<p>First, a distinction between &quot;gravity&quot; and &quot;gravitation&quot; as used in geophysics. &quot;Gravitation&quot; pertains to Newton's law of gravitation. (Nobody uses general relativity to model gravitation for small, lumpy masses such as the Earth, Mars, or the Moon.) &quot;Gravity&quot; pertains to how things appear to fall from the perspective of an observer fixed with respect to the rotating Earth. Thus gravity includes gravitational acceleration (inward, more or less toward the center of the Earth) and centrifugal acceleration (outward, away from the Earth's rotation axis). This question is asking about gravity rather than gravitation.</p>
<blockquote>
<p>As I understand to compute vector of gravity not saficient to compute normal to elipsoid, but we need to computer normal to geoid (by definition geoid is a surface to which the force of gravity is everywhere perpendicular).</p>
</blockquote>
<p>It's the other way around. The geoid is a calculated surface. Before the satellite era, one of the key sets of inputs to the calculation of the geoid were observations of deviations of local gravity, particularly the deflection of the vertical. Which way gravity pointed at several locations gave clues as to the local shape of the geoid.</p>
<p>Observing the orbits of satellites provides a global measure Earth's gravitational field. Satellites have been specially constructed for this job, most recently, GRACE and GOCE. The Earth's gravitational field is published in terms of spherical harmonics coefficients. The gravitational vector for a point in space on or above the Earth's surface can be calculated directly from these coefficients. Vectorially adding in the centrifugal acceleration due to the Earth's rotation results in the gravity vector. The position also yields the nominal gravity vector assuming an ellipsoidal Earth.</p>
<blockquote>
<p>So I have some questions:</p>
<ol>
<li>How to compute normal to geoid?</li>
</ol>
</blockquote>
<p>As noted above, a geoid is not needed. Modern models of the geoid are calculated from those same spherical harmonic coefficients (plus Earth rotation) that used needed to calculate the gravitational acceleration and gravity.</p>
<blockquote>
<ol start="2">
<li>What is the maximum discrepancy of angle between elipsoid normal and geoid normal?</li>
</ol>
</blockquote>
<p>The technical term is &quot;deflection of the vertical&quot; (with variations). Per Hirt et al., it's up to 100 seconds of arc, about 10 kilometers south of the peak Annapurna II. This is a calculated value based on various satellite models (which are a bit coarse) coupled with digital terrain maps coupled with some more hairy mathematics to create fine-scale models.</p>
<p><a href="http://espace.library.curtin.edu.au/cgi-bin/espace.pdf?file=/2014/03/04/file_1/195882" rel="noreferrer">C. Hirt, et al., “New ultrahigh‐resolution picture of Earth's gravity field,” <em>Geophysical Research Letters,</em> 40.16 (2013): 4279-4283.</a></p>

CC BY-SA 3.0
2016-02-17T16:24:32.650Z
Thanks for detailed answer. What do you mean by `nominal gravity vector` in `The position also yields the nominal gravity vector assuming an ellipsoidal Earth`? Also what about this comment `The normal to geoid does not give you the direction of gravity on
land.  In calculating the geoid separation, a fudge is added to the
gravity field to account for the mass of the terrain above the geoid
(the zeta-to-N correction)` here http://lists.maptools.org/pipermail/proj/2012-October/006453.html . Also see my update about practical application.
mrgloom
CC BY-SA 3.0
2016-02-17T16:57:16.313Z
There's a fairly simple formula for the Earth gravity as a function of geodetic latitude. The direction is also simple; it's normal to the reference ellipsoid. That's the nominal gravity vector. It's the direction that is of interest here, because that's how deflection of the vertical is referenced.
David Hammen
CC BY-SA 3.0
2016-02-17T17:01:44.950Z
Regarding the comment that "the normal to the geoid does not give the direction of gravity on land": That is correct, and it is yet another reason you don't want to use the normal to the geoid. That's one of many of the reasons I tried to steer you away from that concept.
David Hammen
CC BY-SA 3.0
2016-02-17T15:03:34.220Z
2016-02-17T15:03:34.220Z
2020-06-18T08:25:19.090Z
David Hammen
<p>As I understand to compute vector of gravity not saficient to compute normal to elipsoid, but we need to computer normal to <a href="https://en.wikipedia.org/wiki/Geoid" rel="noreferrer">geoid</a> (by definition geoid is a surface to which the force of gravity is everywhere perpendicular).</p>

<p>So I have some questions:</p>

<ol>
<li><p>How to compute normal to geoid?</p></li>
<li><p>What is the maximum discrepancy of angle between elipsoid normal and geoid normal?</p></li>
</ol>

