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Torque and RPM of Gears

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Presentation on theme: "Torque and RPM of Gears "β€” Presentation transcript:

1 Torque and RPM of Gears 𝑅𝑃𝑀 2 𝑅𝑃𝑀 1 = 𝑁 1 𝑁 2 = 𝑑 1 𝑑 2 = 𝑇 1 𝑇 2

2 DISCLAIMER & USAGE The content of this presentation is for informational purposes only and is intended for students attending Louisiana Tech University only. The authors of this information do not make any claims as to the validity or accuracy of the information or methods presented. Any procedures demonstrated here are potentially dangerous and could result in damage and injury. Louisiana Tech University, its officers, employees, agents and volunteers, are not liable or responsible for any injuries, illness, damage or losses which may result from your using the materials or ideas, or from your performing the experiments or procedures depicted in this presentation. The Living with the Lab logos should remain attached to each slide, and the work should be attributed to Louisiana Tech University. If you do not agree, then please do not view this content. boosting application-focused learning through student ownership of learning platforms

3 Torque Torque is the moment applied about the longitudinal axis of a body (typically called a shaft). Free Body Diagram of the Gear: x y What is Rx for loading shown? P d r T Rx = 0 Rx F Ry What if another Load, P, was applied? To keep the system in equilibrium, a resisting torque is applied to the shaft. Rx = -P Shaft torque constrains rotation Bearing constrains translation

4 Gears Toothed members that transmit rotary motion from one shaft to another. The servos used with the Arduino contain a series of spur gears which transfer motion between parallel shafts. Pitch diameter is the effective diameter of the gear. If gear 1 has 30 teeth and gear 2 has 10 teeth, then the upper gear will turn ____ times when the lower gear turns ____ times. If gear 1 turns one revolution per minute and gear 2 turns three revolutions per minute, can you develop a relationship between revolutions per minute and gear teeth? d1 GEAR 1 GEAR 2 1 d2 3 3 π‘Ÿπ‘’π‘£π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›π‘  π‘π‘’π‘Ÿ π‘šπ‘–π‘›π‘’π‘‘π‘’ 1 π‘Ÿπ‘’π‘£π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘› π‘π‘’π‘Ÿ π‘šπ‘–π‘›π‘’π‘‘π‘’ = 30 π‘‘π‘’π‘’π‘‘β„Ž 10 π‘‘π‘’π‘’π‘‘β„Ž πΊπ‘’π‘Žπ‘Ÿ 2 π‘Ÿπ‘’π‘£π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›π‘  π‘π‘’π‘Ÿ π‘šπ‘–π‘›π‘’π‘‘π‘’ πΊπ‘’π‘Žπ‘Ÿ 1 π‘Ÿπ‘’π‘£π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›π‘  π‘π‘’π‘Ÿ π‘šπ‘–π‘›π‘’π‘‘π‘’ = π‘‡π‘’π‘’π‘‘β„Ž π‘œπ‘“ πΊπ‘’π‘Žπ‘Ÿ 1 π‘‡π‘’π‘’π‘‘β„Ž π‘œπ‘“ πΊπ‘’π‘Žπ‘Ÿ 2 𝑅𝑃𝑀 2 𝑅𝑃𝑀 1 = # π‘œπ‘“ π‘‘π‘’π‘’π‘‘β„Ž 𝑖𝑛 π‘”π‘’π‘Žπ‘Ÿ 1 # π‘œπ‘“ π‘‘π‘’π‘’π‘‘β„Ž 𝑖𝑛 π‘”π‘’π‘Žπ‘Ÿ 2 = 𝑑 1 𝑑 2

