at the right time) fatal bites. Well never know for sure, but reasoning by analogy with modern carnivores, it seems likely that these theropods aimed for their victims necks and bellies, where a strong bite would cause the most damage. Claws. Some carnivorous dinosaurs (like Baryonyx) were equipped with large, powerful claws on their front hands, which they used to slash at prey, while others (like Deinonychus and its fellow raptors) had single, oversized, curved claws on their hind feet. Its unlikely that a dinosaur could have killed prey with its claws alone; these weapons were probably also used to grapple with opponents and keep them in a death grip. (Bear in mind, however, that huge claws dont necessarily connote a carnivorous diet; the big-clawed Deinocheirus, for example, was a confirmed vegetarian.) Eyesight and smell. The most advanced predators of the Mesozoic Era (like the human-sized Troodon) were equipped with large eyes and relatively advanced binocular vision, which made it easier for them to zero in on prey, especially when hunting by night. Some carnivores also possessed an advanced sense of smell, which enabled them to scent prey from far off (though its also possible that this adaptation was used to home in on already-dead, rotting carcasses). Momentum. Tyrannosaurs were built like battering rams, with enormous heads, thick bodies, and powerful hind legs. Short of delivering a fatal bite, an attacking Daspletosaurus could knock its victim silly, provided it had the element of surprise on its side and a sufficient head of steam. Once the unlucky Stegosaurus was lying on its side, stunned and confused, the hungry theropod could move in for the quick kill. Speed. Speed was an adaptation shared equally by predators and prey, a good example of an evolutionary arms race. Since they were smaller and more lightly built than tyrannosaurs, raptors and dino-birds were especially quick, which created an evolutionary incentive for the plant-eating ornithopods they hunted to run faster as well. As a rule, carnivorous dinosaurs were capable of short bursts of high speed, while herbivorous dinosaurs could sustain a slightly less brisk pace for a longer period of time. Bad breath. This may sound like a joke, but paleontologists believe that the teeth of some tyrannosaurs were shaped so as to purposely accumulate shreds of dead tissue. As these shreds rotted, they bred dangerous bacteria, meaning any non-fatal bites inflicted on other dinosaurs would result in infected, gangrenous wounds. The unlucky plant-eater would drop dead in a few days, at which point the responsible Carnotaurus (or any other predator in the immediate vicinity) chowed down on its carcass. Defensive Dinosaur Weapons Tails. The long, flexible tails of sauropods and titanosaurs had more than one function: they helped to counterbalance these dinosaurs equally long necks, and their ample surface area may have helped dissipate excess heat. However, its also believed that some of these behemoths could lash their tails like whips, delivering stunning blows to approaching predators. The use of tails for defensive purposes reached its apex with the ankylosaurs, or armored dinosaurs, which evolved heavy, macelike growths at the ends of their tails that could crush the skulls of unwary raptors. Armor. Until the knights of medieval Europe learned to forge metallic armor, no creatures on earth were more impervious to attack than Ankylosaurus and Euoplocephalus (the latter even had armored eyelids). When attacked, these ankylosaurs would plop down onto the ground, and the only way they could be killed was if a predator managed to flip them onto their backs and dig into their soft underbellies. By the time the dinosaurs went extinct, even titanosaurs had evolved a light armored coating, which may have helped fend off pack attacks by packs of smaller raptors. Sheer bulk. One of the reasons sauropods and hadrosaurs attained such enormous sizes is that full-grown adults would have been virtually immune to predation: not even a pack of adult Alioramus could hope to take down a 20-ton Shantungosaurus. The downside to this, of course, was that predators shifted their attention to easier-to-pick-off babies and juveniles, meaning that out of a clutch of 20 or 30 eggs laid by a female Diplodocus, only one or two might manage to reach adulthood. Camouflage. The one feature of dinosaurs that rarely (if ever) fossilizes is their skin color--so well never know if Protoceratops sported zebra-like stripes, or if Maiasauras mottled skin made it difficult to see in dense underbrush. However, reasoning by analogy with modern prey animals, it would be very surprising indeed if hadrosaurs and ceratopsians didnt sport some kind of camouflage to cloak them from the attention of predators Speed. As mentioned above, evolution is an equal-opportunity employer: as the predatory dinosaurs of the Mesozoic Era become faster, so do their prey, and vice-versa. While a 50-ton sauropod couldnt have run very fast, the average hadrosaur could rear up onto its hind legs and beat the bipedal retreat in response to danger, and some smaller plant-eating dinosaurs may have been capable of sprinting at 30 or 40 (or possibly 50) miles per hour while being chased. Hearing. As a general rule, predators are endowed with superior sight and smell, while prey animals possess acute hearing (so they can run away if they hear a threatening rustle in the distance). Based on an analysis of their crested skulls, it seems likely that some duck-billed dinosaurs (like Parasaurolophus and Charonosaurus) could bellow to each other over long distances, so an individual hearing the footsteps of an approaching tyrannosaur would be able to warn the herd. Intra-Species Dinosaur Weapons Horns. The fearsome-looking horns of Triceratops may only have been secondarily intended to warn away a hungry T. Rex. The position and orientation of ceratopsian horns lead paleontologists to conclude that their main purpose was in dueling with other males for dominance in the herd or breeding rights. Of course, unlucky males might be wounded, or even killed, in this process--researchers have unearthed numerous dinosaur bones bearing the marks of intra-species combat. Frills. The giant head ornaments of ceratopsian dinosaurs served two purposes. First, oversized frills made these plant-eaters look bigger in the eyes of hungry carnivores, which might opt to concentrate on smaller fare instead. And second, if these frills were brightly colored, they could have been used to signal the desire to fight during mating season. (Frills may also have had yet another purpose, as their large surface areas helped to dissipate and absorb heat.) Crests. Not quite a weapon in the classic sense, crests were protrusions of bone most often found on duck-billed dinosaurs. These backward-pointing growths would have been useless in a fight, but they may well have been employed to attract females (theres evidence that the crests of some Parasaurolophus males were larger than those of the females). As mentioned above, its also likely that some duck-billed dinosaurs funneled air through these crests as a way of signaling to others of their kind. Skulls. This peculiar weapon was unique to the family of dinosaurs known as pachycephalosaurs (thick-headed lizards). Pachycephalosaurs like Stegoceras and Sphaerotholus sported up to a foot of bone on the tops of their skulls, which they presumably used to head-butt one another for dominance in the herd and the right to mate. Theres some speculation that pachycephalosaurs may also have butted the flanks of approaching predators with their thickened domes.