CN112598976B - Experiment teaching device and method for measuring three-dimensional space coordinate of center of gravity of object - Google Patents

Experiment teaching device and method for measuring three-dimensional space coordinate of center of gravity of object Download PDF

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CN112598976B
CN112598976B CN202011499255.9A CN202011499255A CN112598976B CN 112598976 B CN112598976 B CN 112598976B CN 202011499255 A CN202011499255 A CN 202011499255A CN 112598976 B CN112598976 B CN 112598976B
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李达
李书卉
孙凯
陈晓东
刘增利
邢怀念
金立强
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Dalian University of Technology
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Abstract

An experimental teaching device and a measuring method for measuring a three-dimensional space coordinate of the gravity center of an object belong to the technical field of gravity center position measurement. The principle is simple and convenient to operate, the gravity center measuring device is suitable for measuring the gravity centers of irregular objects with various sizes and weights, although the gravity centers are measured by buoyancy in the device and a ship inclination test, the difference is that the initial state of the ship inclination test needs to be in a positive floating state, and the teaching device does not need to be in the positive floating state. The teaching device can supplement the teaching content of the current theoretical mechanics experiment and is suitable for teaching of the theoretical mechanics experiment.

Description

Experiment teaching device and method for measuring three-dimensional space coordinate of center of gravity of object
Technical Field
The invention relates to an experimental teaching device and a determination method for determining three-dimensional space coordinates of the gravity center of an object, belongs to the technical field of gravity center position measurement, and particularly relates to a gravity center determination test for an irregular object.
Background
The gravity center position of the object has very important influence on the hoisting, transportation and use of the object. The centre of gravity of the vessel has a very important influence on the stability of the vessel. For example, the deviation of the position of the center of gravity of the ship body can cause the ship to have larger trim, and influence the floating state, rapidness and wave resistance of the ship; meanwhile, the ship can transversely swing due to the overhigh gravity center, and the use performance of the ship is influenced.
The center of gravity of the mobile crane has very important influence on the weight of an object lifted by the crane, the counterweight of the crane and the working environment of the crane. For example, a horizontal deviation of the position of the center of gravity of the crane can cause a longitudinal tipping risk during the hoisting process of the crane, and an excessively high center of gravity of the crane can cause a transverse tipping risk during the transverse wind load and the horizontal movement of the hoisted object of the crane.
The gravity center position of machine tools and other mechanical components and large objects such as containers has very important significance for hoisting in the process of object transportation.
At present, the methods for measuring the gravity center of an object mainly comprise a weighing method and a suspension method. The embodiment of the weighing method is as follows: weighing an object by taking four supporting points, measuring the gravity center position of the object in the horizontal direction according to a moment balance principle, then inclining or turning the object by 90 degrees, repeating the operation, and measuring the position of the three-dimensional coordinate of the gravity center space of the object according to the moment balance principle; the embodiment of the suspension method is as follows: a rope is used for hanging the object, the gravity center is on the extension line of the suspension line according to the force balance principle, the object is hung by replacing the hanging point position, and the intersection point of the two lines is the gravity center position. Therefore, the weighing method and the hanging method have obvious advantages for objects which have small mass and can be inclined and laid down at will, and have certain limitation for objects with larger volume and weight.
The invention provides a simple experimental teaching device and a simple experimental teaching method for measuring three-dimensional space coordinates of the gravity center of an object, which have simple principle and convenient operation, are suitable for measuring the gravity centers of irregular objects with various sizes and weights, and are different from ship inclination tests in that the initial state of the ship inclination tests is in a positive floating state and the teaching device is not in the positive floating state although the device and the ship inclination tests adopt buoyancy to measure the gravity centers. The teaching device can supplement the teaching content of the current theoretical mechanics experiment and is suitable for teaching of the theoretical mechanics experiment.
Disclosure of Invention
The invention aims to provide an experimental teaching device and a testing method for measuring the three-dimensional space coordinate of the gravity center of an object, which have simple structure and popular and easy understanding and supplement the current theoretical mechanics experimental teaching.
The technical scheme adopted by the application is as follows: an experimental teaching device for measuring the three-dimensional space coordinate of the gravity center of an object comprises a water tank, a test box and a lifting frame, wherein the lifting frame is provided with two upright posts on a base, the top ends of the upright posts are connected with a horizontal rod, the end parts of the horizontal rod are connected through a cross rod, and a threaded rod is arranged at the joint of the horizontal rod and the upright posts; a rotary hand wheel is arranged at one end of the threaded rod, a rotary pulley is arranged in the middle of the threaded rod, one end of a steel wire rope is arranged on the rotary pulley, and the other end of the steel wire rope is connected with a force sensor and a lifting hook through a guide sliding block arranged on the cross rod; the object to be measured acts on the lifting hook; a test box is arranged below the lifting hook, and a water tank is arranged outside the test box; the upper end of the water tank is provided with an upper flow guide port, and the lower end of the water tank is provided with a water outlet with a ball valve switch; the force sensor is connected with the force display through a data transmission line.