5 Gear Analysis T2 T1 F F Rx2 Ry2 x y d1
Every force has an equal and opposite force (Newton’s 2nd Law) F d2 T2 Rx2 Ry2 Analysis of Gear 1: Analysis of Gear 2: 𝐹 π‘₯1 =𝐹+ 𝑅 π‘₯1 =0 𝐹 π‘₯2 =βˆ’πΉ+ 𝑅 π‘₯2 =0 𝑅 π‘₯1 =βˆ’πΉ 𝑅 π‘₯2 =𝐹 𝐹 𝑦1 = 𝑅 𝑦1 =0 𝐹 𝑦1 = 𝑅 𝑦2 =0 𝑅 𝑦1 =0 𝑅 𝑦2 =0 𝑀 π‘π‘’π‘›π‘‘π‘’π‘Ÿ1 = βˆ’π‘‡ 1 +πΉβˆ™ 𝑑 1 2 =0 𝑀 π‘π‘’π‘›π‘‘π‘’π‘Ÿ2 = βˆ’π‘‡ 2 +πΉβˆ™ 𝑑 2 2 =0 + + 𝑇 1 =πΉβˆ™ 𝑑 1 2 𝑇 2 =πΉβˆ™ 𝑑 2 2

6 Gear Analysis T2 T1 F Rx2 Ry2 x y d1 d2 Gear 1: Gear 2: 𝑅 π‘₯1 =βˆ’πΉ
𝑅 π‘₯2 =𝐹 𝑅 𝑦1 =0 𝑅 𝑦2 =0 𝑇 1 =πΉβˆ™ 𝑑 1 2 𝑇 2 =πΉβˆ™ 𝑑 2 2 This is the same F. Solve for F with both equation and set them equal to each other. 2βˆ™ 𝑇 1 𝑑 1 =𝐹 2βˆ™ 𝑇 2 𝑑 2 =𝐹 𝑅𝑃𝑀 2 𝑅𝑃𝑀 1 = 𝑁 1 𝑁 2 = 𝑑 1 𝑑 2 = 𝑇 1 𝑇 2 2βˆ™ 𝑇 1 𝑑 1 = 2βˆ™ 𝑇 2 𝑑 2 # of teeth is often represented with N. 𝑁 1 𝑁 2 is the same as # π‘œπ‘“ π‘‘π‘’π‘’π‘‘β„Ž 𝑖𝑛 π‘”π‘’π‘Žπ‘Ÿ 1 # π‘œπ‘“ π‘‘π‘’π‘’π‘‘β„Ž 𝑖𝑛 π‘”π‘’π‘Žπ‘Ÿ 2 𝑇 1 𝑇 2 = 𝑑 1 𝑑 2

7 Class Problem: Consider the gear train below
Class Problem: Consider the gear train below. Assume the gears are supported by bearings and that all gears have compatible teeth. Find the RPM of gears 2, 3 and 4 (gears 2 and 3 are joined like many of your servo gears). Find torque transmitted through gears 2, 3, and 4. Gear Teeth RPM Torque (in-lbs) 1 10 3600 2 30 3 8 4 40 𝑅𝑃𝑀 2 𝑅𝑃𝑀 1 = 𝑁 1 𝑁 2 = 𝑑 1 𝑑 2 = 𝑇 1 𝑇 2 1200 30 1200 30 240 150 a. 𝑅𝑃𝑀 2 𝑅𝑃𝑀 1 = 𝑁 1 𝑁 2 b. 𝑇 2 𝑇 1 = 𝑁 2 𝑁 1 𝑅𝑃𝑀 2 = 𝑁 1 𝑁 2 𝑅𝑃𝑀 1 𝑇 2 = 𝑁 2 𝑁 1 𝑇 1 𝑅𝑃𝑀 2 = =1200 =𝑅𝑃𝑀 3 𝑇 2 = π‘–π‘›βˆ™π‘™π‘π‘  =30 π‘–π‘›βˆ™π‘™π‘π‘  =𝑇 3 Gear 2 and 3 are connected and so they must turn at the same speed 𝑇 4 = 𝑁 4 𝑁 3 𝑇 3 = π‘–π‘›βˆ™π‘™π‘π‘  𝑅𝑃𝑀 4 = 𝑁 3 𝑁 4 𝑅𝑃𝑀 3 = =150π‘–π‘›βˆ™π‘™π‘π‘  =240

8 Gear Summary Points to Remember:
1 Points to Remember: Bigger gear turns slower than smaller gear. Bigger gear has more torque than smaller gear. Both gears on a compound gear turn at the same rate and transmit the same torque. 𝑅𝑃𝑀 2 𝑅𝑃𝑀 1 = 𝑁 1 𝑁 2 = 𝑑 1 𝑑 2 = 𝑇 1 𝑇 2


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