Athlete or Machine?www.raeng.org.uk/athleteormachine Presented by Dominic Nolan. Education Programme Manager. The Royal Academy of Engineering
CHALLENGE Make a model of a bob skeleton sled See how far you can launch a Barbie! Present an answer to the question: Athlete or Machine? Which is more important in the sport of bob skeleton?
Bob Skeleton • 1500m track • 150 m vertical drop • 143 km/h (40 m/s, 89 mph) • Athletes times differ by tenths of seconds • Rules for sled’s dimensions, mass and materials • 33 – 43 kg sled • Amy Williams - Olympic gold 2010 • www.youtube.com
Make a 1:5 bob skeleton sled • Make the runners by bending the metal rod • Attach runners to pod with cable ties • Make sled’s launch tube using acetate sheet, tape and a plastic nose cone (check that it fits onto the pump’s launch tube) • Fix the launch tube to the pod with double-sided sticky pads
Factors • Weight • The athlete’s shape • The athlete’s position • Aerodynamic lift • Steering • Clothing and equipment • Starting • Corners • Ergonomics (how the body fits a product) • Track incline (the slope down the length of the track) • Friction on the ice • Aerodynamic drag (air resistance) • Tuning the characteristics of the skeleton • Material choice • Sled runners
Gravity (g) = 9.81 m/s2 Max speed if all PE transferred into KE (diagram not to scale) Mass (m) of athlete and sled = 97kg 1450m Vertical drop of track (h) = 152m Amy Williams max speed Energy transfer Potential Energy (PE) = m x g x h Change in PE for our athlete and sled = 144 639 Joules (J) Kinetic Energy (KE) = ½ x m x v2 0.5 x 97 kg x (40.23 x 40.23) = 78495 J Why isn’t the all of the athlete’s and sled’s potential energy transferred into kinetic energy?
Which two forces resist the forward movement of the athlete and sled down the track? friction aerodynamic drag (air resistance)
Friction force Friction is a force that resists the movement of two surfaces against each other. Which combinations provide a lot or a little friction? rubber / concrete felt / wood rubber / rubber steel / ice steel / wood A little friction A lot of friction rubber / rubber (1.16) steel / ice (0.03) rubber / concrete (1.02) steel / wood (0.2 - 0.6) felt / wood (0.22)
Calculating friction force • Ff = x m x g • Ff = ………………………… • = Mu, the coefficient of friction (steel on ice = 0.03). m = Mass (kg). g = The acceleration due to the gravity, which is 9.81 m/s2. What is the friction force acting on the runners of a bob skeleton sled and athlete with the combined mass of 97 kg (athlete = 68 kg, sled = 29 kg)?
CD = 0.42 CD = 1.05 CD = 0.47 CD = 0.5 Aerodynamic drag force The resistance provided by the air passing over a shape is a force called aerodynamic drag. Which shapes have a higher or lower coefficient of drag? Lower CD Higher CD
Calculating drag force • FDRAG = ½ x x CDx Afx V2 • FDRAG = …………………………. • = 1.2 kg/m3 (density of air) CD = 0.45 (drag coefficient of athlete and sled) • Af = 0.139 m2 (frontal area of athlete and sled) • V = 40 m/s (velocity) Calculate the drag force acting on the athlete and sled as they travel down the track at 40 m/s?
What is the total force resisting the forward movement of the athlete and her sled down the track? FTOTAL = …………………………………… Between which velocities is friction force dominant? ……………………………………………….. Between which velocities is drag force dominant? ……………………………………………….. You can compare the two forces on the graph here. 88.56N 80 70 60 50 Force in Newtons (N) 40 30 20 10 0 25 30 35 40 45 5 10 15 20 Speed in metres/second (m/s)
Prove that it is better to be heavy and narrow when competing in The sport of bob skeleton. ATHLETE 1 Total mass: 97 kg Af: 0.139 m2 ATHLETE 2 Total mass: 100 kg Af: 0.129 m2
Athlete or Machine? • Which is more important in the sport of bob skeleton? • Discuss this question with your partner/team • Present your answer to the rest of the group