A method for measuring an experimental teaching device for measuring the three-dimensional space coordinate of the gravity center of an object comprises the following steps:
a. closing a ball valve switch in the water tank, adding liquid into the water tank, and placing the test box into the water tank; the test box is a regular rectangular box body, the length is a, the width is b, and the height is h, so the gravity center position of the test box is obtained according to theoretical calculation; at the moment, the test box should be floating on the water surface, and the longitudinal scales of the test box are all h1Then mass m of the test chamber1＝ρabh1；
b. The irregular object to be tested is put into the test box by the hook, and the mass M of the object to be tested is measured by the force display3Moving the object to be measured downwards to enable the test box to be approximately floated on the water surface, wherein the gravity center of the test box, the gravity center of the object to be measured and the floating center are on the same straight line, recording the relative placement position of the object to be measured, and the gravity center position of the object to be measured in the horizontal direction is superposed with the gravity center position of the box body;
sinking the test box and recording the longitudinal scale h of the test box2(ii) a Mass m of the object to be measured3Calculating to obtain the density rho of the liquid;
further determining the floating center position B, x of the whole body of the test box and the tested object21＝a/2y21＝b/2z21＝h 22; x in the center of gravity G of the measured object mass M3＝a/2y3B/2, vertical coordinate z of center of gravity3Unknown;
d. horizontally moving the object to be tested along the x direction for a moving distance of e, enabling the test box to be inclined in an equal volume, and enabling the longitudinal scales of the test box to be h3、h4、h5、h6(ii) a If h3＜h4＜h6＜h5Then h is6-h3＝h5-h4；
Volume of water to be drained
V＝abh3+(ab(h4-h3))/2+(lb(h5-h4))/2
The floating center coordinate is the gravity center coordinate of the volume of the drained water;
from the center of gravity, the formula G ═ M1G1+M2G2+M3G3)/(M1+M2+M3) Is calculated to obtain
Floating center position B1
Figure GDA0002871381060000031
Figure GDA0002871381060000032
Figure GDA0002871381060000033
e. Establishing a space coordinate system by taking the bottom edge of the test box as an x axis and the central line of the test box as a z axis; by translating the object to be measured only in the x-direction, the formula G ═ M is synthesized from the center of gravity2G2+M3G3)/(M2+M3) Therefore, the tested object is only translated along the x direction, the gravity center of the test box and the tested object is only changed along the x direction, and the y and z coordinates are kept unchanged;
f. because the mass of the test box for translating the tested object and the tested heavy object is not changed, the test box can be inclined in the same volume along the x direction, and two buoyancy action lines are intersected at a point M before and after the test box is inclined; establishing a rectangular coordinate system by taking the bottom edge of the test box as an x axis and the central line as a z axis, wherein the coordinate system rotates along with the rotation of the test box; according to the moment balance principle, the following steps are carried out:
W×e＝ρV×GM×sinθ
in the formula: w is the weight of the measured object, e is the translation distance of the measured object, V is the buoyancy, namely the weight of the drained water, and theta is the inclination angle of the test box and can be known by the scale of the test box immersed in the water; calculating to obtain a GM distance;
because the tested object only moves in the x direction, the gravity center position of the whole test box and the tested object only changes in the x direction; establishing a coordinate system by taking the bottom edge as an x axis and the central line symmetry axis as a z axis, and determining that GG exists1Perpendicular to BM and with a vertical foot of G, BB1Perpendicular to BM and with a foot B;
then: x is the number ofG1＝xB1-B1G1×sinθ
In the formula: x is the number ofG1Is the x coordinate of the center of gravity of the test box and the object to be tested, xB1Is the x coordinate of the floating center of the test box and the measured object, theta is the inclination angle of the test box, and can be known by the scale of the test box immersed in water; calculating to obtain B1G1Distance:
the distance to the BG available is further calculated: BG ═ B1G1×cosθ
Further calculating to obtain zG1The distance of (c): z is a radical ofG1＝zG＝zB+BG
zG1The vertical coordinate of the whole of the test box and the tested object is obtained;
according to the formula:
zG1＝(M2z2+M323)/(M2+M3)
