Lesson Plan

August Anguish at Andersonville (Grades 9-12)

Thousands of graves are arranged in long rows, decorated with American flags.
The graves of August 1864.

NPS/C. Barr

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Grade Level:
Ninth Grade-Twelfth Grade
Algebra, Civil War, History, Mathematics, Military and Wartime History
1 class period, 45-60 minutes
Group Size:
Up to 36
National/State Standards:
CCGPS Mathematics I (9-12)MM1D3. 
CCGPS Mathematics (9-12)MM1G3.
CCGPS Mathematics II (9-12) MM2D1.
CCPGS Mathematics II (9-12) MM2P1.


In August of 1864, over 32,000 American soldiers were held captive at Andersonville. Mismanagement and overcrowding lead to a high rate of suffering and death among the prisoners. Understanding the scale of suffering is a challenge; this activity uses math activities and word problems to explore the darkest moment of the prisons history.


Students will use the Pythagorean Theorem to solve problems.

Students will be able to convert from one unit of measurement to another.

Students will be able to apply appropriate mathematical strategies to solve problems.


After the breakdown of the prison exchange system in mid-1863, both Union and Confederate forces were forced to hold prisoners for an extended period of time. The Confederacy first held Union prisoners in the capital city of Richmond, VA, but this placed a tremendous strain on the city's resources. So in early 1864 the Confederacy built a new prison in rural Georgia – Camp Sumter, which quickly became known as Andersonville.

By the summer of 1864 Andersonville was incredibly overcrowded. Approximately 33,000 prisoners were held in a space intended for 10,000 and diseases related to sanitation and nutrition were rampant. By the time the prison ceased operation 45,000 men passed through the prison gates, and 13,000 of them died, making Andersonville the deadliest ground in American history.

It can sometimes be difficult for students to grasp to scale of suffering at Andersonville. This activity provides an opportunity for students to apply math skills and get a better sense of the overcrowding and death that occurred in August of 1864, when the prison was at its maximum population.   


Photocopies of materials that may be needed are included in this lesson and will be provided by the teacher.



An answer key is provided:

1. What was the distance between the southwest corner and northeast corner of the prison?
a. The Pythagorean Theorem (a2 + b2 = c2) is used to solve this problem.
b. The dimension of the stockade "rectangle" is 1,620 ft. x 779 ft. (a = length and b = width).
c. 1,6202 + 7792 = c2
d. 2,624,400 + 606,841 = c2
e. 3,231,241 = c2
f. c2 = 3,231,241
g. c = √3,231,241
h. c = 1797.57
i. The distance between the two corners of the stockade was approximately 1798 feet.

2. If the total population of the prison increased by 10% on September 1st, what would the new population be? If on September 1st there was a prisoner uprising and the guards opened fire with rifles and cannons killing 25% of the prisoners and severely injuring 10% of those who were not immediately killed, then what would the new population be and how many prisoners would need medical attention? (Students will need to use the data from the daily returns, August 1864)
a. The total population of the prison on August 31st needs to be multiplied by 10%., giving 34,862 prisoners on September 1st.
b. 25% of the population needs to be subtracted from the number found in step "a." This will give the new population total. So 34,862 (prisoners September 1st) x 0.25 (the percent that was killed) = 8,716 (total prisoners that were killed). 34,862 (prisoners September 1st) - 8,716 (prisoners that were killed) = 26,146 (the new prison population).
c. 10% of the population needs to be subtracted from the number found in step "b." This will give you the number of prisoners that need medical attention. 26,146 (the new/remaining prison population) x 0.10 (the percent that were severely injured) = 2,615 prisoners severely injured and need medical attention.

