Acceleration Estimation Routines and Selection
Example 1: Acceleration estimation of a rotating body:
There is a spindle with a diameter of 10 cm. To measure the vibration of the corresponding housing of the bearing, the rotational speed range is 0 to 3000 revolutions per minute (rpm). The vibration of the housing is about 20 microns.
Frequency f=3000/60=50hz
Multiply the square of (2πf) by 20μm/9.8 = 0.2g
Acceleration = (2*3.14*50)2*20*10-6/9.8 = 0.2g
The diameter of the measured shaft shell is 10cm, the light weight of the selected DTS0155, full-scale 7g
Multiply the square of (2πf) by 20μm/9.8 = 0.2g
Acceleration = (2*3.14*50)2*20*10-6/9.8 = 0.2g
The diameter of the measured shaft shell is 10cm, the light weight of the selected DTS0155, full-scale 7g
Example 2: Estimation of Free Fall Acceleration:
10KG object, when it falls freely from a height of 1.5 meters above the ground, the speed at the moment of reaching the ground can be obtained by the formula V^2=2gH V=√(2gH)=√(2*10*1.5)=5.47 meters/second After reaching the ground, it will collide with the ground and have a certain collision time. This time is related to the hardness of the object and the condition of the ground (hard ground or loose soil). Take the general situation, the collision time is taken as t = 0.1 seconds, then by (F-mg) = ma and V = a * t F = mg + (mV / t) = 10 * 10 + (10 * 5.47 / 0.1) = 647 cattle, which is the impact between the object and the ground, has 647 cows. (equivalent to a force of 65 kilograms) 1. Take 0.1 seconds = 100 milliseconds for the soil surface t, and the acceleration a = V/t = 5.47/0.1 = 54.7 (m/s2) because 1g = 9.8 (m/s2) = 10 ( m/s2), so a=5.47g2. Take 0.01 second for cement floor t=10 milliseconds, acceleration a=V/t=5.47/0.01=547(m/s2) because 1g=9.8(m/s2)=10 (m/s2), so a = 54.7g In order to suit the test of two kinds of situations, choose DTS0103T full scale 100g Example 3: Estimation of Free-fall Acceleration:
20KG object, when it is free to fall from a height of 100 meters above the ground, the speed when it reaches the ground can be obtained by the formula V^2=2gH V=√(2gH)=√(2*10*100)=44.72 meters/second After reaching the ground, it will collide with the ground and have a certain collision time. This time is related to the hardness of the object and the condition of the ground (hard ground or loose soil). Take the general situation, the collision time is taken as t = 0.1 seconds, then by (F-mg) = ma and V = a * t F = mg + (mV / t) = 20 * 10 + (20 * 44.72 / 0.1) = 9144 cattle that is the impact between the object and the ground has 9144 cattle. (equivalent to 914 kilograms of force) 1. Take 0.1 seconds for mud surface t = 100 milliseconds, acceleration a = V/t = 44.72/0.1 = 447.2 (m/s2) because 1g = 9.8 (m/s2) = 10 ( m/s2), so a = 44.72g2. Take 0.01 second for cement floor t = 10 milliseconds, acceleration a = V/t = 44.72/0.01 = 4472 (m/s2) because 1g = 9.8 (m/s2) = 10 (m/s2), so a=447.2g Select DTS0102T full scale 500g to suit the test of both conditions