in the formula: m2For testing the weight of the box, z2Is the vertical coordinate of the center of gravity of the test box, M3Is the weight of the object to be measured, z3The gravity center vertical coordinate of the measured object is obtained; the vertical coordinate z of the measured object can be determined3So as to calculate the measured object M3The barycentric coordinates of (a);
g. if the test box and the object to be tested are not floating on the liquid surface in the step b, the gravity center of the test box and the object to be tested is G1The floating core is B1(ii) a Assuming that the current non-positive floating state is a positive floating state and the weight distance e is deviated1To obtain G1(ii) a In a similar way, in G1Offset weight distance e on foundation2To obtain G2A state; deviation of weight distance e assuming positive floating state1+e2To obtain G2A state;
because the test box has equal volume deviation, the BG and B on the transverse section can be known from the above reasoning1G1、B2G2Must intersect at a point;
establishing a rectangular coordinate system by taking the bottom edge of the test box as an x axis and the central line as a z axis, wherein the coordinate system rotates along with the rotation of the test box;
according to the moment balance principle, the following steps are carried out:
W×e＝ρV×G1M×sinθ2
in the formula: w is the weight of the object to be measured, e is the translation distance of the object to be measured, V is the buoyancy, i.e. the weight of the drained water, theta2The inclination angle of the test box can be known by the scale of the test box immersed in water;
calculating to obtain G1M distance;
because the tested object only moves in the x direction, the gravity center position of the whole test box and the tested object only changes in the x direction; establishing a coordinate system by taking the bottom edge as an x axis and the central line symmetry axis as a z axis, and determining that GG exists1G2Perpendicular to BM and with a vertical foot of G, BB1B2Perpendicular to BM and with a foot B;
then:
GM＝G1M×cosθ1
xG2＝GM×sin(θ1+θ2)
xG2＝xB2-B2G2×sin(θ1+θ2)
in the formula: x is the number ofG2Is the x coordinate of the gravity center of the test box and the tested object,
xB2is the x coordinate of the floating center of the test box and the tested object,
θ1the initial inclination angle of the test box can be known by the scale of the test box immersed in water;
θ2the inclination angle generated by moving the tested object for the test box can be known by the scale of the test box immersed in water;
then B is known2G2
BG＝B2G2×cos(θ1+θ2)
Then BG can be obtained
zG2＝zG＝zB2+BG
Then further calculate the available zG2The distance of (d);
xG2the x-direction coordinate of the whole of the test box and the tested object is obtained;
zG2the z-direction coordinate of the whole of the test box and the measured object is obtained;
according to the formula:
xG2＝(M2x2+M3x3)/(M2+M3)
zG2＝(M2z2+M3z3)/(M2+M3)
in the formula: m2In order to test the weight of the box,
x2is the lateral coordinate of the center of gravity of the test box,
z2is the vertical coordinate of the gravity center of the test box,
M3is the weight of the object to be measured,
x3is the transverse coordinate of the gravity center of the measured object,
z3is the gravity center vertical coordinate of the measured object,
the transverse coordinate x of the measured object can be determined3And the vertical coordinate z3；
h. Taking the longitudinal section for analysis in the same way: establishing a rectangular coordinate system by taking the bottom edge of the test box as a y axis and the central line as a z axis, wherein the coordinate system rotates along with the rotation of the test box;
according to the moment balance principle, the following steps are carried out: w × h ═ ρ V × G1M×sinβ2
In the formula: w is the weight of the object to be measured,
h is the translation distance of the measured object in the y direction,
v is buoyancy, namely the weight of the drained water,
β2the inclination angle of the test box can be known by the scale of the test box immersed in water;
calculating to obtain G1M distance;
because the tested object only moves in the y direction, the gravity center position of the whole test box and the tested object only changes in the y direction; establishing a coordinate system by taking the bottom edge as the y axis and the central line symmetry axis as the z axis, and determining that GG exists1G2Perpendicular to BM and with a vertical foot of G, BB1B2Perpendicular to BM and with a foot B;
then:
GM＝G1M×cosβ1
yG2＝GM×sin(β1+β2)
according to the formula:
yG2＝(M2y2+M3y3)/(M2+M3)
in the formula: m2To test the weight of the case, y2As the longitudinal coordinate of the center of gravity of the test chamber, M3Is the weight of the object to be measured, y3Is the longitudinal coordinate of the gravity center of the measured object,
the longitudinal coordinate y of the measured object can be determined3(ii) a So that the measured object M is obtained by calculation3The barycentric coordinates of (a).