3. Assuming that each prisoner would have needed 48 square feet (the standard size of modern jail cells) to survive the month of August, what would the new dimensions of the prison need to be? Express your answer in square feet and acres. Use the number of prisoners on hand on August 1st found in the Consolidated Monthly Return to solve this problem. 1 acre = 43,560 square feet
a. The total population of the prison on August 1st needs to be multiplied by 48 square feet. 31,678 (prisoners on hand August 1st) x 48 sq. ft. per prisoner = 1,520,544 sq. ft. needed.
b. The available space INSIDE the deadline must be determined. The prison dimensions (taking into consideration that the prisoners could not occupy the space within the "dead line") were 1,582 ft. (length) x 741 ft. (width) = 1,172,262 sq. ft. (space available inside the "dead line").
c. The amount of additional space needed must be determined using the totals in step "a" and step "b." 1,520,544 sq. ft. (total needed) - 1,172,262 sq. ft. (current space available) = 348,282 sq. ft. (additional space needed).
d. The value found in step "c" should be divided by the current width of the stockade inside the deadline. This will give the length that must be added. 348,282 sq. ft. (space needed) ÷ 741 ft. (the current width of the stockade rectangle inside the "dead line") = ~470 ft. (needed to be added to the LENGTH of the stockade rectangle).
e. The current length of the stockade should be added to the value found in step "d." This will give the total length of the prison that would be needed.   1,620 ft. (the current length of the stockade including "dead line" at both ends) + 470 ft. (length needed to be added) = 2,090 ft. (total length of the new, larger stockade).  The new dimension of the prison stockade would be 2,090 ft. (length) by 779 ft. (width).
f. The new length multiplied by the old (or same) width will give the new total square footage. 7.  2,090 ft. - 38 ft. (for both "dead line" spaces) x 779 ft. - 38 ft. ("dead line") = 1,520,532 sq. ft. of the 1,520,544 sq. ft. calculated as being needed.
g. The acre to square feet ratio needs to be used to determine acreage.There are 43,560 sq. ft. in an acre. 1,520,544 sq. ft. (amount of space needed) ÷ 43,560 (sq. ft. per acre) = 34.91 total acres occupied by the new stockade, with 8.41 of those acres added to the original 26.5 acres.

4. If a particular virus infected the prison population and had a 100% mortality rate, then how many days would it take for 50% of the total population to die? Use the population of the prison as of August 31st and assume that no more prisoners are added to the population because of the virus outbreak. Also, assume that once a prisoner is contagious he has 24 hours to live and will infect exactly two more prisoners before death.

a.  Divide the total number of prisoners on August 31st by two. This number represents 50% of the prison population. 31,693 (prisoners on hand August 31st) x 0.50 = 15,847 (half the prison population as of August 31st).
b.  Multiply 1 by 2. These numbers represent the host (the first person to be infected by the virus), and the two individuals that the host spreads the virus to. Keep multiplying by two (since each carrier infects exactly two more people before death) until you reach a number that is equal to, or more than, the number found in step "a." 1 (the original virus host) x 2 (prisoners infected by host) = 2 (total infected after host dies). 2 (total infected at this point) x 2 (the number the infected will each infect) = 4 (total infected after the original two die). 4 (total infected at this point) x 2 (the number the infected will each infect) = 8 (total infected after the original 4 die). Continue to multiply by two (the number that each infected person will infect) until AT LEAST 15,847 (half the prison population) is reached.
c. Count the number of times you multiplied by two. Each time this is done represents one day. After multiplying by two 14 times you will reach 16,384. This is at least half the prison population.
d.Add 1 to the number found in step "c." This represents the host and the 24 hour period that he was infected.1 (the prisoner/host who lived for 24 hours) + 14 (the number of times multiplied by two) = 15 days it would have taken for the virus to wipe out at least half the prison population.

Park Connections

The prisoners buried in sections E and F in Andersonville National Cemetery died over a 32 day period from the end of July to the end of August 1864. These sections are a powerful visual representation of the scale of suffering in Andersonville when the prison was at its height.   


1. Students should imagine themselves as prisoners at Andersonville. Their job is to write a letter to Abraham Lincoln describing the conditions they are forced to live in and giving a detailed report of prisoner numbers (in the hospital, deaths per day, etc.) and living space (dimensions of the prison and square feet per prisoner) using the Consolidated Returns of the Month of August.

2.    Angola prison in Louisiana (the largest maximum security in the United State) houses 5,000 inmates and occupies approximately 18,000 acres (not all of this acreage is used entirely for the prisoners). Andersonville held nearly 32,000 prisoners and occupied 26.5 acres. Calculate the two different prisoner-to-acre ratios and compare them.


Representative Sample
Mortality Rate
Linear Functions

Last updated: April 14, 2015