<p><a href="https://i.sstatic.net/jwtBH.gif" rel="noreferrer"><img src="https://i.sstatic.net/jwtBH.gif" alt="enter image description here"></a></p>

<p><a href="https://i.sstatic.net/5hkBm.png" rel="noreferrer"><img src="https://i.sstatic.net/5hkBm.png" alt="enter image description here"></a></p>

<p><strong>Update:</strong></p>

<p>About practical application of gravity normal:</p>

<p>I have some tall object that stand on Earth (like Leaning Tower of Pisa) and I can obtain GPS coordinates of it top and bottom points by using photos taken from air and applying photogrammetry methods. Then from GPS coordinates of top and bottom of object I can get vector in ECEF coordinates and find angle betwen this vector and gravity vector. </p>

<p>Also my question:</p>

<p>If difference in angle between ellipsoid normal and geoid normal("deflection of the vertical") is about 100 seconds of arc (0,0277778 degree) is it safe to use ellipsoid normal for this practical application?</p>

<p><strong>Update 2:</strong></p>

<p>First:</p>

<p>I'm trying to understand if normal to ellipsoid is sufficient for practical application.
In real life object is mounted by using <a href="https://en.wikipedia.org/wiki/Spirit_level" rel="noreferrer">spirit level</a> and on this picture we can see that theoretically I need normal to geoid, but if deflection of vertical small enough I can use normal to ellipsoid as approximation.</p>

<p><a href="https://i.sstatic.net/zhCyA.gif" rel="noreferrer"><img src="https://i.sstatic.net/zhCyA.gif" alt="enter image description here"></a></p>

<p>Second:</p>

<p>I'm trying to estimate the error of angle measure, I have some error in calculating of GPS positions of my object top and bottom points say epsilon_1, also I have some error in calculating gravity vector because of I'm using normal to elispoid say epsilon_2 and I'm trying to understand how epsilon_1 and epsilon_2 affects resulting angle between object vector and gravity vector. So I need to derive the formula that will give me measurement error of angle using measurement error of GPS points and gravity vector(epsilon_1 and epsilon_2).</p>

CC BY-SA 3.0
2016-02-17T17:28:18.950Z
With regard to your update, first off, that 100 seconds of arc applies only to the Himalaya, and only to a few spots. There are a few spots in the Andes where deflection of the vertical exceeds 60 arc seconds, in the Alps and Rockies, it's even less yet. On flat land, there are a few spots with a 40 arc second deflection. In most places, it's less than 10. Even the 100 arc seconds is a very small deflection. Then again, you didn't say what your sensitivity is.
David Hammen
CC BY-SA 3.0
2016-02-17T18:34:41.093Z
If by "sensitivity" you mean measurement error I haven't it beforehand I need to derive the formula that will give me measurement error of angle using measurement error of GPS points and gravity vector.Also see update 2.
mrgloom
<geophysics><models><gravity>
Compute direction of gravity vector
mrgloom
# Compute direction of gravity vector

(Asked by: username_0 on 2016-02-17)

As I understand to compute vector of gravity not saficient to compute normal to elipsoid, but we need to computer normal to [geoid](https://en.wikipedia.org/wiki/Geoid) (by definition geoid is a surface to which the force of gravity is everywhere perpendicular).

So I have some questions:

1. How to compute normal to geoid?
2. What is the maximum discrepancy of angle between elipsoid normal and geoid normal?

[![enter image description here](https://i.sstatic.net/jwtBH.gif)](https://i.sstatic.net/jwtBH.gif)

[![enter image description here](https://i.sstatic.net/5hkBm.png)](https://i.sstatic.net/5hkBm.png)

**Update:**

About practical application of gravity normal:

I have some tall object that stand on Earth (like Leaning Tower of Pisa) and I can obtain GPS coordinates of it top and bottom points by using photos taken from air and applying photogrammetry methods. Then from GPS coordinates of top and bottom of object I can get vector in ECEF coordinates and find angle betwen this vector and gravity vector.

Also my question:

If difference in angle between ellipsoid normal and geoid normal("deflection of the vertical") is about 100 seconds of arc (0,0277778 degree) is it safe to use ellipsoid normal for this practical application?

**Update 2:**

First:

I'm trying to understand if normal to ellipsoid is sufficient for practical application.
In real life object is mounted by using [spirit level](https://en.wikipedia.org/wiki/Spirit_level) and on this picture we can see that theoretically I need normal to geoid, but if deflection of vertical small enough I can use normal to ellipsoid as approximation.