The teaching device comprises a frame mechanism, a hoisting mechanism, a gravity measuring mechanism and a gravity center testing mechanism. The frame mechanism comprises a base, an upright post and a horizontal rod, wherein the base is fixed on the operating table, the upright post is fixed on the base, and the horizontal rod is fixed on the upright post. The lifting mechanism comprises a rotary hand wheel, a threaded rod, a rotary bearing, a rotary pulley, a guide pulley, a steel wire rope and a lifting hook, wherein the rotary bearing is fixed on the stand column, the threaded rod penetrates through the bearing, one end of the threaded rod is connected with the rotary hand wheel, the rotary pulley is fixed in the middle of the threaded rod, the guide pulley is fixed in the middle of the horizontal rod, one end of the steel wire rope is connected with the lifting hook, one end of the steel wire rope is connected with the rotary pulley, and the threaded rod is provided with a locking device. The quality measuring mechanism comprises a force sensor, a force display and a data transmission line, wherein one end of the force sensor is connected with one end of the lifting hook and is connected with the steel wire rope, and the force sensor is connected with the force display through the data transmission line. The gravity center measuring mechanism mainly comprises a water tank and a test box, wherein the water tank is a rectangular box body with an opening at the upper end, the upper end of the water tank is provided with a flow guide port, the lower end of the water tank is provided with a water outlet, and the flow guide port can be connected with a hose to guide the water tank into the water tank; the test box is a regular rectangular box body, and graduated scales are arranged on the periphery of the test box, so that the depth of the test box immersed in water can be read. The invention has the beneficial effects that: the method has simple principle and convenient operation, and is suitable for measuring the gravity centers of irregular objects with various sizes and weights. Although the device and the ship inclination test both adopt buoyancy to measure the gravity center, the difference is that the initial state of the ship inclination test needs to be in a positive floating state, and the teaching device does not need to be in the positive floating state. The teaching device can supplement the teaching content of the current theoretical mechanics experiment and is suitable for teaching of the theoretical mechanics experiment.
Drawings
FIG. 1 is a block diagram of an experimental teaching device for determining three-dimensional spatial coordinates of the center of gravity of an object.
FIG. 2 is a block diagram of an experimental teaching device for determining three-dimensional spatial coordinates of the center of gravity of an object.
FIG. 3 is a volume segmentation of the test chamber from the submerged portion of the object under test.
FIG. 4 is a diagram of a spatial coordinate system with the bottom of the test chamber being the x-axis and the centerline of the test chamber being the z-axis.
FIG. 5 is a schematic representation of the lines of buoyancy effect before and after the test chamber is tilted.
FIG. 6 is a transverse cross-sectional view of the test chamber with the object under test not floating on the surface of the liquid.
FIG. 7 is a longitudinal cross-sectional view of the test chamber and the object under test not floating on the surface of the liquid.
In the figure: 1. the device comprises a water tank, 2, a test box, 3, a tested object, 4, a lifting frame, 4a, a base, 4b, a stand column 5, a rotary hand wheel, 6, a cross rod, 7, a rotary pulley, 8, a threaded rod, 9, a horizontal rod, 10, a steel wire rope, 11, a guide pulley, 12, a force sensor, 13, a lifting hook, 14, a force display, 15, a data transmission line, 16, an upper diversion port, 17, a water outlet, 18 and a ball valve switch.
Detailed Description
The invention will be further described with reference to the accompanying drawings in which:
1. the water tank 1 is a rectangular tank body with an opening at the upper end, is made of transparent glass fiber reinforced plastics, liquid can be filled in the water tank 1, a water outlet is formed in the lower end of the water tank, the liquid in the water tank can be drained after testing is finished, a flow guide opening is formed in the upper end of the water tank, and the flow guide opening can be connected with a hose, so that the liquid is prevented from overflowing out of the water tank.
2. The test box 2 is a regular rectangular box body, the length is a, the width is b, the height is h, and graduated scales are arranged around the test box, so that the depth of the test box immersed in water can be read.
3. The device comprises a rotary hand wheel 5, a cross rod 6, a rotary pulley 7, a threaded rod 8, a horizontal rod 9, a steel wire rope 10, a guide pulley 11, a force sensor 12 and a lifting hook 13.
4. The force sensor 12 is connected to a force display 14 via a data transmission line 15, which displays the weight measured by the force sensor.
5. The ball valve switch 18 in the tank 1 is closed and liquid is added to the tank 1 and the test chamber 2 is placed in the tank. Because the material of the test box 2 is uniform and the external dimension is regular, the gravity center position of the test box can be obtained according to theoretical calculation. At the moment, the test box should be floating on the water surface, and the longitudinal scales of the test box are all h1. The mass m of the test chamber1＝ρabh1。
6. At this time, the irregular object to be measured 3 is put into the test box 2 by the hook 13, and the force display will measure the mass M of the object to be measured 33And moving the object to be tested to make the test box approximately float on the water surface, wherein the gravity center of the test box, the gravity center of the object to be tested and the floating center are on the same straight line, recording the relative placement position of the object to be tested, the gravity center position of the object to be tested in the horizontal direction coincides with the gravity center position of the box body, at the moment, the test box 2 sinks, and recording the longitudinal scale h of the test box2. Mass m of the object to be measured3＝ρabh2-h1And calculating the density rho of the obtained liquid. The floating center position Bx of the whole body of the test box and the tested object can be determined according to the definition of the floating center21，y21，z21，x21＝a/2y21＝b/2z21＝h2/2
The center of gravity Gx of the mass M of the measured object3、y3，z3x3＝a/2y3B/2, vertical coordinate z of center of gravity only3Is unknown.
7. The tested object is horizontally moved along the x direction by a moving distance e, the test box is inclined by the same volume as the tested object because the quality of the object in the test box is not changed, and the longitudinal scales of the test box are h3、h4、h5、h6If h is a plane because the horizontal plane is a plane3<h4<h6<h5And h6-h3 is h5-h 4. Therefore, the volume of the test box and the 2+3 underwater part of the tested object, namely the volume of the drained water, can be divided into a cuboid and two right-angle triangular prisms (as shown in figure 3). The floating core position is the gravity center position of the drained water, so that:
volume of water displaced
V＝abh3+(ab(h4-h3))/2+(lb(h5-h4))/2
The floating center coordinate is the gravity center coordinate of the volume of the drained water.
From the center of gravity, the formula G ═ M1G1+M2G2+M3G3)/(M1+M2+M3) The position B1x of the floating center can be obtained by calculation22，y22，z22
Figure GDA0002871381060000081
Figure GDA0002871381060000082
Figure GDA0002871381060000083
8. The bottom edge of the test box is taken as an x axis, the middle line of the test box is taken as a z axis, and a space coordinate system is established and is shown in figure 4. Since the measured object is only translated in the x-directionThe object, as shown in the figure, has a center of gravity, and the formula G ═ M2G2+M3G3)/(M2+M3) Therefore, the measured object is only translated along the x direction, the gravity center of the test box and the measured object is only changed along the x direction, and the y and z coordinates are kept unchanged.
9. Because the mass of the test box for translating the tested object and the tested heavy object is not changed, the test box 2 can be inclined in the same volume along the x direction, and two buoyancy action lines are intersected at a point M before and after the inclination, as shown in figure 5.
And establishing a rectangular coordinate system by taking the bottom edge of the test box as an x axis and the central line as a z axis, wherein the coordinate system rotates along with the rotation of the test box.
According to the moment balance principle, the following steps are carried out:
W×e＝ρV×GM×sinθ
in the formula: w is the weight of the object 3 to be measured, known as
e is the translation distance of the measured object, known
V is the weight of the water discharged by buoyancy, known as
Theta is the inclination angle of the test box and can be known by the scale of the test box immersed in water.
Therefore, the GM distance is calculated.
Because the tested object 3 only generates x-direction translation, the gravity center position of the whole 2+3 of the test box and the tested object only changes in the x direction. The bottom edge is still taken as an x axis, the central line symmetry axis is taken as a z axis, and GG exists when a coordinate system is established1Perpendicular to BM and with a vertical foot of G, BB1Perpendicular to BM and the foot is B.
Then:
xG1＝xB1-B1G1×sinθ
in the formula: x is the number ofG1For the x-coordinate of the centre of gravity of the test chamber 2 and the object 3 to be tested, it is known
xB1For the x-coordinate of the floating center of the test chamber 2 and the object 3 to be tested, it is known
Theta is the inclination angle of the test box and can be known by the scale of the test box immersed in water.
Therefore, B is obtained by calculation1G1Distance.
BG＝B1G1×cosθ
Further calculating the distance of the available BG
zG1＝zG＝zB+BG
Then further calculate the available zG1The distance of (c).
zG1I.e. the vertical coordinate of the test box 2 and the measured object 3 as a whole.
According to the formula:
zG1＝(M2z2+M3z3)/(M2+M3)
in the formula: m2To test the weight of the case 2, it is known
z2For the vertical coordinates of the centre of gravity of the test chamber 2, it is known
M3Is the weight of the object 3 to be measured, is known
z3For the vertical barycentric coordinate of the measured object 3 to be solved
The vertical coordinate z of the measured object can be determined3
So that the measured object M is obtained by calculation3Center of gravity coordinates of
10. Further, as in step 6, the test chamber and the object to be tested are not floating on the liquid surface. E.g. center of gravity G1The floating core is B1At this time, we can assume that the current non-positive floating state is the positive floating state and offset by the weight distance e1To obtain G1(ii) a In a similar way, in G1Offset weight distance e on foundation2To obtain G2State, we can assume that the floating state shifts the weight distance e1+e2To obtain G2Status. Because the test box has equal volume deviation, the BG and B on the transverse section can be known from the above reasoning1G1、B2G2Must intersect at a point M, as shown in fig. 6.
And establishing a rectangular coordinate system by taking the bottom edge of the test box as an x axis and the central line as a z axis, wherein the coordinate system rotates along with the rotation of the test box.