[![enter image description here](https://i.sstatic.net/zhCyA.gif)](https://i.sstatic.net/zhCyA.gif)

Second:

I'm trying to estimate the error of angle measure, I have some error in calculating of GPS positions of my object top and bottom points say epsilon\_1, also I have some error in calculating gravity vector because of I'm using normal to elispoid say epsilon\_2 and I'm trying to understand how epsilon\_1 and epsilon\_2 affects resulting angle between object vector and gravity vector. So I need to derive the formula that will give me measurement error of angle using measurement error of GPS points and gravity vector(epsilon\_1 and epsilon\_2).

**Tags:** geophysics, models, gravity

> **Comments:**
> **username_1**: With regard to your update, first off, that 100 seconds of arc applies only to the Himalaya, and only to a few spots. There are a few spots in the Andes where deflection of the vertical exceeds 60 arc seconds, in the Alps and Rockies, it's even less yet. On flat land, there are a few spots with a 40 arc second deflection. In most places, it's less than 10. Even the 100 arc seconds is a very small deflection. Then again, you didn't say what your sensitivity is.  
> **username_0**: If by "sensitivity" you mean measurement error I haven't it beforehand I need to derive the formula that will give me measurement error of angle using measurement error of GPS points and gravity vector.Also see update 2.  

---

## 1 answers

### Answer by username_1 on 2016-02-17. Score: 5 | Accepted answer

First, a distinction between "gravity" and "gravitation" as used in geophysics. "Gravitation" pertains to Newton's law of gravitation. (Nobody uses general relativity to model gravitation for small, lumpy masses such as the Earth, Mars, or the Moon.) "Gravity" pertains to how things appear to fall from the perspective of an observer fixed with respect to the rotating Earth. Thus gravity includes gravitational acceleration (inward, more or less toward the center of the Earth) and centrifugal acceleration (outward, away from the Earth's rotation axis). This question is asking about gravity rather than gravitation.

> As I understand to compute vector of gravity not saficient to compute normal to elipsoid, but we need to computer normal to geoid (by definition geoid is a surface to which the force of gravity is everywhere perpendicular).

It's the other way around. The geoid is a calculated surface. Before the satellite era, one of the key sets of inputs to the calculation of the geoid were observations of deviations of local gravity, particularly the deflection of the vertical. Which way gravity pointed at several locations gave clues as to the local shape of the geoid.

Observing the orbits of satellites provides a global measure Earth's gravitational field. Satellites have been specially constructed for this job, most recently, GRACE and GOCE. The Earth's gravitational field is published in terms of spherical harmonics coefficients. The gravitational vector for a point in space on or above the Earth's surface can be calculated directly from these coefficients. Vectorially adding in the centrifugal acceleration due to the Earth's rotation results in the gravity vector. The position also yields the nominal gravity vector assuming an ellipsoidal Earth.

> So I have some questions:
>
> 1. How to compute normal to geoid?

As noted above, a geoid is not needed. Modern models of the geoid are calculated from those same spherical harmonic coefficients (plus Earth rotation) that used needed to calculate the gravitational acceleration and gravity.

> 2. What is the maximum discrepancy of angle between elipsoid normal and geoid normal?

The technical term is "deflection of the vertical" (with variations). Per Hirt et al., it's up to 100 seconds of arc, about 10 kilometers south of the peak Annapurna II. This is a calculated value based on various satellite models (which are a bit coarse) coupled with digital terrain maps coupled with some more hairy mathematics to create fine-scale models.

[C. Hirt, et al., “New ultrahigh‐resolution picture of Earth's gravity field,” *Geophysical Research Letters,* 40.16 (2013): 4279-4283.](http://espace.library.curtin.edu.au/cgi-bin/espace.pdf?file=/2014/03/04/file_1/195882)

> **Comments:**
> **username_0**: Thanks for detailed answer. What do you mean by `nominal gravity vector` in `The position also yields the nominal gravity vector assuming an ellipsoidal Earth`? Also what about this comment `The normal to geoid does not give you the direction of gravity on land.  In calculating the geoid separation, a fudge is added to the gravity field to account for the mass of the terrain above the geoid (the zeta-to-N correction)` here http://lists.maptools.org/pipermail/proj/2012-October/006453.html . Also see my update about practical application.  
> **username_1**: There's a fairly simple formula for the Earth gravity as a function of geodetic latitude. The direction is also simple; it's normal to the reference ellipsoid. That's the nominal gravity vector. It's the direction that is of interest here, because that's how deflection of the vertical is referenced.  
> **username_1**: Regarding the comment that "the normal to the geoid does not give the direction of gravity on land": That is correct, and it is yet another reason you don't want to use the normal to the geoid. That's one of many of the reasons I tried to steer you away from that concept.  

---