According to the moment balance principle, the following steps are carried out:
W×e＝ρV×G1M×sinθ2
in the formula: w is the weight of the object 3 to be measured, known as
e is the translation distance of the measured object, known
V is the weight of the water discharged by buoyancy, known as
θ2The inclination angle of the test box can be known by the scale of the test box immersed in water.
Therefore, G can be obtained by calculation1And M distance.
Because the tested object 3 only generates x-direction translation, the gravity center position of the whole 2+3 of the test box and the tested object only changes in the x direction. The bottom edge is still taken as an x axis, the central line symmetry axis is taken as a z axis, and GG exists when a coordinate system is established1G2Perpendicular to BM and with a vertical foot of G, BB1B2Perpendicular to BM and the foot is B.
Then:
GM＝G1M×cosθ1
xG2＝GM×sin(θ1+θ2)
xG2＝xB2-B2G2×sin(θ1+θ2)
in the formula: x is the number ofG2For finding the x coordinate of the center of gravity of the test box 2 and the object 3 to be measured
xB2For the x-coordinate of the floating center of the test chamber 2 and the object 3 to be tested, it is known
θ1For the initial tilt angle of the test chamber, the scale of the test chamber immersed in water is known
θ2The angle of inclination produced by the test chamber moving the object to be tested is known from the immersion scale of the test chamber in water, as is known
Then B is known2G2
BG＝B2G2×cos(θ1+θ2)
Then BG can be obtained
zG2＝zc＝zB2+BG
Then further calculate the available zG2The distance of (c).
xG2I.e. the x-coordinate of the test box 2 and the object 3 to be tested as a whole.
zG2I.e. the z-coordinate of the test box 2 and the measured object 3 as a whole.
According to the formula:
xG2＝(M2x2+M3x3)/(M2+M3)
zG2＝(M2z2+M3z3)/(M2+M3)
in the formula: m2To test the weight of the case 2, it is known
x2For the transverse coordinates of the center of gravity of the test chamber 2, it is known
z2For the vertical coordinates of the centre of gravity of the test chamber 2, it is known
M3Is the weight of the object 3 to be measured, is known
x3For the transverse coordinates of the center of gravity of the measured object 3
z3For the vertical barycentric coordinate of the measured object 3 to be solved
The transverse coordinate x of the measured object can be determined3Vertical coordinate z3。
11. Similarly, a longitudinal cross-section was taken for analysis as shown in FIG. 7.
And establishing a rectangular coordinate system by taking the bottom edge of the test box as a y axis and the central line as a z axis, wherein the coordinate system rotates along with the rotation of the test box.
According to the moment balance principle, the following steps are carried out: w × h ═ ρ V × G1M×sinβ2
In the formula: w is the weight of the object 3 to be measured, known as
h is the distance of the measured object in the y direction, known as
V is the weight of the water discharged by buoyancy, known as
β2The inclination angle of the test box can be known by the scale of the test box immersed in water.
Therefore, G can be obtained by calculation1And M distance.
Due to the quiltThe object 3 to be tested only translates in the y direction, so the gravity center position of the whole 2+3 of the test box and the object to be tested only changes in the y direction. The bottom edge is still used as the y axis, the central line symmetry axis is used as the z axis, and GG exists when a coordinate system is established1G2Perpendicular to BM and with a vertical foot of G, BB1B2Perpendicular to BM and the foot is B.
Then:
GM＝G1M×cosβ1
yG2＝GM×sin(β1+β2)
according to the formula:
yG2＝(M2y2+M3y3)/(M2+M3)
in the formula: m2To test the weight of the case 2, it is known
y2For the longitudinal coordinates of the center of gravity of the test chamber 2, it is known
M3Is the weight of the object 3 to be measured, is known
y3The gravity center longitudinal coordinate of the measured object 3 is to be solved;
the longitudinal coordinate y of the measured object can be determined3
So that the measured object M is obtained by calculation3The barycentric coordinates of (a).

Claims (1)

1. A method for measuring an experimental teaching device of a three-dimensional space coordinate of the gravity center of an object comprises a water tank (1), a test box (2) and a lifting frame (4), wherein the lifting frame (4) adopts a base (4a) on which two upright posts (4b) are arranged, the top ends of the upright posts (4b) are connected with a horizontal rod (9), the end part of the horizontal rod (9) is connected through a cross rod (6), and a threaded rod (8) is arranged at the connection part of the horizontal rod (9) and the upright posts (4 b); a rotary hand wheel (5) is arranged at one end, extending out of the threaded rod (8), a rotary pulley (7) is arranged in the middle of the threaded rod (8), one end of a steel wire rope (10) is arranged on the rotary pulley (7), and the other end of a guide pulley (11) arranged on the cross rod (6) is connected with a force sensor (12) and a lifting hook (13); the object to be measured (3) acts on the lifting hook (13);
a test box (2) is arranged below the lifting hook (13), and a water tank (1) is arranged on the outer side of the test box (2); the upper end of the water tank (1) is provided with an upper flow guide opening (16), and the lower end of the water tank (1) is provided with a water outlet (17) with a ball valve switch (18); the force sensor (12) is connected with the force display (14) through a data transmission line (15);
the method is characterized by comprising the following steps:
a. after a ball valve switch (18) in the water tank (1) is closed, adding liquid into the water tank (1), and placing the test box (2) into the water tank (1); the test box (2) is a regular rectangular box body, the length is a, the width is b, and the height is h, so that the gravity center position of the test box (2) is obtained according to theoretical calculation; at the moment, the test box should be floating on the water surface, and the longitudinal scales of the test box are all h1Then mass m of the test chamber1＝ρabh1；
b. The irregular object to be measured (3) is placed in the test box (2) by the lifting hook (13), and the mass M of the object to be measured (3) is measured by the force display (14)3Moving the measured object (3) to enable the test box to approximately float on the water surface, wherein the gravity center of the test box, the gravity center of the measured object and the floating center are on the same straight line, recording the relative placement position of the measured object (3), and the gravity center position of the measured object in the horizontal direction is superposed with the gravity center position of the box body;
sinking the test box (2), recording the longitudinal scale h of the test box2(ii) a Mass m of the object to be measured3＝ρab(h2-h1) Calculating to obtain the density rho of the liquid;
further determining the floating center position B (x) of the whole body of the test box and the tested object21，y21，z21)，x21＝a/2y21＝b/2z21＝h22; the center of gravity G (x) of the mass M of the measured object3、y3，z3) In x3＝a/2y3B/2, vertical coordinate z of center of gravity3Unknown;
d. the measured object (3) is horizontally moved along the x direction, the moving distance is e, the test box (2) is inclined by the same volume, and the longitudinal scales of the test box (2) are h respectively3、h4、h5、h6(ii) a If h3<h4<h6<h5Then h is6-h4＝h5-h3(ii) a Volume of water to be drained
V＝abh3+(ab(h4-h3))/2+(lb(h5-h4))/2
The floating center coordinate is the gravity center coordinate of the volume of the drained water;
from the center of gravity, the formula G ═ M1G1+M2G2+M3G3)/(M1+M2+M3) The position of the floating center B1 (x) can be obtained by calculation22，y22，z22)
Figure FDA0003197652510000021
Figure FDA0003197652510000022
Figure FDA0003197652510000023
e. Establishing a space coordinate system by taking the bottom edge of the test box (2) as an x axis and the central line of the test box as a z axis; by translating the object to be measured only in the x-direction, the formula G ═ M is synthesized from the center of gravity2G2+M3G3)/(M2+M3) Therefore, the tested object is only translated along the x direction, the gravity center of the test box and the tested object is only changed along the x direction, and the y and z coordinates are kept unchanged;
f. because the mass of the test box for translating the tested object and the tested heavy object is not changed, the test box (2) can be inclined in the same volume along the x direction, and two buoyancy action lines are intersected at a point M before and after the inclination; establishing a rectangular coordinate system by taking the bottom edge of the test box as an x axis and the central line as a z axis, wherein the coordinate system rotates along with the rotation of the test box;
according to the moment balance principle, the following steps are carried out:
W×e＝ρV×GM×sinθ
in the formula: w is the weight of the measured object (3), e is the translation distance of the measured object, V is the buoyancy, namely the weight of the drained water, and theta is the inclination angle of the test box and can be known by the scale of the test box immersed in the water; calculating to obtain a GM distance;
because the tested object (3) only translates in the x direction, the center of gravity of the whole test box and the tested object only changes in the x direction; establishing a coordinate system by taking the bottom edge as an x axis and the central line symmetry axis as a z axis, and determining that GG exists1Perpendicular to BM and with a vertical foot of G, BB1Perpendicular to BM and with a foot B;
then: x is the number ofG1＝xB1-B1G1×sinθ
In the formula: x is the number ofG1Is the x coordinate of the gravity center of the test box (2) and the measured object (3), xB1Is the x coordinate of the floating center of the test box (2) and the tested object (3), theta is the inclination angle of the test box, and can be known by the scale of the test box immersed in water; calculating to obtain B1G1Distance:
the distance to the BG available is further calculated: BG ═ B1G1×cosθ
Further calculating to obtain zG1Distance z ofG1＝zG＝zB+BG
zG1Namely the vertical coordinate of the whole of the test box (2) and the tested object (3);
according to the formula:
zG1＝(M2z2+M3z3)/(M2+M3)
in the formula: m2For testing the weight of the box (2), z2Is the vertical coordinate of the gravity center of the test box (2), M3Is the weight of the object (3) to be measured, z3Is the gravity center vertical coordinate of the measured object (3); the vertical coordinate z of the measured object can be determined3So as to calculate the measured object M3The barycentric coordinates of (a);
g. if the test box and the object to be tested are not floating on the liquid surface in the step b, the gravity center of the test box and the object to be tested is G1The floating core is B1(ii) a Assuming that the current non-positive floating state is a positive floating state and the weight distance e is deviated1To obtain G1(ii) a In a similar way, in G1Offset weight on foundationDistance e2To obtain G2A state; offset weight distance e in positive floating state1+e2To obtain G2A state;
because the test box has equal volume deviation, the BG and B on the transverse section can be known from the above reasoning1G1、B2G2Must intersect at a point;
establishing a rectangular coordinate system by taking the bottom edge of the test box as an x axis and the central line as a z axis, wherein the coordinate system rotates along with the rotation of the test box;
according to the moment balance principle, the following steps are carried out:
W×e＝ρV×G1M×sinθ2
in the formula: w is the weight of the object (3) to be measured, e is the translation distance of the object to be measured, V is the buoyancy, i.e. the weight of the drained water, theta2The inclination angle of the test box can be known by the scale of the test box immersed in water;
calculating to obtain G1M distance;
because the tested object (3) only translates in the x direction, the center of gravity of the whole test box and the tested object only changes in the x direction; establishing a coordinate system by taking the bottom edge as an x axis and the central line symmetry axis as a z axis, and determining that GG exists1G2Perpendicular to BM and with a vertical foot of G, BB1B2Perpendicular to BM and with a foot B;
then:
GM＝G1M×cosθ1
xG2＝GM×sin(θ1+θ2)
xG2＝xB2-B2G2×sin(θ1+θ2)
in the formula: x is the number ofG2Is the x coordinate of the gravity center of the test box (2) and the tested object (3),
xB2is the x coordinate of the floating center of the test box (2) and the object to be tested (3),
θ1the initial inclination angle of the test box can be known by the scale of the test box immersed in water;
θ2the inclination angle generated by moving the object to be tested for the test box can be known from the scale of the test box immersed in water(ii) a Then B is known2G2
BG＝B2G2×cos(θ1+θ2)
Then BG can be obtained
zG2＝zG＝zB2+BG
Then further calculate the available zG2The distance of (d);
xG2namely the x-direction coordinate of the whole of the test box (2) and the tested object (3);
zG2namely the z-direction coordinate of the whole of the test box (2) and the tested object (3);
according to the formula:
xG2＝(M2x2+M3x3)/(M2+M3)
zG2＝(M2z2+M3z3)/(M2+M3)
in the formula: m2In order to test the weight of the box (2),
x2is the lateral coordinate of the gravity center of the test box (2),
z2is the gravity center vertical coordinate of the test box (2),
M3is the weight of the object (3) to be measured,
x3is the horizontal coordinate of the gravity center of the measured object (3),
z3is the gravity center vertical coordinate of the measured object (3),
the transverse coordinate x of the measured object can be determined3And the vertical coordinate z3；
h. Taking the longitudinal section for analysis in the same way: establishing a rectangular coordinate system by taking the bottom edge of the test box as a y axis and the central line as a z axis, wherein the coordinate system rotates along with the rotation of the test box;
according to the moment balance principle, the following steps are carried out: w × h ═ ρ V × G1M×sinβ2
In the formula: w is the weight of the object (3) to be measured,
h is the translation distance of the measured object in the y direction,
v is buoyancy, namely the weight of the drained water,
β2the inclination angle of the test box can be known by the scale of the test box immersed in water;
calculating to obtain G1M distance;
because the tested object (3) only translates in the y direction, the gravity center position of the whole test box and the tested object only changes in the y direction; establishing a coordinate system by taking the bottom edge as the y axis and the central line symmetry axis as the z axis, and determining that GG exists1G2Perpendicular to BM and with a vertical foot of G, BB1B2Perpendicular to BM and with a foot B;
then:
GM＝G1M×cosβ1
yG2＝GM×sin(β1+β2)
according to the formula:
yG2＝(M2y2+M3y3)/(M2+M3)
in the formula: m2To test the weight of the box (2), y2Is the longitudinal coordinate of the center of gravity of the test box (2), M3Is the weight of the object (3) to be measured, y3Is the longitudinal coordinate of the gravity center of the measured object (3),
the longitudinal coordinate y of the measured object can be determined3(ii) a So that the measured object M is obtained by calculation3The barycentric coordinates of (a).
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